NCERT Solutions Class 11 Maths Chapter-3 (Trigonometric Functions)

NCERT Solutions Class 11 Maths from class 11th Students will get the answers of Chapter-3 (Trigonometric Functions) . This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

NCERT Question-Answer

Class 11 Mathematics

Chapter-3 (Trigonometric Functions)

Questions and answers given in practice

Chapter-3 (Trigonometric Functions)

Exercise 3.1

Question 1:

Find the radian measures corresponding to the following degree measures:
(i) 25°
(ii) – 47° 30′
(iii) 240°
(iv) 520°
Answer. $\begin{array}{c}\text{(i)}{25}^{\circ }\\ \text{We know that}{180}^{\circ }=\pi \text{radian}\\ \therefore {25}^{\circ }=\frac{\pi }{180}\times 25\text{radian}=\frac{5\pi }{36}\text{radian}\end{array}$ $\begin{array}{c}\text{(ii)}-{47}^{\circ }{30}^{\prime }\\ -{47}^{\circ }{30}^{\prime }=-47\frac{1}{2}\\ =\frac{-95}{2}\end{array}$ Since ${180}^{\circ }=\pi$ radian $\begin{array}{c}\text{(iii)}{240}^{\circ }\\ \text{We know that}{180}^{\circ }=\pi \text{radian}\\ \therefore {240}^{\circ }=\frac{\pi }{180}\times 240\text{radian}=\frac{4}{3}\pi \text{radian}\end{array}$ $\begin{array}{c}\text{(iv)}{520}^{\circ }\\ \text{We know that}{180}^{\circ }=\pi \text{radian}\\ \therefore {520}^{\circ }=\frac{\pi }{180}\times 520\text{radian}=\frac{26\pi }{9}\text{radian}\end{array}$

Q2. Find the degree measures corresponding to the following radian measures (Use $\pi =\frac{22}{7})$. $\begin{array}{cccc}\text{(i)}\frac{11}{16}& \text{(ii)}-4& \text{(iii)}\frac{5\pi }{3}& \text{(iv)}\frac{7\pi }{6}\end{array}$

Answer. $\begin{array}{c}\text{(i)}\frac{11}{16}\\ \text{We know that}\pi \text{radian}={180}^{\circ }\end{array}$ $\begin{array}{cc}\therefore \frac{11}{16}\mathrm{radain}& =\frac{180}{\pi }\times \frac{11}{16}\text{degree}=\frac{45\times 11}{\pi \times 4}\text{degree}\\ & =\frac{45\times 11\times 7}{22\times 4}\text{deg ree}=\frac{315}{8}\text{degree}\end{array}$ $\begin{array}{cc}=39\frac{3}{8}\text{deg ree}& \\ ={39}^{\circ }+\frac{3\times 60}{8}\text{min utes}[1^{\circ }={60}^{\circ }]& \\ ={39}^{\circ }+{22}^{\prime }+\frac{1}{2}\text{min utes}& \\ ={39}^{\circ }{22}^{\prime }{30}^{\prime \prime }& [1^{\prime }={60}^{\prime \prime }]\end{array}$ $\begin{array}{c}\left(ii\right)-4\\ \text{We know that}\pi \text{radian}={180}^{\circ }\\ -4\text{radian}=\frac{180}{\pi }\times (-4)\text{deg ree}=\frac{180\times 7(-4)}{22}\text{degree}\\ =\frac{-2520}{11}\text{deg ree}=-229\frac{1}{11}\text{degree}\end{array}$ $=-{229}^{\circ }+\frac{1\times 60}{11}\text{min utes}[1^{\circ }={60}^{\circ }]$ $=-{229}^{\circ }{5}^{12}{27}^n[1^{\prime }={60}^{\prime \prime }]$ $\begin{array}{c}\text{(iii)}\frac{5\pi }{3}\\ \text{We know that}\pi \text{radlan}={180}^{\circ }\\ \therefore \frac{5\pi }{3}\text{radian}=\frac{180}{\pi }\times \frac{5\pi }{3}\text{degree}={300}^{\circ }\end{array}$ $\begin{array}{c}\text{(iv)}\frac{7\pi }{6}\\ \text{We know that}\pi \text{radian}={180}^{\circ }\\ \therefore \frac{7\pi }{6}\text{radian}=\frac{180}{\pi }\times \frac{7\pi }{6}={210}^{\circ }\end{array}$

