NCERT Solutions for Class 11 Maths Chapter 3 - Trigonometric Functions
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Exercise 3.1
1. Find the radian measure corresponding to following degree measures:
(i) 25°
(ii) – 47° 30′
(iii) 240°
(iv) 520°
$$\textbf{Sol.}\space(\text{i})\space 25\degree=25×\frac{\pi}{180}\text{rad}=\frac{5\pi}{36}\text{radian}\\\bigg(\because 1\degree=\frac{\pi}{180}\text{rad}\bigg)\\\text{(ii)}\space-47\degree30'=-\bigg(47+\frac{30}{60}\bigg)^\degree\\=-\bigg(47+\frac{1}{2}\bigg)^\degree\\\bigg[\because 1'=\bigg(\frac{1}{60}\bigg)^\degree\bigg]\\=-\bigg(\frac{94+1}{2}\bigg)^\degree=-\frac{95\degree}{2}$$
$$=-\frac{95}{2}×\frac{\pi}{180}\space\text{radian}\\=-\frac{19\pi}{72}\text{radian}$$
$$\text{(iii)\space} 240\degree=240×\frac{\pi}{180}\text{radian}\\=\frac{4\pi}{3}\text{radian}\\\text{(iv) 520 = 520}×\frac{\pi}{180}\text{radian}\\=\frac{26\pi}{9}\text{radian}$$
2. Find the degree corresponding to the following radian measure
$$\bigg(\textbf{use}\space\pi=\frac{\textbf{22}}{\textbf{7}}\bigg).$$
$$\textbf{(i)}\space\bigg(\frac{\textbf{11}}{\textbf{16}}\bigg)$$
(ii) – 4
$$\textbf{(iii)\space}\frac{\textbf{5}\pi}{\textbf{3}}\\\textbf{(iv)}\space\frac{\textbf{7}\pi}{\textbf{6}}$$
Sol. (i)
$$\bigg(\frac{11}{16}\bigg)^c=\bigg(\frac{11}{16}×\frac{180}{\pi}\bigg)^\degree\\=\frac{11×180×7}{16×22}\\=\bigg(\frac{11}{8}×\frac{90}{22}×7\bigg)^{\degree}\\=\bigg(\frac{45×7}{8}\bigg)^{\degree}\\\bigg[\because 1^c=\bigg(\frac{180}{\pi}\bigg)^\degree\bigg]\\=\bigg(\frac{315}{8}\bigg)^\degree=39\degree+\bigg(\frac{3}{8}\bigg)^\degree\\=39\degree+\frac{3}{8}×60'$$
$$=39\degree+\bigg(\frac{45}{2}\bigg)' (\because 1\degree=60')\\=39\degree + 22\frac{1'}{2}\\=39\degree+22'+\frac{1×60''}{2}\\=39\degree+22'+30''$$
= 39° 22′ 30′′ (∵ 1′ = 60′′)
$$\text{(ii)\space} -4^c=-\bigg(4×\frac{180}{\pi}\bigg)^\degree\\=-\bigg(4×\frac{180}{22}×7\bigg)\\\bigg[\because 1\degree=\bigg(\frac{180}{\pi}\bigg)^\degree\bigg]\\=-\bigg(\frac{2×1260\degree}{11}\bigg)=-\bigg(\frac{2520}{11}\bigg)^\degree\\=-\bigg[229\degree+\bigg(\frac{1}{11}\bigg)^\degree\bigg]\\=-\bigg(229\degree+\frac{1}{11}×60'\bigg)\space(\because 1\degree=60')$$
$$=-\bigg(229\degree+\frac{1}{11}×60'\bigg)(\because 1\degree=60')\\=-\bigg(229\degree+5'+\bigg(\frac{5}{11}\bigg) '\bigg)\\=-\bigg(229\degree+5'+\frac{5}{11}×60''\bigg)(\because 1'=60'')$$
= – (229° 5′ 27′′) (approx)
$$\text{(iii)\space}\bigg(\frac{5\pi}{3}\bigg)^c=\bigg(\frac{5\pi}{3}×\frac{180}{\pi}\bigg)^\degree\\=300\degree$$
Hence, 12π radians will be turn in 1 s.
$$\text{(iv)\space}\bigg(\frac{7\pi}{6}\bigg)^c=\bigg[\frac{7\pi}{6}×\frac{180}{\pi}\bigg]^\degree=210\degree$$
3. A wheel makes 360 revolutions in 1 min. Through how many radians does it turn in one second.
Sol. In 1 min, number of revolutions of wheel = 360
i.e., In 60 s, number of revolutions of wheel = 360
$$\therefore\space\text{In 1 s, number of revolutions of wheel}=\frac{360}{60}\\=6$$
In 1 revolutions, angle = 360° = 2π
In 6 revolutions, angle = 2π × 6 = 12π radians.
4. Find the degree measure of the angle subtended at the centre of the circle of radius 100 cm by an arc of length 22 cm.
$$\bigg(\textbf{Use}\space\pi=\frac{\textbf{22}}{\textbf{7}}\bigg)$$
Sol. Here, length of arc l = 22 cm and radius r = 100 cm
$$\text{Using the formula}\space\theta=\frac{l}{r},\text{we have}\\\Rarr\space\theta=\frac{22}{100}=\frac{11}{50}\text{radians}\space\\\bigg(\because \theta=\frac{l}{r}\bigg)\\=\bigg(\frac{11}{50}×\frac{180}{\pi}\bigg)^{\degree}\\=\bigg(\frac{11×18×7}{5×22}\bigg)^{\degree}\\=\bigg(\frac{63}{5}\bigg)^\degree\\=12\degree+\bigg(\frac{3}{5}\bigg)^\degree$$
$$= 12\degree + \frac{3}{5}×60'\space (\because 1\degree = 60')$$
= 12° 36′
Hence, the angle formed by arc at the centre is 12°36′.
5. In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.
Sol. Given, diameter = 40 cm
$$\therefore\space \text{Radius CA = AB}=\frac{\text{diameter}}{2}\\=\frac{40}{2}=20\space\text{cm}$$
Also, chord AB = 20 cm
Now, we have all the three sides of ΔABC equal so, it is an equilateral triangle.
$$\Rarr\space Q=\angle\text{ACB}=60\degree\\=60×\frac{\pi}{180}\space\text{rad}\\\text{Now, using the formula}\space\theta=\frac{l}{r}\\ 60×\frac{\pi}{180}=\frac{\text{AB}}{20}\\\Rarr\space\text{AB}=60×20×\frac{\pi}{180\degree}=\frac{20\pi}{2}\text{cm}$$
Hence, the length of minor arc of the chord is
$$\frac{20 \pi}{2}\space\text{cm}.$$
6. If in two circles, arcs of the same length subtend angles of 60° and 75° at the centre, find the ratio of their radii.
Sol.
$$\text{Using the formula}\space\theta=\frac{l}{r}\\\text{For first circle,}\space\theta_1=\frac{l}{r_1}\\60×\frac{\pi}{180}=\frac{l}{r_1}\space\text{...(i)}$$
And for second circle
$$\theta_2=\frac{l}{r_2}\qquad\\75×\frac{\pi}{180}=\frac{l}{r_2}\space\text{...(ii)}$$
Dividing equation (i) by equation (ii), we get
$$\frac{60×\frac{\pi}{180}}{75×\frac{\pi}{180}}=\frac{\frac{l}{r_1}}{\frac{l}{r_2}}\\\Rarr\space \frac{r_2}{r_1}=\frac{4}{5}$$
⇒ r1 : r2 = 5 : 4
Hence, the ratio of their radii is 5 : 4.
7. Find the angle in radian through which a pendulum swings if its length is 75 cm and the tip describes an arc of length:
(a) 10 c m
(b) 15 cm
(c) 21 cm
Sol. (a) $$\theta=\frac{10}{75}=\frac{2}{15}\space\text{radians}$$
$$\text{(b)\space}\theta=\frac{15}{75}=\frac{1}{5}\space\text{radians}$$
$$\text{(c)\space}\theta=\frac{21}{75}=\frac{7}{25}\space\text{radians}$$
Exercise 3.2
Find the values of other five trigonometric functions in Exercises 1 to 5:
$$\textbf{1. cos x =}-\frac{\textbf{1}}{\textbf{2}},\textbf{where x lies in third}\\ \textbf{quadrant.}$$
$$\textbf{Sol.}\space\text{cos x}=-\frac{1}{2}$$
Given that x lies in third quadrant.
$$\text{i.e.}\space \pi\lt x\lt\frac{3\pi}{2}$$
We have sin2 x + cos2 x = 1
⇒ sin2 x = 1 – cos2 x
$$\Rarr\space\text{sin}^{2}x=1-\bigg(-\frac{1}{2}\bigg)^{2}\\\Rarr\space\text{sin}^{2}x=1-\frac{1}{4}=\frac{3}{4}\\\Rarr\space\text{sin x}=\pm\frac{\sqrt{3}}{2}$$
∵ In third quadrant sin x in negative, so we will leave its positive value.
$$\text{i.e.,\space sin x}=-\frac{\sqrt{3}}{2}\\\text{Now,}\space \text{tan x}=\frac{\text{sin}\space x}{\text{cos}\space x}=\frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}}=\sqrt{3},\\\text{cot x}=\frac{1}{\text{tan x}}=\frac{1}{\sqrt{3}}\\\text{and\space sec x}=\frac{1}{\text{cos} \space x}=-2,\\\text{cosec x}=\frac{1}{ \text{sin} \space x}=-\frac{2}{\sqrt{3}}$$
$$\textbf{2. sin x}=\frac{\textbf{3}}{\textbf{5}},\textbf{where x lies in}\\\space \textbf{second quadrant.}$$
$$\textbf{Sol.}\space\text{sin x}=\frac{3}{5}$$
Given that x lies in second quadrant.