Question 3:
A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?
Answer:
Number of revolutions made by the wheel in 1 minute = 360
∴ Number of revolutions made by the wheel in 1  second = $\frac{360}{60}=$6 Hence, 
In one complete revolution, the wheel turns an angle of 2π radian.
Hence, in 6 complete revolutions, it will turn an angle of 6 × 2π radian, i.e., 12π radian
Thus, in one second, the wheel turns an angle of 12π radian.
Question 4:
Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm y an arc of length 22 cm (Use $\pi =\frac{22}{7}$).
Answer. We know that in a circle of radius r unit, if an arc of length l unit subtends and angle $\theta$ radian at the centre, then $\theta =\frac{1}{r}$ Therefore, for r = 100 cm, l = 22 cm, we have $\begin{array}{cc}\theta & =\frac{22}{100}\text{radian}=\frac{180}{\pi }\times \frac{22}{100}\text{deg ree}=\frac{180\times 7\times 22}{22\times 100}\text{deg ree}\\ & =\frac{126}{10}\text{deg ree}=12\frac{3}{5}\text{deg ree}={12}^{\circ }{36}^{\circ }[1^{\circ }={60}^{\prime }]\end{array}$ Thus, the required angle is ${12}^{\circ }{36}^{\prime }$.

Question 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.
Answer. Diameter of the circle = 40 cm Radius of the circle = $\frac{40}{2}cm=20cm$ Let AB be a chord (length = 20 cm) of the circle.  $\begin{array}{c}\text{In}△OAB,OA=OB=\text{Radius of circle}=20cm\\ \text{Also,}AB=20cm\\ \text{Thus,}△OAB\text{is an equilateral triangle.}\end{array}$ $\therefore \theta ={60}^{\circ }=\frac{\pi }{3}$ We know that in a circle of radius r unit, if an arc of length l unit subtends an angle $\theta$ radian at the centre, then $\theta =\frac{l}{r}$ $\frac{\pi }{3}=\frac{\widetilde{AB}}{20}\Rightarrow \widetilde{AB}=\frac{20\pi }{3}cm$ Thus, the length of the minor arc of the chord is $\frac{20\pi }{3}cm$.

Question 6:
If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.

Answer. Let the radii of the two circles be $r_1\text{and}{r}_2$ . Let an arc of length l subtend an angle of ${60}^{\circ }$ at the centre of the circle of radius $r_1$ , while let an arc of length l subtend an angle of ${60}^{\circ }$ at the centre of the circle of radius $r_2$. Now, ${60}^{\circ }=\frac{\pi }{3}\text{radian}\text{and}{75}^{\circ }=\frac{5\pi }{12}$ We know that in a circle of radius r unit, if an arc of length l unit subtends an angle $\theta$ radian at the centre, then $\theta =\frac{l}{r}\text{or}l=r\theta$ $\begin{array}{c}\therefore l=\frac{r_1\pi }{3}\text{and}l=\frac{r_{2}5\pi }{12}\\ \Rightarrow \frac{r_1\pi }{3}=\frac{r_{2}5\pi }{12}\\ \Rightarrow r_1=\frac{r_{2}5}{4}\\ \Rightarrow \frac{r_1}{r_2}=\frac{5}{4}\end{array}$ Thus, the ratio of the radii is 5 : 4.

Question 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
(i) 10 cm
(ii) 15 cm
(iii) 21 cm.
Answer. We know that in a circle of radius r unit, if an arc of length l unit subtends an angle $\theta$ radian at the centre, then $\theta =\frac{l}{r}$. It is given that r = 75 cm (i) Here, l = 10 cm $\theta =\frac{10}{75}\text{radian}=\frac{2}{15}$ radian (ii) Here, l = 15 cm $\theta =\frac{15}{75}\text{radian}=\frac{1}{5}$ radian (iii) Here, l = 21 cm $\theta =\frac{21}{75}\text{radian}=\frac{7}{25}$ radian

Exercise 3.2

Question 1:
Find the values of other five trigonometric functions if cos x = – \(\frac{1}{2}\) x lies in third quadrant.
Answer:

Question 2:
Find the values of other five trigonometric functions if sin $x=\frac{3}{5}$ , x  lies in second quadrant.
Answer. $\begin{array}{c}\sin x=\frac{3}{5}\\ cosecx=\frac{1}{\sin x}=\frac{1}{\left(\frac{3}{5}\right)}=\frac{5}{3}\\ {\sin }^{2}x+{\cos }^{2}x=1\\ \Rightarrow {\cos }^{2}x=1-{\sin }^{2}x\\ \Rightarrow {\cos }^{2}x=1-{\left(\frac{3}{5}\right)}^2\end{array}$ $\begin{array}{c}\Rightarrow {\cos }^{2}x=1-\frac{9}{25}\\ \Rightarrow {\cos }^{2}x=\frac{16}{25}\\ \Rightarrow \cos x=\pm \frac{4}{5}\end{array}$ Since x lies in the 2nd quadrant, the value of cos x will be negative $\begin{array}{c}\therefore \cos x=-\frac{4}{5}\\ \sec x=\frac{1}{\cos x}=\frac{1}{(-\frac{4}{5})}=-\frac{5}{4}\\ \tan x=\frac{\sin x}{\cos x}=\frac{\left(\frac{3}{5}\right)}{\cos x}=\frac{\left(\frac{3}{5}\right)}{\cos x}=\frac{(-\frac{4}{5})}{(-\frac{4}{5})}=-\frac{3}{4}\\ \cot x=\frac{1}{\tan x}=-\frac{4}{3}\end{array}$.

Question 3:
Find the values of other five trigonometric functions if cot x $=\frac{3}{4}$ , x lies in third quadrant.
Answer. $\begin{array}{c}\cot x=\frac{3}{4}\\ \tan x=\frac{1}{\cot x}=\frac{1}{\left(\frac{3}{4}\right)}=\frac{4}{3}\\ 1+{\tan }^{2}x={\sec }^{2}x\end{array}$ $\begin{array}{c}\Rightarrow 1+{\left(\frac{4}{3}\right)}^2={\sec }^{2}x\\ \Rightarrow 1+\frac{16}{9}={\sec }^{2}x\\ \Rightarrow \frac{25}{9}={\sec }^{2}x\\ \Rightarrow \sec x=\pm \frac{5}{3}\end{array}$ Since, x lies in the 3rd quadrant, the value of sec x will be negative. $\begin{array}{c}\therefore \sec x=-\frac{5}{3}\\ \cos x=\frac{1}{\sec x}=\frac{1}{(-\frac{5}{3})}=-\frac{3}{5}\\ \tan x=\frac{\sin x}{\cos x}\\ \Rightarrow \frac{4}{3}=\frac{\sin x}{\left(\frac{-3}{5}\right)}\end{array}$ $\begin{array}{c}\Rightarrow \sin x=\left(\frac{4}{3}\right)\times \left(\frac{-3}{5}\right)=-\frac{4}{5}\\ cosecx=\frac{1}{\sin x}=-\frac{5}{4}\end{array}$

Question 4:
Find the values of other five trigonometric functions if  $\sec x=\frac{13}{5}$ , x lies in fourth quadrant.
Answer. $\begin{array}{c}\sec x=\frac{13}{5}\\ \cos x=\frac{1}{\sec x}=\frac{1}{\left(\frac{13}{5}\right)}=\frac{5}{13}\\ {\sin }^{2}x+{\cos }^{2}x=1\\ \Rightarrow {\sin }^{2}x=1-{\cos }^{2}x\end{array}$ $\begin{array}{c}\Rightarrow {\sin }^{2}x=1-{\left(\frac{5}{13}\right)}^2\\ \Rightarrow {\sin }^{2}x=1-\frac{25}{169}=\frac{144}{169}\\ \Rightarrow \sin x=\pm \frac{12}{13}\end{array}$ Since x lies in the 4th quadrant, the value of sin x will be negative. $\begin{array}{c}\therefore \sin x=-\frac{12}{13}\\ cosecx=\frac{1}{\sin x}=\frac{1}{(-\frac{12}{13})}=-\frac{13}{12}\\ \tan x=\frac{\sin x}{\cos x}=\frac{\left(\frac{-12}{13}\right)}{\cos x}=\frac{\left(\frac{-12}{13}\right)}{\left(\frac{5}{13}\right)}=-\frac{12}{5}\\ \cot x=\frac{1}{\tan x}=\frac{1}{(-\frac{12}{5})}=-\frac{12}{5}\end{array}$