$$i.e.,\space \frac{\pi}{2}\lt x\lt\pi$$
$$\because\space \text{sin}^{2}x+\text{cos}^{2}x=1\\\Rarr\space \text{cos}^{2}x=1-\text{sin}^{2}x\\1-\bigg(\frac{3}{5}\bigg)^{2}=1-\frac{9}{25}\\=\frac{25-9}{25}=\frac{16}{25}\\\Rarr\space\text{cos x}=\pm\frac{4}{5}$$
∵ In second quadrant, cos x is negative, so we will leaves its positive value.
$$\text{i.e.\space}\text{cos x}=-\frac{4}{5}\\\Rarr\space\text{tan x}=\frac{\text{sin x}}{\text{cos} x}=\frac{\frac{3}{5}}{-\frac{4}{5}}=-\frac{3}{4},\\\text{cot x =}\frac{1}{\text{tan x}}=-\frac{4}{\text{3}}, \\\text{sec x =}\frac{1}{\text{cos x}}=-\frac{5}{4}\\\text{and\space cosec x =}\frac{1}{\text{sin x}}=\frac{5}{3}$$
$$\textbf{3. cot x}=\frac{\textbf{3}}{\textbf{4}}, \textbf{where x lies in third quadrant.}$$
$$\textbf{Sol.}\space \text{cot x}=\frac{3}{4}$$
Given that x lies in third quadrant.
$$\text{i.e.,}\space \pi\lt x\lt\frac{3\pi}{2}\\\Rarr\space\text{tan x}=\frac{1}{\text{cot x}}=\frac{4}{3}\\\because\space \text{sec}^{2}x=\text{tan}^{2}x+1=\frac{16}{9}+1\\\text{sec}^{2}x=\frac{25}{9}\\\Rarr\space \text{sec x}=\pm\frac{5}{3}$$
∵ x lies in third quadrant, so we take only its negative value.
$$\text{i.e.,}\space\text{sec x = −}\frac{5}{3}\\\Rarr\space\text{cos x = −}\frac{3}{5}\\\because\space\text{tan x}=\frac{\text{sin x}}{\text{cos x}}\\\Rarr\space\frac{4}{3}=\frac{\text{sin x}}{-\frac{3}{5}}\\\Rarr\space \text{sin x}=-\frac{4}{5}\\\text{and cosec x}=-\frac{5}{4}.$$
$$\textbf{4. sec x}=\frac{\textbf{13}}{\textbf{5}},\textbf{where x lies in }\\\textbf{fourth quadrant.}\\\textbf{Sol.}\space\text{sec x}=\frac{13}{5}$$
Given that x lies in fourth quadrant.
$$\text{i.e.}\space\frac{3\pi}{2}\lt x\lt2\pi\\\Rarr\space\text{cos x}=\frac{5}{13}$$
∵ sin2 x + cos2 x = 1
$$\Rarr\space \text{sin}^{2}x=1-\text{cos}^{2}x=1-\bigg(\frac{5}{13}\bigg)^{2}\\=1-\frac{25}{169}=\frac{169-25}{169}\\=\frac{144}{169}=\bigg(\frac{12}{13}\bigg)^{2}\\\Rarr\space\text{sinx}=\pm\frac{12}{13}$$
Since, x lies in fourth quadrant, so cos x and sec x will be positive other function will take negative sign.
$$\text{i.e.,\space}\text{sin x}=-\frac{12}{13}\\\text{tan x}=\frac{\text{sin x}}{\text{cos x}}=\frac{-\bigg(\frac{12}{13}\bigg)}{\bigg(\frac{5}{13}\bigg)}=-\frac{12}{5}\\\Rarr\space\text{cosec x}=\frac{1}{\text{sin x}}=\frac{-13}{12},\\\text{cot x}=\frac{1}{\text{tan x}}=-\frac{5}{12}$$
$$\textbf{5. tan x}=-\frac{\textbf{5}}{\textbf{12}}, \textbf{where x lies in}\\\textbf{ second quadrant.}\\\textbf{Sol.}\space\text{tan x =}-\frac{5}{12}$$
Given that x lies in second quadrant, so, sin x, cosec x will be psotive. Other functions will take negative sign
$$\text{i.e.,}\space \frac{\pi}{2}\lt x \lt\pi\\\text{cot x}=\frac{1}{\text{tan x}}=-\frac{12}{5}\\\text{Now, \space}\text{sec}^{2}x=1+\text{tan}^{2}x=1+\bigg(-\frac{5}{12}\bigg)^{2}\\=1+\frac{25}{44}=\frac{169}{144}\\\text{sec}^{2}x=\bigg(\frac{13}{12}\bigg)^{2}\\\Rarr\space\text{sec x}=\pm\frac{13}{12}$$
∵ x lies in second quadrant, so we take negative sign.
$$\therefore\space\text{sec x}=-\frac{13}{12}\\\Rarr\space\text{cos x}=\frac{1}{\text{sec x}}=\frac{-12}{13}\\\text{Now,}\space\text{tan x}=\frac{\text{sin x}}{\text{cos x}}\\\Rarr\space -\frac{5}{12}=\frac{\text{sin x}}{-\frac{12}{13}}\\\Rarr\space \text{sin x}=\frac{5}{13}\\\text{and cosec x}=\frac{1}{\text{sin x}}=\frac{13}{5}.$$
Find the values of trigonometric functions in Exercises 6 to 10.
6. sin 765°.
$$\textbf{Sol.}\space\text{sin 765\degree}=\text{sin}(360×2+45)\degree\\=\text{sin 45\degree}=\frac{1}{\sqrt{2}}\\\lbrack\because \text{sin (2}n\pi + \theta) = sin \theta\rbrack$$
7. cosec (– 1410°).
Sol. cosec (– 1410°) = – cosec (1410°)
[∵ cosec (– θ) = – cosec θ]
= – cosec (360 × 4 – 30)°
= – (– cosec 30°) = cosec 30° = 2
[∵ cosec (2nπ – θ) = – cosec θ]
$$\textbf{8. tan}\space\frac{\textbf{19}\pi}{\textbf{3}}.$$
$$\textbf{Sol.}\space\text{tan}\frac{19\pi}{3}=\text{tan}\bigg(6\pi+\frac{\pi}{3}\bigg)\\=\text{tan}\bigg(2\pi×3+\frac{\pi}{3}\bigg)\\=\text{tan}\frac{\pi}{3}=\sqrt{3}\\\lbrack\therefore\space\text{tan(2n}\pi+\theta)=\text{tan}\space\theta\rbrack$$
$$\textbf{9. sin}\bigg(\frac{\normalsize-\textbf{11}\pi}{\textbf{3}}\bigg)\\\textbf{Sol.\space} \text{sin}\bigg(\frac{-11\pi}{3}\bigg)=-\text{sin}\bigg(\frac{11\pi}{3}\bigg)\\\lbrack\because \text{sin}(-\theta)=-\text{sin}\space\theta\rbrack\\=-\text{sin}\bigg(4\pi-\frac{\pi}{3}\bigg)=-\text{sin}\bigg(2\pi×2-\frac{\pi}{3}\bigg)\\=-\bigg(-\text{sin}\frac{\pi}{3}\bigg)=\text{sin}\bigg(\frac{\pi}{3}\bigg)=\frac{\sqrt{3}}{2}\\\lbrack\because \text{sin(2n}\pi-\theta)=-\text{sin}\theta\rbrack$$
$$\textbf{10. cot}\bigg(\frac{\normalsize-\textbf{15}\pi}{\textbf{4}}\bigg)\\\textbf{Sol.}\space\text{cot}\bigg(-\frac{15\pi}{4}\bigg)=-\text{cot}\bigg(\frac{15\pi}{4}\bigg)\\\lbrack\because \text{cot}(-\theta)=-\text{cot}\space\theta\rbrack\\=-\text{cot}\bigg(4\pi-\frac{\pi}{4}\bigg)\\=-\text{cot}\bigg(2\pi×2-\frac{\pi}{4}\bigg)\\=-\bigg(-\text{cot}\frac{\pi}{4}\bigg)=\text{cot}\frac{\pi}{4}=1$$
[∵ cot (2nπ – θ) = – cot θ]
Exercise 3.3
$$\textbf{1. Prove that: \space sin}^{2}\frac{\pi}{\textbf{6}}+\textbf{cos}^{2}\frac{\pi}{\textbf{3}}-\textbf{tan}^{2}\frac{\pi}{\textbf{4}}=-\frac{\textbf{1}}{\textbf{2}}\\\textbf{Sol.}\space \text{sin}^{2}\frac{\pi}{6}+\text{cos}^{2}\frac{\pi}{3}-\text{tan}^{2}\frac{\pi}{4}=-\frac{1}{2}\\\text{L.H.S.}=\text{sin}^{2}\frac{\pi}{6}+\text{cos}^{2}\frac{\pi}{3}-\text{tan}^{2}\frac{\pi}{4}\\=\bigg(\frac{1}{2}\bigg)^{2}+\bigg(\frac{1}{2}\bigg)^{2}-(1)^{2}\\=\frac{1}{4}+\frac{1}{4}-1\\=\frac{1}{2}-1\\=-\frac{1}{2}=\text{R.H.S}$$
∴ L.H.S. = R.H.S.