Question 5:Find the values of other five trigonometric functions if

$\tan x=-\frac{5}{12}$ , x lies in second quadrant.
Answer. $\begin{array}{c}\tan x=-\frac{5}{12}\\ \cot x=\frac{1}{\tan x}=\frac{1}{(-\frac{5}{12})}=-\frac{12}{5}\\ 1+{\tan }^{2}x={\sec }^{2}x\\ \Rightarrow 1+{(-\frac{5}{12})}^2={\sec }^{2}x\end{array}$ $\begin{array}{c}\Rightarrow 1+\frac{25}{144}={\sec }^{2}x\\ \Rightarrow \frac{169}{144}={\sec }^{2}x\\ \Rightarrow \sec x=\pm \frac{13}{12}\end{array}$ Since x lies in the 2nd quadrant, the value of sec x will be negative. $\begin{array}{c}\therefore \sec x=-\frac{13}{12}\\ \cos x=\frac{1}{\sec x}=\frac{1}{(-\frac{13}{12})}=-\frac{12}{13}\\ \tan x=\frac{\sin x}{\cos x}\\ \Rightarrow -\frac{5}{12}=\frac{\sin x}{(-\frac{12}{13})}\end{array}$ $\begin{array}{c}\Rightarrow \sin x=(-\frac{5}{12})\times (-\frac{12}{13})=\frac{5}{13}\\ cosecx=\frac{1}{\sin x}=\frac{1}{\left(\frac{5}{13}\right)}=\frac{13}{5}\end{array}$

Question 6:
Find the value of the trigonometric function sin 765°.
Ans:
It is known that the values of sin x repeat after an interval of 2π or 360°.

Question 7:
Find the value of the trigonometric function cosec (- 1410°)
Ans:
It is known that the values of cosec x repeat after an interval of 2π or 360°.
∴ cosec (- 1410°) = cosec (- 1410° + 4 x 360°)
= cosec (- 1410° + 1440°)
= cosec 30° = 2.
Question 8:
Find the value of the trigonometric function in $\tan \frac{19\pi }{3}$.
Ans:

It is known that the values of tan x repeat after an interval 
of $\pi \text{or}{180}^{\circ }$ . $\therefore \tan \frac{19\pi }{3}=\tan 6\frac{1}{3}\pi =\tan (6\pi +\frac{\pi }{3})=\tan \frac{\pi }{3}=\tan {60}^{\circ }=\sqrt{3}$.

Question 9:
Find the value of the trigonometric function in $\sin (-\frac{11\pi }{3})$.

Ans .It is known that the values of cot x repeat after an interval of 

$\pi \text{or}{360}^{\circ }$ . $\therefore \sin (-\frac{11\pi }{3})=\sin (-\frac{11\pi }{3}+2\times 2\pi )=\sin \left(\frac{\pi }{3}\right)=\frac{\sqrt{3}}{2}$

Question 10:
Find the value of the trigonometric function  in $\cot (-\frac{15\pi }{4})$.
Ans:

It is known that the values of cot x repeat after an interval of

$\pi \text{or}{180}^{\circ }$ . $\therefore \cot (-\frac{15\pi }{4})=\cot (-\frac{15\pi }{4}+4\pi )=\cot \frac{\pi }{4}=1$.

Exercise 3.3

Question 1:

Prove that: ${\sin }^2\frac{\pi }{6}+{\cos }^2\frac{\pi }{3}-{\tan }^2\frac{\pi }{4}=-\frac{1}{2}$

Answer. $\begin{array}{c}\text{L. H.S.}={\sin }^2\frac{\pi }{6}+{\cos }^2\frac{\pi }{3}-{\tan }^2\frac{\pi }{4}\\ ={\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2-(1{)}^2\\ =\frac{1}{4}+\frac{1}{4}-1=-\frac{1}{2}\\ =\text{R.H.S.}\end{array}$

Q2. Prove that: $2{\sin }^2\frac{\pi }{6}+cose{c}^2\frac{7\pi }{6}{\cos }^2\frac{\pi }{3}=\frac{3}{2}$
Ans