Hence Proved.
$$\textbf{2. Prove that:}\space\textbf{2 sin}^{2}\frac{\pi}{\textbf{6}}+\textbf{cosec}^{2}\frac{\textbf{7}\pi}{\textbf{6}}\textbf{cos}^{2}\frac{\pi}{\textbf{3}}=\frac{\textbf{3}}{\textbf{2}}\\\textbf{Sol.\space}2\text{sin}^{2}\frac{\pi}{6}+\text{cosec}^{2}\frac{7\pi}{6}\text{cos}^{2}\frac{\pi}{3}=\frac{3}{2}\\\text{L.H.S.} =2\text{sin}^{2}\frac{\pi}{6}+\text{cosec}^{2}\frac{7\pi}{6}\text{cos}^{2}\frac{\pi}{3}\\=2×\bigg(\frac{1}{2}\bigg)^{2}+\text{cosec}^{2}\bigg(\pi+\frac{\pi}{6}\bigg)\text{cos}^{2}\frac{\pi}{3}\\=2×\frac{1}{4}+\text{cosec}^{2}\frac{\pi}{6}\text{cos}^{2}\frac{\pi}{3}\\\lbrack\because\space \text{cosec}(\pi+\theta)=-\text{cosec}\space\theta\rbrack\\=\frac{1}{2}+(2)^{2}×\bigg(\frac{1}{2}\bigg)^{2}=\frac{1}{2}+1=\frac{3}{2}=\text{R.H.S.}$$
∴ L.H.S. = R.H.S. Hence Proved.
$$\textbf{3. Prove that: cot}^{2}\frac{\pi}{\textbf{6}}+\textbf{cosec}\frac{\textbf{5}\pi}{\textbf{6}}\\+\textbf{3 tan}^{2}\frac{\pi}{\textbf{6}}=\textbf{6}\\\textbf{Sol.}\space \text{cot}^{2}\frac{\pi}{6}+\text{cosec}\frac{5\pi}{6}+3\text{tan}^{2}\frac{\pi}{6}=6\\\text{L.H.S = }\text{cot}^{2}\frac{\pi}{6}+\text{cosec}\frac{5\pi}{6}+3\text{tan}^{2}\frac{\pi}{6}\\=(\sqrt{3})^{2}+\text{cosec}\bigg(\pi-\frac{\pi}{6}\bigg)+3\bigg(\frac{1}{\sqrt{3}}\bigg)^{2}\\=3+\text{cosec}\frac{\pi}{6}+3×\frac{1}{3}$$
[∵ cosec (π – θ) = cosec θ]
= 3 + 2 + 1 = 6 = R.H.S.
∴ L.H.S. = R.H.S. Hence Proved.
$$\textbf{4. Prove that: 2 sin}^{2}\space\frac{\textbf{3}\pi}{\textbf{4}}+2\textbf{cos}^{2}\frac{\pi}{\textbf{4}}+\\2\textbf{sec}^{2}\frac{\pi}{\textbf{3}}\textbf{=10}\\\textbf{Sol.\space}2\space\text{sin}^{2}\frac{3\pi}{4}+2\text{cos}^{2}\frac{\pi}{4}+2\space\text{sec}^{2}\frac{\pi}{3}=10\\\text{L.H.S}=2\space\text{sin}^{2}\frac{3\pi}{4}+2\text{cos}^{2}\frac{\pi}{4}+2\text{sec}^{2}\frac{\pi}{3}\\=2\space\text{sin}^{2}\frac{\pi}{4}+2×\bigg(\frac{1}{\sqrt{2}}\bigg)^{2}+2(2)^{2}$$
[∵ sin (π – θ) = sin θ]
$$= 2×\bigg(\frac{1}{\sqrt{2}}\bigg)^{2}+2×\frac{1}{2}×4\\=2×\frac{1}{2}+1+8$$
= 1 + 1 + 8 = 10 = R.H.S.
L.H.S. = R.H.S. Hence Proved.
5. Find the value of (i) sin 75°, (ii) tan 15°.
Sol. (i) sin 75° = sin (45° + 30°)
[∵ sin (A + B) = sin A cos B + cos A sin B]
= sin 45° cos 30° + cos 45° sin 30°
$$=\frac{1}{\sqrt{2}}×\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}×\frac{1}{2}=\frac{\sqrt{3}+1}{2\sqrt{2}}\\=\frac{\sqrt{2}(\sqrt{3}+1)}{4}$$
(ii) tan 15° = tan (45° – 30°)
$$\bigg(\because\space\text{tan}(A-B)=\frac{\text{tan A - tan B}}{1 + \text{tan A tan B}}\bigg)\\=\frac{\text{tan}\space45-\text{tan}\space30\degree}{1+\text{tan 45\degree tan 30\degree}}=\frac{1-\frac{1}{\sqrt{3}}}{1+1×\frac{1}{\sqrt{3}}}\\=\frac{\sqrt{3}-1}{\sqrt{3}+1}\\=\frac{\sqrt{3}-1}{\sqrt{3}+1}×\frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{(\sqrt{3}-1)^{2}}{3-1}\\=\frac{3+1-2\sqrt{3}}{2}=\frac{4-2\sqrt{3}}{2}\\=\frac{2(2-\sqrt{3})}{2}=2-\sqrt{3}$$
6. Prove that:
$$\textbf{cos}\bigg(\frac{\pi}{\textbf{4}}-x\bigg)\textbf{cos}\bigg(\frac{\pi}{\textbf{4}}-\textbf{y}\bigg)-\\\textbf{sin}\bigg(\frac{\pi}{\textbf{4}}-\textbf{x}\bigg)\textbf{sin}\bigg(\frac{\pi}{\textbf{4}}-\textbf{y}\bigg)\\\textbf{= sin (x + y)}$$
Sol.
$$\text{cos}\bigg(\frac{\pi}{4}-x\bigg)\text{cos}\bigg(\frac{\pi}{4}-y\bigg)-\\\text{sin}\bigg(\frac{\pi}{4}-x\bigg)\text{sin}\bigg(\frac{\pi}{4}-y\bigg)\\=\text{sin}(x+y)\\\because\space\text{L.H.S. = cos}\bigg(\frac{\pi}{4}-x\bigg)\text{cos}\bigg(\frac{\pi}{4}-y\bigg)\\-\space\text{sin}\bigg(\frac{\pi}{4}-x\bigg)\text{sin}\bigg(\frac{\pi}{4}-y\bigg)\\\text{Let}\space\frac{\pi}{4}-x=\text{A}\space\text{and}\space\frac{\pi}{4}-y=\text{B}$$
Then, L.H.S. = cos A cos B – sin A cos B
= cos (A + B)
$$=\text{cos}\bigg(\frac{\pi}{4}-x + \frac{\pi}{4}-y\bigg)\\=\text{cos}\bigg(\frac{\pi}{2}-x-y\bigg)=\text{cos}\bigg[\frac{\pi}{2}-(x+y)\bigg]$$
= sin (x + y) = R.H.S.
∴ L.H.S. = R.H.S. Hence Proved.
$$\textbf{7. Prove that:}\frac{\textbf{tan}\bigg(\frac{\pi}{\textbf{4}}+\textbf{x}\bigg)}{\textbf{tan}\bigg(\frac{\pi}{\textbf{4}}-\textbf{x}\bigg)}\\=\bigg(\frac{\textbf{1}+\textbf{tan}\space \textbf{x}}{\textbf{1}- \textbf{tan} \space \textbf{x}}\bigg)^{2}$$
$$\textbf{Sol.}\space\frac{\text{tan}\bigg(\frac{\pi}{4}+x\bigg)}{\text{tan}\bigg(\frac{\pi}{4}-x\bigg)}=\bigg(\frac{1+\text{tan x}}{1-\text{tan x}}\bigg)^{2}\\\text{L.H.S.}=\frac{\text{tan}\bigg(\frac{\pi}{4}+x\bigg)}{\text{tan}\bigg(\frac{\pi}{4}-x\bigg)}\\=\frac{\text{tan}\space\frac{\pi}{4}+\text{tan x}}{1-\text{tan}\frac{\pi}{4}\space\text{tan x}}×\frac{1+\text{tan}\frac{\pi}{4}\text{tan x}}{\text{tan}\frac{\pi}{4}-\text{tan x}}\\\bigg(\because \text{tan(A+B)}=\frac{\text{tan A}+\text{tan B}}{1-\text{tan A}\text{tan B}}\text{and}\\\text{tan(A-B)}=\frac{\text{tan A - tan B}}{1 + \text{tan A tan A}}\bigg)$$
$$=\frac{1+\text{tan x}}{1-\text{tan x}}×\frac{1+\text{tan x}}{1-\text{tan x}}\\=\bigg(\frac{1+\text{tan x}}{1 - \text{tan x}}\bigg)^{2}=\text{R.H.S}$$
∴ L.H.S. = R.H.S. Hence Proved.
$$\textbf{8. Prove that:}\space\frac{\textbf{cos}(\pi+\textbf{x})\textbf{cos}(-\textbf{x})}{\textbf{sin}(\pi-\textbf{x})\textbf{cos}\bigg(\frac{\pi}{2}+\textbf{x}\bigg)}\\=\textbf{cot}^{2}x.\\\textbf{Sol.}\space\frac{\text{cos}(\pi+x)\text{cos}(-x)}{\text{sin}(\pi-x)\text{cos}\bigg(\frac{\pi}{2}+x\bigg)}=\text{cot}^{2}x\\\text{L.H.S}=\frac{\text{cos}(\pi+x)\text{cos}(-x)}{\text{sin}(\pi-x)\text{cos}\bigg(\frac{\pi}{2}+x\bigg)}\\=\frac{(-\text{cos x})(\text{cos x})}{(\text{sin x})(\text{-sin x})}$$
$$\begin{pmatrix}\because\space\text{cos}(\pi+\theta)=-\text{cos}\space\theta\\\text{cos}(-\theta)=\text{cos}\space\theta\\\text{sin}(\pi-\theta)=\text{sin}\space\theta\\\text{cos}(\frac{\pi}{2}+\theta)=-\text{sin}\space\theta\end{pmatrix}\\=\frac{\text{cos}^{2}x}{\text{sin}^{2}x}=\text{cot}^{2}x=\text{R.H.S}$$
∴ L.H.S. = R.H.S. Hence Proved.
9. Prove that:
$$\textbf{cos}\bigg(\frac{\textbf{3}\pi}{\textbf{2}}+\textbf{x}\bigg)\textbf{cos}(\textbf{2}\pi+\textbf{x})\\\bigg[\textbf{cot}\bigg(\frac{\textbf{3}\pi}{\textbf{2}}-\textbf{x}\bigg)+\textbf{cot}(\textbf{2}\pi+\textbf{x})\bigg]=\textbf{1}$$
$$\textbf{Sol.}\space\text{L.H.S.}=\text{cos}\bigg(\frac{3\pi}{2}+x\bigg)\text{cos}(2\pi+x)\\\bigg[\text{cot}\bigg(\frac{3\pi}{2}-x\bigg)+ \text{cot}(2\pi+x)\bigg]=1$$
= (sin x) (cos x) [tan x + cot x]
$$=\text{sin x cos x}\bigg[\frac{\text{sin x}}{\text{cos x}}+\frac{\text{cos x}}{\text{sin x}}\bigg]\\\begin{bmatrix}\because\space \text{cot}\bigg(\frac{3\pi}{2}-\theta\bigg)=\text{tan}\space\theta\\\text{cos}\bigg(\frac{3\pi}{2}+\theta\bigg)=\text{sin}\space\theta\end{bmatrix}\\=\text{sin x cos x}\bigg[\frac{\text{sin}^{2}x+\text{cos}^{2}x}{\text{cos x sin x}}\bigg]$$
= 1 = R.H.S.
L.H.S. = R.H.S. Hence Proved.
10. Prove that:
sin (n + 1) x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x.
Sol. sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x
L.H.S. = sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n+ 2)x
= cos [(n + 2)x – (n + 1)x
[By the formula cos (A – B) = cos A cos B + sin A cos B]
= cos (nx + 2x – nx – x)
= cos x = R.H.S.
∴ L.H.S. = R.H.S. Hence Proved.
11. Prove that:
$$\textbf{cos}\bigg(\frac{\textbf{3}\pi}{\textbf{4}}+\textbf{x}\bigg)-\textbf{cos}\bigg(\frac{\textbf{3}\pi}{\textbf{4}}-\textbf{x}\bigg)=-\sqrt{\textbf{2}}\space\textbf{sin x}.$$
Sol.
$$\text{cos}\bigg(\frac{3\pi}{4}+x\bigg)-\text{cos}\bigg(\frac{3\pi}{4}-x\bigg)=-\sqrt{2}\space\text{sin x}\\\text{L.H.S}=\text{cos}\bigg(\frac{3\pi}{4}+x\bigg)-\text{cos}\bigg(\frac{3\pi}{4}-x\bigg)\\=-\text{2 sin}\frac{\frac{3\pi}{4}+x+\frac{3\pi}{4}-x}{2}\text{sin}\frac{\frac{3\pi}{4}+x-\frac{3\pi}{4}+x}{2}\\\begin{bmatrix}\because\space\text{cos A - cos B}\\=-2\space\text{sin}\bigg(\frac{\text{A+B}}{2}\bigg)\text{sin}\bigg(\frac{\text{A-B}}{2}\bigg) \end{bmatrix}\\=-2\space\text{sin}\frac{3\pi}{4}\text{sin x}\\=-2\space\text{sin }\bigg(\pi-\frac{\pi}{4}\bigg)\text{sin x}\\=-\text{2 sin}\space\frac{\pi}{4}\text{sin x}=-2×\frac{1}{\sqrt{2}}\space\text{sin x}$$
$$=-\sqrt{2}\space\text{sin x}=\text{R.H.S}$$
∴ L.H.S. = R.H.S. Hence Proved.
12. Prove that: sin2 6x – sin2 4x = sin 2x sin 10x.
Sol. sin2 6x – sin2 4x = sin 2x sin 10x
L.H.S. = sin2 6x – sin2 4x
= sin (6x + 4x) sin (6x – 4x)
= sin 10x sin 2x = R.H.S.
∴ L.H.S. = R.H.S. Hence Proved.
13. Prove that: cos2 2x – cos2 6x = sin 4x sin 8x.
Sol. cos2 2x – cos2 6x = sin 4x sin 8x
∵ L.H.S. = cos2 2x – cos2 6x
= (1 – sin2 2x) – (1 – sin2 6x)
(∵ sin2 A + cos2 A = 1)
= sin2 6x – sin2 2x
= sin (6x + 2x) sin (6x – 2x)
[∵ sin2 A – sin2 B = sin (A + B) sin (A – B)]
= sin 8x sin 4x = R.H.S.
∴ L.H.S. = R.H.S. Hence Proved.
14. Prove that:
sin 2x + 2 sin 4x + sin 6x = 4 cos2 x sin 4x.
Sol. sin 2x + 2 sin 4x + sin 6x = 4 cos2 x sin 4x
L.H.S. = sin 2x + 2 sin 4x + sin 6x
= sin 6x + sin 2x + 2 sin 4x
$$= 2\space\text{sin}\frac{6x+2x}{2}\text{cos}\frac{6x-2x}{2}+2\text{sin} 4x\\ \begin{bmatrix}\because\space\text{sin A + sin B}\\=2\space \text{sin}\bigg(\frac{\text{A+B}}{2}\bigg)\text{cos}\bigg(\frac{\text{A-B}}{2}\bigg)\end{bmatrix}$$
= 2 sin 4x cos 2x + 2 sin 4x
= 2 sin 4x (cos 2x + 1)
= 2 sin 4x (2 cos2 x – 1 + 1)
= 2 sin 4x 2 cos2 x
= 4 cos2 x sin 4x = R.H.S.
∴ L.H.S. = R.H.S. Hence Proved.
15. Prove that:
cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x).
Sol. cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x).
L.H.S. = cot 4x (sin 5x + sin 3x)
$$= \text{cot 4x.2sin}\space\frac{5x+3x}{2}\space\text{cos}\frac{5x-3x}{2}\\\bigg[\because \text{sin} C + \text{sin} D =2 \text{sin}\frac{\text{C+D}}{2}\text{cos}\frac{\text{C-D}}{2} \bigg]\\=\frac{\text{cos 4x}}{\text{sin} 4x}×2 \text{sin 4x cos x}$$
= 2 cos 4x cos x
R.H.S. = cot x (sin 5x – sin 3x)
$$=\text{cot x}\space.\space2\space\text{cos}\frac{5x+3x}{2}\text{sin}\frac{5x+3x}{2}\\\bigg[\because \text{sin C - sin D = 2 cos}\frac{\text{C+D}}{2}\text{sin}\frac{\text{C-D}}{2}\bigg]\\=\frac{\text{cos x}}{\text{sin x}}×\text{2 cos 4x sin x}$$
= 2 cos 4x cos x
∴ L.H.S. = R.H.S. Hence Proved.
$$\textbf{16. Prove that:}\space\frac{\textbf{cos 9x- cos 5x}}{\textbf{sin 17x - sin 3x} }\\=-\frac{\textbf{sin 2x}}{\textbf{cos 10x}}\\\qquad\textbf{Sol.}\space\frac{\text{cos 9x - cos 5x}}{\text{sin 17 x - sin 3x}}=-\frac{\text{sin 2x}}{\text{cos 10x}}\\\text{L.H.S.}=\frac{\text{cos}\space 9x- \text{cos}\space5x}{\text{sin 17x - sin 3x}}\\=\frac{-2\space\text{sin}\frac{9x+5x}{2}\text{sin}\frac{9x-5x}{2}}{2 \space\text{cos}\frac{17x+3x}{2} \text{sin}\frac{17x-3x}{2}}\\ \begin{bmatrix}\because \text{cos A - cos B }\\=-2\text{sin}\frac{A+B}{2}\text{sin}\frac{\text{A-B}}{2}\\\text{sin A- sin B}\\=\text{2 cos}\frac{A+B}{2}\text{sin}\frac{A-B}{2}\end{bmatrix}$$
$$=\frac{\text{-sin 7x}\space\text{sin 2x}}{\text{cos 10 x sin 7x}}=-\frac{\text{sin}\space2x}{\text{cos}\space10 x}$$
= R.H.S.