Q3. Prove that: ${\cot }^2\frac{\pi }{6}+cosec\frac{5\pi }{6}+3{\tan }^2\frac{\pi }{6}=6$
Answer. $\begin{array}{c}\text{L.H.S.}={\cot }^2\frac{\pi }{6}+cosec\frac{5\pi }{6}+3{\tan }^2\frac{\pi }{6}\\ =(\sqrt{3}{)}^2+\cos \mathrm{cc}(\pi -\frac{\pi }{6})+3{\left(\frac{1}{\sqrt{3}}\right)}^2\end{array}$ $\begin{array}{c}=3+\cos ec\frac{\pi }{6}+3\times \frac{1}{3}\\ =3+2+1=6\\ =RHS\end{array}$ 
Q4. Prove that: $2{\sin }^2\frac{3\pi }{4}+2{\cos }^2\frac{\pi }{4}+2{\sec }^2\frac{\pi }{3}=10$

Answer. $L.H.S=2{\sin }^2\frac{3\pi }{4}+2{\cos }^2\frac{\pi }{4}+2{\sec }^2\frac{\pi }{3}$ $\begin{array}{c}=2{\left\{\sin (\pi -\frac{\pi }{4})\right\}}^2+2{\left(\frac{1}{\sqrt{2}}\right)}^2+2(2{)}^2\\ =2{\left\{\sin \frac{\pi }{4}\right\}}^2+2\times \frac{1}{2}+8\\ =2{\left(\frac{1}{\sqrt{2}}\right)}^2+1+8\\ =1+1+8\\ =10\\ =\text{R.H.S}\end{array}$

Question 5:
Find the value of: (i) sin 75°,
(ii) tan 15°

Answer. $\begin{array}{c}\left(i\right)\sin {75}^{\circ }=\sin ({45}^{\circ }+{30}^{\circ })\\ =\sin {45}^{\circ }\cos {30}^{\circ }+\cos {45}^{\circ }\sin {30}^{\circ }\\ \left[\sin \right(x+y)=\sin x\cos y+\cos x\sin y]\\ =\left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right)+\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right)\\ =\frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}=\frac{\sqrt{3}+1}{2\sqrt{2}}\end{array}$ (ii) $\tan {15}^{\circ }=\tan ({45}^{\circ }-{30}^{\circ })$ $=\frac{\tan {45}^{\circ }-\tan {30}^{\circ }}{1+\tan {45}^{\circ }\tan {30}^{\circ }}\left[\tan \right(x-y)=\frac{\tan x-\tan y}{1+\tan x\tan y}]$ $\begin{array}{c}=\frac{1-\frac{1}{\sqrt{3}}}{1+1\left(\frac{1}{\sqrt{3}}\right)}=\frac{\sqrt{3}-1}{\frac{\sqrt{3}}{\sqrt{3}}}\\ =\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{(\sqrt{3}-1{)}^2}{(\sqrt{3}+1)(\sqrt{3}-1)}=\frac{3+1-2\sqrt{3}}{(\sqrt{3}{)}^2-(1{)}^2}\\ =\frac{4-2\sqrt{3}}{3-1}=2-\sqrt{3}\end{array}$

Q6. Prove the following: $\cos (\frac{\pi }{4}-x)\cos (\frac{\pi }{4}-y)-\sin (\frac{\pi }{4}-x)\sin (\frac{\pi }{4}-y)=\sin (x+y)$
Answer. $\begin{array}{c}\cos (\frac{\pi }{4}-x)\cos (\frac{\pi }{4}-y)-\sin (\frac{\pi }{4}-x)\sin (\frac{\pi }{4}-y)\\ =\frac{1}{2}\left[2\cos (\frac{\pi }{4}-x)\cos (\frac{\pi }{4}-y)\right]+\frac{1}{2}[-2\sin (\frac{\pi }{4}-x)\sin (\frac{\pi }{4}-y)]\\ =\frac{1}{2}[\cos \{(\frac{\pi }{4}-x)+(\frac{\pi }{4}-y)\}+\cos \{(\frac{\pi }{4}-x)-(\frac{\pi }{4}-y)\}]\\ +\frac{1}{2}[\cos \{(\frac{\pi }{4}-x)+(\frac{\pi }{4}-y)\}-\cos \{(\frac{\pi }{4}-x)-(\frac{\pi }{4}-y)\}]\end{array}$ $\left[\begin{array}{c}\because 2\cos A\cos B=\cos (A+B)+\cos (A-B)\\ -2\sin A\sin B=\cos (A+B)-\cos (A-B)\end{array}\right]$ $\begin{array}{cc}& =2\times \frac{1}{2}\left[\cos \{(\frac{\pi }{4}-x)+(\frac{\pi }{4}-y)\}\right]\\ & =\cos [\frac{\pi }{2}-(x+y\left)\right]\\ & =\sin (x+y)\\ & =R.H.S\end{array}$Q7. Prove the following: $\frac{\tan (\frac{\pi }{4}+x)}{\tan (\frac{\pi }{4}-x)}={\left(\frac{1+\tan x}{1-\tan x}\right)}^2$
Answer. It is known that $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\text{and}\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$ $\frac{\tan (\frac{\pi }{4}+x)}{\tan (\frac{\pi }{4}-x)}=\frac{\left(\frac{\tan \frac{\pi }{4}+\tan x}{1-\tan \frac{\pi }{4}\tan x}\right)}{\left(\frac{\tan \frac{\pi }{4}-\tan x}{1+\tan \frac{\pi }{4}\tan x}\right)}=\frac{\left(\frac{1+\tan x}{1-\tan x}\right)}{\left(\frac{1-\tan x}{1+\tan x}\right)}={\left(\frac{1+\tan x}{1-\tan x}\right)}^2=R.H.S.$
Q8. Prove the following: $\frac{\cos (\pi +x)\cos (-x)}{\sin (\pi -x)\cos (\frac{\pi }{2}+x)}={\cot }^{2}x$

Answer. $\begin{array}{cc}\text{L.H.S.}& =\frac{\cos (\pi +x)\cos (-x)}{\sin (\pi -x)\cos (\frac{\pi }{2}+x)}\\ & =\frac{[-\cos x]\left[\cos x\right]}{\left(\sin x\right)(-\sin x]}\\ & =\frac{-{\cos }^{2}x}{-{\cot }^{2}x}\\ & ={\cot }^{2}x\\ & =R.H.S\end{array}$

Q9. Prove the following: $\cos (\frac{3\pi }{2}+x)\cos (2\pi +x)[\cot (\frac{3\pi }{2}-x)+\cot (2\pi +x\left)\right]=1$

Answer. L.H.S $=\cos (\frac{3\pi }{2}+x)\cos (2\pi +x)[\cot (\frac{3\pi }{2}-x)+\cot (2\pi +x\left)\right]$ $\begin{array}{c}=\sin x\cos x[\tan x+\cot x]\\ =\sin x\cos x(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x})\\ =\left(\sin x\cos x\right)\left[\frac{{\sin }^{2}x+{\cos }^{2}x}{\sin x\cos x}\right]\\ =1=R.H.S.\end{array}$
Q10.Prove that: sin (n + 1) x sin (n + 2) x + cos (n + 1) x cos (n + 2) x = cos x
Answer. L.H.S = $\sin (n+1)x\sin (n+2)x+\cos (n+1)x\cos (n+2)x$ $=\frac{1}{2}\left[2\sin \right(n+1\left)x\sin \right(n+2)x+2\cos (n+1\left)x\cos \right(n+2\left)x\right]$ $=\frac{1}{2}\left[\begin{array}{c}\cos \left\{\right(n+1)x-(n+2\left)x\right\}-\cos \left\{\right(n+1)x+(n+2\left)x\right\}\\ +\cos \left\{\right(n+1)x+(n+2\left)x\right\}+\cos \left\{\right(n+1)x-(n+2\left)x\right\}\end{array}\right]$ $\left[\begin{array}{c}\because -2\sin A\sin B=\cos (A+B)-\cos (A-B)\\ 2\cos A\cos B=\cos (A+B)+\cos (A-B)\end{array}\right]$ $\begin{array}{c}=\frac{1}{2}\times 2\cos \left\{\right(n+1)x-(n+2\left)x\right\}\\ =\cos (-x)=\cos x=R.HS\end{array}$

Q11. Prove the following: $\cos (\frac{3\pi }{4}+x)-\cos (\frac{3\pi }{4}-x)=-\sqrt{2}\sin x$