∴ L.H.S. = R.H.S. Hence Proved.
$$\textbf{17. Prove that:}\space\frac{\textbf{sin 5x+ sin 3x}}{\textbf{cos 5x + cos 3x}}=\textbf{tan 4x.}$$
$$\textbf{Sol.}\space\frac{\text{sin 5x + sin 3x}}{\text{cos 5x + cos 3x}}=\text{tan 4x}\\\text{L.H.S}=\frac{\text{sin 5x + sin 3x}}{\text{cos 5x + cos 3x}}\\=\frac{2\space\text{sin} \frac{5x+3x}{2}\text{cos}\frac{5x-3x}{2}}{2 \text{cos}\frac{5x+3x}{2}\text{cos}\frac{5x-3x}{2}}\\\begin{bmatrix}∵ \space \text{cos A+cos B}\\ 2\space\text{cos}\frac{\text{A+B}}{2}\text{cos}\frac{\text{A-B}}{2}\\\text{and \space sin A + sin B}\\=2\text{sin}\frac{\text{A+B}}{2}\text{cos}\frac{\text{A-B}}{2}\end{bmatrix}\\=\frac{\text{sin 4x cos x}}{\text{cos 4x cos x}}$$
= tan 4x = R.H.S.
∴ L.H.S. = R.H.S. Hence Proved.
$$\textbf{18. Prove that:}\space\frac{\textbf{sin x- sin y}}{\textbf{cos x + cos y}}\\=\textbf{tan}\frac{\textbf{x-y}}{\textbf{2}}$$
$$\textbf{Sol.}\space\frac{\text{sin x- sin y}}{\text{cos x- cos y}}=\text{tan}\frac{x-y}{2}\\\text{L.H.S}=\frac{\text{sin x - siny}}{\text{cos x + cos y}}\\=\frac{\text{2 cos}\frac{x+y}{2}\text{sin}\frac{x-y}{2}}{2\text{cos}\frac{x+y}{2}\text{cos}\frac{x-y}{2}}\\\begin{bmatrix}\because\space \text{cos A + cos B}\\2 \text{cos}\frac{A+B}{2}\text{cos}\frac{A-B}{2}\\\text{and}\space \text{sinA}-\text{sinB}\\ = 2 \text{cos}\frac{\text{A+B}}{2}\text{sin}\frac{\text{A-B}}{2}\end{bmatrix}\\=\text{tan}\frac{x-y}{2} =\text{R.H.S.}$$
∴ L.H.S. = R.H.S. Hence Proved.
$$\textbf{19. Prove that:}\space\frac{\textbf{sin x + sin 3x}}{\textbf{cos x + cos 3x}}\\=\textbf{tan 2x.}\\\textbf{Sol.}\space\frac{\text{sin x + sin 3x}}{\text{cos x + cos 3x}}\\=\frac{\text{sin 3x + sin x}}{\text{cos 3x + cos x}}\\=\frac{2\space\text{sin}\frac{3x+x}{2}\text{cos}\frac{3x-x}{2}}{2\space\text{cos}\frac{3x+x}{2}\text{cos}\frac{3x-x}{2}}\\$$
$$\begin{bmatrix}\because\space \text{cos A + cos B = 2 cos}\frac{\text{A+B}}{2}\text{cos}\frac{\text{A-B}}{2}\\\text{and}\space\text{sin A + sin B = 2 sin}\frac{\text{A+B}}{2}\text{cos}\frac{\text{A-B}}{2}\end{bmatrix}\\=\frac{\text{sin 2x cos x}}{\text{cos 2x cosx}}=\text{tan 2x = R.H.S}$$
∴ L.H.S. = R.H.S. Hence Proved.
$$\textbf{20. Prove that:}\frac{\textbf{sin x - sin 3x}}{\textbf{sin}^{2}\textbf{x}-\textbf{cos}^{2}\textbf{x}}=2\textbf{sin x}\\\textbf{Sol.\space}\frac{\text{sin x- sin 3x}}{\text{sin}^{2}\text{x}-\text{cos}^{2}\text{x}}=\text{2 sin x}\\\text{L.H.S}=\frac{\text{sin x- sin 3x}}{\text{sin}^{2}x- \text{cos}^{2}x}\\=\frac{\text{sin 3x - sin x}}{\text{cos}^{2}x-\text{sin}^{2}x}\\=\frac{2\space\text{cos}\frac{3x+x}{2}\text{sin}\frac{3x-x}{2}}{\text{cos 2x}}\\\begin{bmatrix}\because \text{sin A - sin B = 2\text{cos}}\frac{\text{A+B}}{2}\text{sin}\frac{\text{A-B}}{2}\\\text{cos}^{2}x-\text{sin}^{2}x=\text{cos 2x}\end{bmatrix}\\=\frac{\text{2 cos 2x sin x}}{\text{cos 2x}}=\text{2 sin x= R.H.S}$$
∴ L.H.S. = R.H.S.
$$\textbf{21. Prove that:}\space\frac{\textbf{cos 4x} + \textbf{cos 3x} + \textbf{cos 2x}}{\textbf{sin 4x} + \textbf{sin 3x} + \textbf{sin 2x}}=\\\textbf{cot 3x}\\\textbf{Sol.}\space\frac{\text{cos 4x + cos 3x + cos 2x}}{\text{sin 4x + sin 3x + sin 2x}}\\=\text{cot 3x}\\\text{L.H.S}=\frac{\text{cos 4x + cos 3x + cos 2x}}{\text{sin 4x + sin 3x + sin 2x}}\\=\frac{\text{(cos 4x + cos 2x) +}\space\text{cos 3x}}{\text{(sin 4x + sin 2x) + sin 3x}}\\=\frac{\text{2 cos}\frac{4x+2x}{2}\text{cos}\frac{4x-2x}{2} + \text{cos 3x} }{\text{ 2 sin}\frac{4x+2x}{2}\text{cos}\frac{4x-2x}{2} + \text{sin 3x}}\\\begin{bmatrix}\because \text{cos A + cos B = 2 cos}\frac{\text{A+B}}{2}\text{cos}\frac{\text{A-B}}{2}\\\because\text{sin A + sin B = 2 sin}\frac{\text{A+B}}{2}\text{cos}\frac{\text{A-B}}{2}\end{bmatrix}\\=\frac{\text{2 cos 3x cos x + cos 3x}}{\text{2 sin 3x cos x + sin 3x}}$$
= cot 3x = R.H.S.
∴ L.H.S. = R.H.S. Hence Proved.
22. Prove that:
cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1.
Sol. cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1
Now, cot 3x = cot (2x + x)
$$\frac{\text{cot 3x}}{1}=\frac{\text{cot 2x cot x - 1}}{\text{cot 2x + cot x}}\\\bigg[\because\space\text{cot(A+B)}=\frac{\text{cot A cot B - 1}}{\text{cot B + cot A}}\bigg]$$
⇒ cot 3x cot 2x + cot 3x cot x = cot 2x cot x – 1
⇒ cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1.
Hence Proved.
$$\textbf{23. Prove that: 4x}=\frac{\textbf{4 tan x}(1 - \textbf{tan}^{2}\textbf{x})}{\textbf{1-6 tan}^{2}x + \textbf{tan}^{4}\textbf{x}}\\\textbf{Sol.}\space\text{tan 4x}=\frac{\text{4 tan x(1- tan}^{2}\text{x})}{1-6\text{tan}^{2}x + \text{tan}^{4}\textbf{x}}$$
L.H.S. = tan 4x = tan 2(2x)
$$=\frac{\text{2 tan 2x}}{1- \text{tan}^{2}\space 2x}=\frac{2\frac{\text{2 tan x}}{\text{1- tan}^{2}x}}{1-\bigg(\frac{\text{2 tan x}}{\text{1- tan}^{2}x}\bigg)^{2}}\\\begin{bmatrix}\because\space\text{tan 2A}=\frac{\text{2 tan A}}{1- \text{tan}^{2}\textbf{A}}\end{bmatrix}\\=\frac{\text{4 tan x}}{\text{1- tan}^{2}x}×\frac{(1- \text{tan}^{2}x)^{2}}{(1-\text{tan}^{2}x)^{2}-4\text{tan}^{2}x}\\=\frac{\text{4 tan x(1- tan})^{2}x}{1+\text{tan}^{4}x-2\text{tan}^{2}x-4\text{tan}^{2}x}\\=\frac{4 \text{tan} x(1- \text{tan}^{2}x)}{1-6 \text{tan}^{2}x + \text{tan}^{4}x}=\text{R.H.S}$$
∴ L.H.S. = R.H.S.
24. Prove that: cos 4x = 1 – 8 sin2 x cos2 x.
Sol. cos 4x = 1 – 8 sin2 x cos2 x
L.H.S. = cos 4x = 1 – 2 sin2 2x
= 1 – 2 (sin 2x)2
[∵ cos 2x = 1 – 2 sin2 x]
= 1 – 2(2 sin x cos x)2
= 1 – 8 sin2 x cos2 x = R.H.S.
∴ L.H.S. = R.H.S. Hence Proved.
25. Prove that:
cos 6x = 32 cos6 x – 48 cos4 x + 18 cos2x – 1.
Sol. cos 6x = 32 cos6 x – 48 cos4 x + 18 cos2 x – 1
L.H.S. = cos 6x = cos 2(3x) = 2 cos2 3x – 1
= 2(cos 3x)2 – 1 (∵ cos 2θ = 2 cos2 θ – 1)
= 2(4 cos3 x – 3 cos x)2 – 1
(∴ cos 3θ = 4 cos3 θ – 3 cos θ)
= 2(16 cos6 x + 9 cos2 x – 24 cos4 x) – 1
= 32 cos6 x – 48 cos4 x + 18 cos2 x – 1
= R.H.S.
∴ L.H.S. = R.H.S. Hence Proved.
Exercise 3.4
Find the principal and general solutions of the following equations:
$$\textbf{1. tan x =}\sqrt{\textbf{3}}\\\textbf{Sol.}\space\text{tan x}=\sqrt{3}\\\text{tan x}=\text{tan}\frac{\pi}{3}\text{or tan}\bigg(\pi + \frac{\pi}{3}\bigg)\\=\text{tan}\frac{\pi}{3}\space\text{or tan}\bigg(\pi + \frac{\pi}{3}\bigg)\\=\text{tan}\frac{\pi}{3}\text{or tan}\frac{4\pi}{3}\\\Rarr\space x=\frac{\pi}{3}\text{or}\frac{4\pi}{3}\\\text{Hence, principal value is x =}\frac{\pi}{3}.$$
We know that, if tan x = tan α, then
General solution is x = nπ + α
$$\therefore\space\text{x = nπ +}\frac{\pi}{3}\space\space\text{where (n ∈ Z)}$$
2. sec x = 2.
Sol. sec x = 2
$$\Rarr\space\text{cos x =}\frac{1}{2}\\\Rarr\space\text{cos x = cos}\frac{\pi}{3}\text{or cos}\bigg(2\pi-\frac{\pi}{3}\bigg)$$
(∵ cos x is positive in 1st and 4th quadrant)
$$x=\frac{\pi}{3}\text{or}\frac{5\pi}{3}\\\text{Here, principal value is x =}\frac{\pi}{3},\\\text{we know that, if cos x = cos a, then}\\\text{General solution, x = 2n}\pi ± a\\\therefore\space\text{x = 2n}\pi\pm\frac{\pi}{3}\\\text{where (n ∈ Z)}$$
3. cotx=- $$\sqrt{3}$$
$$\textbf{Sol.}\space\text{cot x}=-\sqrt{3}\\\Rarr\space\text{tan x}=-\frac{1}{\sqrt{3}}\\\Rarr\space\text{tan x = -tan}\frac{\pi}{6}\\\Rarr\text{tan x = tan}\bigg(\pi-\frac{\pi}{6}\bigg)\text{or tan}\\\bigg(2\pi-\frac{\pi}{6}\bigg)$$
[∵ tan x is negative in 2nd and 4th quadrant and tan (π – θ) = – tan θ and tan (2π – θ) = – tan θ]
$$\text{tan x} = \text{tan}\frac{5\pi}{6}\text{or tan}\frac{11 \pi}{6}\\\Rarr\space x=\frac{5\pi}{6}\text{or}\frac{11\pi}{6}\\\text{Here, principal value is x =}\frac{5\pi}{6}$$
We know that, if tan x = tan α, then
General solution, x = nπ + α
$$x=\text{n}\pi+\frac{5\pi}{6}\space\text{where (n ∈ Z)}$$
4. cosec x = – 2
Sol. cosec x = – 2
$$\Rarr\space\text{sin x}=-\frac{1}{2}\\\Rarr\space\text{sin x}=-\text{sin}\frac{\pi}{6}\\\Rarr\space\text{sin x = sin}\bigg(\pi +\frac{\pi}{6}\bigg)\\\text{or sin }\bigg(2\pi-\frac{\pi}{6}\bigg)$$
[∵ sin x is negative in 3rd and 4th quadrant, or sin (π + θ) = sin (2π – θ) = – sin θ]
$$\text{sin x}=\text{sin}\frac{7\pi}{6}\text{or sin}\frac{11 \pi}{6}$$
Here, principal value is
$$x=\frac{7\pi}{6}$$
General solution,
x = nπ + (– 1)n a
$$\text{x = n}\pi + (\normalsize-1)^{n}\frac{7\pi}{6}\space\text{where (n ∈ Z)}$$
Find the general solution for each of the following equations:
5. cos 4x = cos 2x.
Sol. Given, cos 4x = cos 2x
⇒ cos 4x – cos 2x = 0
$$\Rarr\space\text{- 2 sin}\frac{4x+2x}{2}\text{sin}\frac{4x-2x}{2}=0\\\bigg[\because\space \text{cos A - cos B =\space-\text{2 sin}}\bigg(\frac{\text{A+B}}{2}\bigg)\text{sin}\bigg(\frac{\text{A-B}}{2}\bigg)\bigg]$$
⇒ sin 3x sin x = 0
⇒ sin 3x = 0 or sin x = 0
⇒ 3x = nπ or x = nπ
$$\Rarr\space\text{x}=\frac{n\pi}{3}\text{or x=n}\pi$$
where (n ∈ Z)
6. cos 3x + cos x – cos 2x = 0.
Sol. cos 3x + cos x – cos 2x = 0
$$\Rarr\space\text{2 cos}\frac{3x+x}{2}\text{cos}\frac{3x-x}{2}-\text{cos 2x}= 0\\\begin{bmatrix} \because\space\text{cos C+ cos D }\\\text{= 2 cos}\frac{\text{C+D}}{2}\text{cos}\frac{\text{C-D}}{2}\end{bmatrix}$$
⇒ 2 cos 2x cos x – cos 2x = 0
⇒ cos 2x (2 cos x – 1) = 0
⇒ cos 2x = 0 or 2 cos x – 1 = 0
$$\Rarr\space\text{2x}=(2n+1)\frac{\pi}{2}\text{or cos x}=\frac{1}{2}\\\Rarr\space\text{x = (2n + 1)}\frac{\pi}{4}\text{or cos x = cos}\frac{\pi}{3}\\\Rarr\space\text{x=(2n+1)}\frac{\pi}{4}\text{or x = 2n}\pi+\frac{\pi}{3}$$
(n ∈ Z)
7. sin 2x + cos x = 0.
Sol. sin 2x + cos x = 0
⇒ 2 sin x cos x + cos x = 0
(∵ sin 2x = 2 sin x cos x)
⇒ cos x (2 sin x + 1) = 0
$$\Rarr\space\text{cos x = 0 or sin x = −}\frac{1}{2}$$
When cos x = 0
$$\text{Then,}\space x=(2n+1)\frac{\pi}{2}\\\text{When sin x = −}\frac{1}{2},\\\text{Then, sin x = – sin}\frac{\pi}{6}\\\text{sin x = sin}\bigg(\pi+\frac{\pi}{6}\bigg)$$
[∵ sin (π + θ) = – sin θ]
$$\text{sin x = sin}\frac{7\pi}{6}\\\Rarr\space\text{x = n}\pi + (\normalsize-1)^{n}\frac{7\pi}{6}\\\space\text{where (n ∈ Z)} $$
8. sec2 2x = 1 – tan 2x.
Sol. sec2 2x = 1 – tan 2x
1 + tan2 2x = 1 – tan 2x
(∵ sec2 x = 1 + tan2 x)
⇒ tan2 2x + tan 2x = 0
⇒ tan 2x (tan 2x + 1) = 0
⇒ tan 2x = 0 or tan 2x = – 1
When tan 2x = 0, then 2x = nπ
$$\Rarr\space\text{x}=\frac{n\pi}{2}$$
When tan 2x = – 1
$$\text{then, tan 2x = – tan}\frac{\pi}{4}\\\text{tan 2x = tan}\bigg(\pi-\frac{\pi}{4}\bigg)\\=\text{tan}\frac{3\pi}{4}$$
[∴ tan (π – θ) = – tan θ]
∴ General solutions
$$\text{2x = n}\pi + \frac{3\pi}{4}\\\Rarr\space x=\frac{n\pi}{2}+\frac{3\pi}{8}$$
where n ∈ Z
9. sin x + sin 3x + sin 5x = 0.
Sol. sin x + sin 3x + sin 5x = 0
⇒ (sin 5x + sin x) + sin 3x = 0
$$\Rarr\space\text{2 sin}\frac{5x+x}{2}\text{cos}\frac{5x-x}{2}+\text{sin 3x = 0}\\\begin{bmatrix}\because\space\text{sin A + sin B} = \\2\space\text{sin}\bigg(\frac{\text{A+B}}{2}\bigg)\text{cos}\bigg(\frac{\text{A-B}}{2}\bigg)\end{bmatrix}$$
⇒ 2 sin 3x cos 2x + sin 3x = 0
⇒ sin 3x (2 cos 2x + 1) = 0
Either sin 3x = 0 or 2 cos 2x + 1 = 0
When sin 3x = 0,
Then, 3x = nπ
$$\Rarr\space\text{x}=\frac{n\pi}{3}$$
When 2 cos 2x + 1 = 0,
$$\text{Then,}\space\text{cos 2x}=-\frac{1}{2}\\\Rarr\space\text{cos 2x = cos}\frac{\pi}{3}\\\Rarr\space\text{cos 2x =}\space\text{cos}\bigg(\pi-\frac{\pi}{3}\bigg)$$
[∵ cos (π – θ) = – cos θ]
$$\Rarr\space\text{cos 2x = cos}\frac{2\pi}{3}$$
∴ General solutions
$$\text{2x = 2n}\pi + \frac{2\pi}{3}\\\Rarr\space\text{x=n}\pi\pm\frac{\pi}{3}.\space\text{where n ∈ Z}$$
Miscellaneous Exercise
$$\textbf{1. 2 cos}\frac{\pi}{\textbf{3}}\textbf{cos}\frac{\textbf{9}\pi}{\textbf{13}} + \textbf{cos}\frac{\textbf{3}\pi}{\textbf{13}} +\textbf{cos}\frac{\textbf{5}\pi}{\textbf{13}}\textbf{=0}$$
$$\textbf{Sol.}\space\text{2 cos}\frac{\pi}{3}\text{cos}\frac{9\pi}{13} + \text{cos}\frac{3\pi}{13} + \text{cos}\frac{5\pi}{13}=0\\\text{L.H.S.}= \text{2 cos}\frac{\pi}{13}\text{cos}\frac{9\pi}{13} + \text{cos}\frac{3\pi}{13} + \text{cos}\frac{5\pi}{13}\\=\text{cos}\bigg(\frac{9\pi}{13} + \frac{\pi}{13}\bigg) + \text{cos}\bigg(\frac{9\pi}{13}-\frac{\pi}{13}\bigg) + \\\text{cos}\frac{5\pi}{13} + \text{cos}\frac{3\pi}{13}\\\text{(By forumla)}\\=\text{cos}\frac{10\pi}{13} + \text{cos}\frac{8\pi}{13} + \text{cos}\frac{5\pi}{13} + \text{cos}\frac{3\pi}{13}\\=\bigg(\text{cos}\frac{10\pi}{13} + \text{cos}\frac{3\pi}{13}\bigg) + \bigg(\text{cos}\frac{8\pi}{13} + \text{cos}\frac{5\pi}{13}\bigg)$$
$$=\space\text{2 cos}\frac{\frac{10\pi}{13} +\frac{3\pi}{13} }{2}\text{cos}\frac{\frac{10\pi}{13}-\frac{3\pi}{13}}{2}\\+\space\text{2 cos}\frac{\frac{8\pi}{13} + \frac{5\pi}{13}}{2}\text{cos}\frac{\frac{8\pi}{13}-\frac{5\pi}{13}}{2}$$
[∵ 2 cos A cos B = cos (A + B) cos (A – B)]
$$=\text{2 cos}\frac{\pi}{2}\text{cos}\frac{7\pi}{26} + \text{2 cos}\frac{\pi}{2}\text{cos}\frac{3\pi}{26}$$
= 0 + 0 = R.H.S.
∴ L.H.S. = R.H.S. Hence Proved.
2. (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0.
Sol. ∵ L.H.S. = (sin 3x + sin x) sin x + (cos 3x – cos x) cos x
$$=\text{2 sin}\frac{3x+x}{2}\text{cos}\frac{3x-x}{2}\text{sin x}\\-\text{2 sin}\frac{3x+x}{2}\text{sin}\frac{3x-x}{2}\text{cos x}\\\begin{bmatrix}\because\space \text{sinn A + sin B}= \\2\space\text{sin}\frac{\text{A+B}}{2}\text{sin}\frac{\text{A-B}}{2}\end{bmatrix}$$
$$=\text{2 sin}\frac{3x+x}{2}\text{cos}\frac{3x-x}{2}\text{sin x}\\-\text{2 sin}\frac{3x+x}{2}\text{sin}\frac{3x-x}{2}\text{cos x}\\\begin{bmatrix}\because\space \text{sinn A + sin B}= \\2\space\text{sin}\frac{\text{A+B}}{2}\text{cos}\frac{\text{A-B}}{2}\\\text{and}\space\text{cos A - cos B}\\-\text{2 sin}\frac{\text{A+B}}{2}\text{sin}\frac{\text{A-B}}{2}\end{bmatrix}$$
= 2 sin 2x cos x sin x – 2 sin 2x sin x cos x = 0
= R.H.S.
∴ L.H.S. = R.H.S. Hence Proved.
$$\textbf{3. (cos x + cos y)}^2 + \textbf{(sin x – sin y)}^2\\ = \textbf{4 cos}^2\frac{\textbf{x+y}}{\textbf{2}}\\\textbf{Sol.}\space\text{(cos x + cos y)}^2 + \text{(sin x –sin y)}^2 \\= \text{4 cos}^2\frac{\text{x+y}}{2}$$
L.H.S. = (cos x + cos y)2 + (sin x – sin y)2
$$=\bigg(\text{2 cos}\frac{x+y}{2}\text{cos}\frac{x-y}{2}\bigg)^{2} \\+ \bigg(\text{2 cos}\frac{x+y}{2}\text{sin}\frac{x-y}{2}\bigg)^{2}\\\text{(By formulae)}\\=\text{4 cos}^{2}\frac{x+y}{2}\text{cos}^{2}\frac{x-y}{2}\\+\text{4 cos}^{2}\frac{x+y}{2}\text{sin}^{2}\frac{x-y}{2}\\=\text{4 cos}^{2}\frac{x+y}{2}\bigg(\text{cos}^{2}\frac{x-y}{2} +\\\text{sin}^{2}\frac{x-y}{2} \bigg)\\=\text{4 cos}^{2}\frac{x+y}{2}$$
(∵ cos2 x + sin2 y = 1)
= R.H.S.
∴ L.H.S. = R.H.S. Hence Proved.
$$\textbf{4. (cos x – cos y)}^{2} + \textbf{(sin x – sin y)}^2\\ = \textbf{4 sin}^{2}\frac{\textbf{x-y}}{\textbf{2}}\\\textbf{Sol.}\space\text{(cos x – cos y)}^{2} + (\text{sin x - sin y})^{2}\\\text{4 sin}^{2}\space\frac{x-y}{2}$$
∵ L.H.S. = (cos x – cos y)2 + (sin x – sin y)2
$$=\bigg(\text{-2 sin}\frac{x+y}{2}\text{sin}\frac{\text{x-y}}{2}\bigg)^{2} \\+ \bigg(\text{2 cos}\frac{x+y}{2}\text{sin}\frac{x-y}{2}\bigg)^{2}$$
(By formulae)
$$=\text{4 sin}^{2}\space\frac{x+y}{2}\text{sin}^{2}\frac{x-y}{2}\\+\text{4 cos}^{2}\frac{x+y}{2}\text{sin}^{2}\frac{x-y}{2}\\=\text{4 sin}^{2}\frac{x-y}{2}\bigg(\text{sin}^{2}\frac{x+y}{2} + \text{cos}^{2}\frac{x+y}{2}\bigg)\\=\text{4 sin}^{2}\frac{x-y}{2}=\text{R.H.S}$$
(∵ cos2 x + sin2 x = 1)
∴ L.H.S. = R.H.S. Hence Proved.
5. sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x.
Sol. sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x
L.H.S. = sin x + sin 3x + sin 5x + sin 7x = (sin 7x + sin x) + (sin 5x + sin 3x)
$$=\text{2 sin}\frac{7x+x}{2}\text{cos}\frac{7x-x}{2}\\+\text{2 sin}\frac{5x+3x}{2}\text{cos}\frac{5x-3x}{2}\\\begin{bmatrix}\because\space \text{sin A + sin B}\\=\text{2 sin}\bigg(\frac{\text{A+B}}{2}\bigg)\text{cos}\bigg(\frac{\text{A-B}}{2}\bigg)\end{bmatrix}$$
= 2 sin 4x cos 3x + 2 sin 4x cos x
= 2 sin 4x (cos 3x + cos x)
$$=\text{2 sin 4x 2 cos}\frac{3x+x}{2}\text{cos}\frac{3x-x}{2}\\\begin{bmatrix}\because\space\text{cos A + cos B}\\\text{2 cos}\bigg(\frac{\text{A+B}}{2}\bigg)\text{cos}\bigg(\frac{\text{A-B}}{2}\bigg)\end{bmatrix}$$
= 4 sin 4x cos 2x cos x = R.H.S.
∴ L.H.S. = R.H.S. Hence Proved.
$$\textbf{6.}\space\frac{\textbf{(sin 7x + sin 5x) + (sin 9x + sin 3x) }}{\textbf{(cos 7x + cos 5x) + (\text{cos 9x + cos 3x})}}\\=\textbf{tan 6x.}\\\textbf{Sol.}\space\frac{(\text{sin} 7x + \text{sin} 5x) + (\text{sin} 9x + \text{sin} 3x)}{\text{(cos 7x + cos 5x) + (cos 9x + cos 3x)}}\\=\text{tan 6x}\\\text{L.H.S}=\frac{(\text{sin 7x + sin 5x})(\text{sin}\space9x + \text{sin}\space3x)}{(\text{cos 7x + cos 5x}) + (\text{cos}\space9x + \text{cos}\space3x)}\\=\frac{\text{2 sin}\frac{7x+5x}{2}\text{cos}\frac{7x-5x}{2} + \text{2 sin}\frac{9x+3x}{2}\text{cos}\frac{9x-3x}{2}}{\text{2 cos}\frac{7x+5x}{2}\text{cos}\frac{7x-5x}{2} + 2\text{cos}\frac{9x+3x}{2}\text{cos}\frac{9x-3x}{2}}\\\begin{bmatrix}\because\text{sin A + sin B}\\=\text{2 sin}\bigg(\frac{\text{A+B}}{2}\bigg)\text{cos}\bigg(\frac{\text{A-B}}{2}\bigg)\end{bmatrix}$$
$$=\frac{\text{2 sin 6x cos x + 2 sin 6x cos 3x}}{\text{2 cos 6x cos x + 2 cos 6x cos 3x}}\\=\frac{\text{2 sin 6x}(\text{cos x + cos 3x})}{\text{2 cos 6x}(\text{cos x + cos 3x})}$$
= tan 6x = R.H.S.
∴ L.H.S. = R.H.S. Hence Proved.
$$\textbf{7. sin 3x + sin 2x – sin x}\\ \textbf{= 4 sin x cos}\space\frac{\textbf{x}}{\textbf{2}}\textbf{cos}\frac{\textbf{3x}}{\textbf{2}}.$$
$$\textbf{Sol.}\space\text{sin 3x + sin 2x - sin x}\\=\text{4 sin x cos}\frac{x}{2}\text{cos}\frac{3x}{2}$$
L.H.S. = sin 3x + sin 2x – sin x
= (sin 3x – sin x) + sin 2x
$$=\text{2 cos}\frac{3x+x}{2}\text{sin}\frac{3x-x}{2}+\text{sin 2x}\\\begin{bmatrix}\because\space\text{sin C - sin D}\\= \text{2 cos}\frac{\text{C+D}}{2}\text{sin}\frac{\text{C-D}}{2}\end{bmatrix}$$
= 2 cos 2x sin x + 2 sin x cos x
(∵ sin 2x = 2 sin x cos x)
= 2 sin x (cos 2x + cos x)
$$\text{= 2\space sin x 2 cos x}\frac{2x+x}{2}\text{cos}\frac{2x-x}{2}\\\begin{bmatrix}\because\space\text{cos C + cos D} \\= 2 cos\frac{\text{C+D}}{2}\text{cos}\frac{\text{C-D}}{2}\end{bmatrix}\\\text{ = 4 sin x cos}\frac{3x}{2}\text{cos}\frac{x}{2}=\text{R.H.S.}$$
∴ L.H.S. = R.H.S. Hence Proved.
$$\textbf{Find sin}\space\frac{\textbf{x}}{\textbf{2}},\textbf{cos}\frac{\textbf{x}}{\textbf{2}}\textbf{and}\frac{\textbf{x}}{\textbf{2}}\textbf{tan in each} \\\textbf{of the following:}\\\textbf{8. tan x}=-\frac{\textbf{4}}{\textbf{3}}, \textbf{where x in second quadrant.}\\\textbf{Sol.}\space\text{tan x}=-\frac{4}{3}$$
Given that x lies in second quadrant.
$$\text{i.e., }\frac{\pi}{2}\lt x \lt\pi\\\because\space\text{tan x}=\frac{\text{2 tan x}\frac{x}{2}}{\text{1- tan}^{2}\frac{x}{2}}=-\frac{4}{3}\\\therefore\space\text{3 tan x}\frac{x}{2}=-2\bigg(1-\text{tan }^{2}\frac{x}{2}\bigg)\\\text{3 tan}\frac{x}{2}=-2+2 \text{tan}^{2}\frac{x}{2}\\\Rarr\space\text{2 tan}^{2}\frac{x}{2}-3\text{tan}\frac{x}{2}-2=0\\\Rarr\space \text{2 tan}^{2}\frac{x}{2}-(4-1)\text{tan}\frac{x}{2}-2=0\\\Rarr\space\text{2 tan}\frac{x}{2}\bigg(\text{tan}\frac{x}{2}-2\bigg)+1\bigg(\text{tan}\frac{x}{2}-2\bigg)=0\\\Rarr\space\bigg(\text{2 tan}\frac{x}{2}+1\bigg)\bigg(\text{tan}\frac{x}{2}-2\bigg)=0$$
$$\Rarr\space\text{tan}\frac{x}{2}=\text{2 or tan}\frac{x}{2}=-\frac{1}{2}\\\because\space\frac{\pi}{2}\lt x\lt\pi\\\Rarr\space\frac{\pi}{4}\lt\frac{x}{2}\lt\frac{\pi}{2}\\\text{i.e.,}\frac{x}{2}\text{lies in first quadrant.}\\\text{Therefore,}\text{tan}\frac{x}{2}=2=\frac{2}{1}\\=\frac{\text{perpendicular}}{\text{base}}=\frac{\text{AC}}{\text{AB}}$$
Using Pythaogras theorem,
(BC)2 = (AC)2 + (AB)2
⇒ (BC)2 = 4 + 1 = 5
$$\Rarr\space\text{BC}=\sqrt{5}$$
$$\text{Now,}\space\text{sin}\frac{x}{2}=\frac{\text{perpendicular}}{\text{hypotenuse}}\\=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5}\\\therefore\space\text{cos}\frac{x}{2}=\frac{\text{base}}{\text{hypotenuse}}\\=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}$$
$$\textbf{9. cos x}=-\frac{1}{3}, \textbf{where x in third quadrant.}\\\textbf{Sol.}\space\text{cos x}=-\frac{1}{3}$$
Given that x is in third quadrant.
$$\text{i.e.\space}\space\pi\lt x \lt\frac{3\pi}{2}$$
(∵ In third quadrant θ lies between π and 3π/2)
From the formula
$$\text{cos x = 2 cos}^{2}\frac{x}{2}-1\\\Rarr\space\text{2 cos}^{2}\frac{x}{2}\\=1+\text{cos x}=1-\frac{1}{3}\\\Rarr\space \text{2 cos}^{2}\frac{x}{2}=\frac{2}{3}\\\Rarr\space\text{cos}^{2}\frac{x}{2}= \frac{1}{3}\\\Rarr\space\text{cos}\frac{x}{2}=\pm\frac{1}{\sqrt{3}}\\\text{Now,\space}\pi\lt x\lt\frac{3\pi}{2}\\\Rarr\space\frac{\pi}{2}\lt \frac{x}{2}\lt\frac{3\pi}{4}$$
$$\text{i.e.,\space}\frac{x}{2}\space\text{lies in second quadrant,}\\\text{ therefore cos}\space\frac{x}{2}\text{will be negative.}\\\Rarr\space\text{cos}\frac{x}{2}=-\frac{1}{\sqrt{3}}=-\frac{\sqrt{3}}{3}\\\text{Now,\space sin}^{2}\frac{x}{2} =1-\text{cos}^{2}\frac{x}{2}\\\text{sin}^{2}\frac{x}{2}=1-\bigg(-\frac{1}{\sqrt{3}}\bigg)^{2}\\1-\frac{1}{3}=\frac{2}{3}\\\text{sin}\frac{x}{2}=\pm\sqrt{\frac{2}{3}}$$
$$\Rarr\space\text{sin}\frac{x}{2}=\sqrt{\frac{2}{3}}=\frac{\sqrt{6}}{3}\\\text{(\text{As}}\space\frac{x}{2}\space \text{lies in second quadrant})\\\Rarr\space\text{tan}\frac{x}{2}=\frac{\text{sin}\frac{x}{2}}{\text{cos}\frac{x}{2}}=\frac{\sqrt{\frac{2}{3}}}{-\frac{1}{\sqrt{3}}}=-\sqrt{2}$$
$$\textbf{10. sin x =}\frac{\textbf{1}}{\textbf{4}},\textbf{where x is in second quadrant.}\\\textbf{Sol.}\space\because\text{sin x}=\frac{1}{4}\\\because\space\text{sin x}=\frac{\text{2 tan x}\frac{x}{2}}{1+\text{tan}^{2}\frac{x}{2}}\\\therefore\space\frac{1}{4}=\frac{\text{2 tan x}\frac{x}{2}}{1+ \text{tan}^{2}\frac{x}{2}}\\\Rarr\space 1+\text{tan}^{2}\frac{x}{2}=\text{8 tan}\frac{x}{2}\\\Rarr\space\text{tan}^{2}\frac{x}{2}-\text{8 tan}\frac{x}{2}+1=0\\\Rarr\space\text{tan}\frac{x}{2}=\frac{8\pm\sqrt{64-4}}{2}\\\Rarr\space\text{tan}\frac{x}{2}=\frac{8\pm 2\sqrt{15}}{2}$$
$$\Rarr\space\text{tan}\frac{x}{2}=4\pm\sqrt{15}\\\because\space\frac{\pi}{2}\lt2\lt\pi\\\Rarr\space\frac{\pi}{4}\lt\frac{x}{2}\lt\frac{\pi}{2}\\\text{i.e.,}\frac{x}{2}\space\text{lies in Ist quadrant.}\\\Rarr\space\text{tan}\frac{x}{2}=4+\sqrt{15}\\\text{Now,}\space\text{sec}^{2}\frac{x}{2}=1+\text{tan}^{2}\frac{x}{2}\\=1+(4+\sqrt{15})^{2}\\=1+16+15+8\sqrt{15}\\\Rarr\space\text{sec}^{2}\frac{x}{2}=32+8\sqrt{5}\\\Rarr\space\text{sec}^{2}\frac{x}{2}=8(4+\sqrt{15})$$
$$\Rarr\space\text{cos}^{2}\frac{x}{2}=\frac{1}{\text{sec}^{2}\frac{x}{2}}\\=\frac{1}{8(4+\sqrt{15})}×\frac{4-\sqrt{15}}{4-\sqrt{15}}\\\Rarr\space\text{cos}^{2}\frac{x}{2}=\frac{(4-\sqrt{15})}{8}×2\\\Rarr\space\sqrt{\frac{4-\sqrt{15}}{8}}\\=\text{cos}\frac{x}{2}=\sqrt{\frac{2(4-\sqrt{15})}{2×8}}\\=\sqrt{\frac{8-2\sqrt{15}}{16}}$$
$$\text{cos}\frac{x}{2}=\frac{\sqrt{8-2\sqrt{15}}}{4}\\\text{Now,}\space\text{sin}^{2}\frac{x}{2}=1-\text{cos}^{2}\frac{x}{2}\\1-\frac{8-2\sqrt{15}}{16}\\\Rarr\space\text{sin}^{2}\frac{x}{2}=\frac{16-8+2\sqrt{15}}{16}\\\Rarr\space\text{sin}^{2}\frac{x}{2}=\frac{8+2\sqrt{15}}{16}\\\Rarr\space\text{sin}\frac{x}{2}=\frac{\sqrt{8+2\sqrt{15}}}{4}\\\bigg(\because\space\frac{x}{2}\space\text{lies in Ist quadrant}\bigg)$$