Answer. It is known that $\cos A-\cos B=-2\sin \left(\frac{A+B}{2}\right)\cdot \sin \left(\frac{A-B}{2}\right)$ $\therefore$ = $\cos (\frac{3\pi }{4}+x)-\cos (\frac{3\pi }{4}-x)$ $=-2\sin \left\{\frac{(\frac{3\pi }{4}+x)+(\frac{3\pi }{4}-x)}{2}\right\}\cdot \sin \left\{\frac{(\frac{3\pi }{4}+x)-(\frac{3\pi }{4}-x)}{2}\right\}$ $\begin{array}{c}=-2\sin \left(\frac{3\pi }{4}\right)\sin x\\ =-2\sin (\pi -\frac{\pi }{4})\sin x\\ =-2\sin \frac{\pi }{4}\sin x\\ =-2\times \frac{1}{\sqrt{2}}\times \sin x\\ =-\sqrt{2}\sin x\\ =R.H.S\end{array}$

Q12. Prove the following: ${\sin }^{2}6x-{\sin }^{2}4x=\sin 2x\sin 10x$

Answer. It is known that $\sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right),\sin A-\sin B=2\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)$ $\begin{array}{c}\therefore L.H.S.={\sin }^{2}6x-{\sin }^{2}4x\\ =(\sin 6x+\sin 4x)(\sin 6x-\sin 4x)\end{array}$ $=\left[2\sin \left(\frac{6x+4x}{2}\right)\cos \left(\frac{6x-4x}{2}\right)\right][2\cos \left(\frac{6x+4x}{2}\right)\cdot \sin \left(\frac{6x-4x}{2}\right)]$ $\begin{array}{c}=\left(2\sin 5x\cos x\right)\left(2\cos 5x\sin x\right)\\ =\left(2\sin 5x\cos 5x\right)\left(2\sin x\cos x\right)\\ =\sin 10x\sin 2x\end{array}$ = R. H . S.

Question 13:
Prove that: cos2 2x cos2 6x = sin 4x sin 8x
Answer. It is known that $\cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right),\cos A-\cos B=-2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)$ $\begin{array}{c}\therefore L.H\cdot S,={\cos }^{2}2x-{\cos }^{2}6x\\ =(\cos 2x+\cos 6x)(\cos 2x-6x)\end{array}$ $\begin{array}{c}=\left[2\cos \left(\frac{2x+6x}{2}\right)\cos \left(\frac{2x-6x}{2}\right)\right][-2\sin \left(\frac{2x+6x}{2}\right)\sin \frac{(2x-6x)}{2}]\\ =\left[2\cos 4x\cos \right(-2x\left)\right][-2\sin 4x\sin (-2x\left)\right]\end{array}$ $\begin{array}{c}=\left[2\cos 4x\cos 2x\right][-2\sin 4x(-\sin 2x\left)\right]\\ =\left(2\sin 4x\cos 4x\right)\left(2\sin 2x\cos 2x\right)\\ =\sin 8x\sin 4x\\ =R.H.S.\end{array}$

Q14.Prove that: sin 2x + 2 sin 4x + sin 6x = 4 cos2 x sin 4x
Answer. $\begin{array}{c}LH.S.=\sin 2x+2\sin 4x+\sin 6x\\ =[\sin 2x+\sin 6x]+2\sin 4x\\ =\left[2\sin \left(\frac{2x+6x}{2}\right)\left(\frac{2x-6x}{2}\right)\right]+2\sin 4x\end{array}$ $[\because \sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)]$ $\begin{array}{c}=2\sin 4x\cos (-2x)+2\sin 4x\\ =2\sin 4x\cos 2x+2\sin 4x\\ =2\sin 4x(\cos 2x+1)\\ =2\sin 4x(2{\cos }^{2}x-1+1)\\ =2\sin 4x\left(2{\cos }^{2}x\right)\\ =4{\cos }^{2}x\sin 4x\\ =R.H.S\end{array}$
Q15prove that:cot 4x (sinve  5x + sin 3x) = cot x (sin 5x – sin 3x)
Answer. $\begin{array}{c}\text{L. H.S}=\cot 4x(\sin 5x+\sin 3x)\\ =\frac{\cot 4x}{\sin 4x}\left[2\sin \left(\frac{5x+3x}{2}\right)\cos \left(\frac{5x-3x}{2}\right)\right]\\ [\because \sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)]\end{array}$ $\begin{array}{c}=\left(\frac{\cos 4x}{\sin 4x}\right)\left[2\sin 4x\cos x\right]\\ =2\cos 4x\cos x\\ R.4.5.=\cot x(\sin 5x-\sin 3x)\\ =\frac{\cos x}{\sin x}\left[2\cos \left(\frac{5x+3x}{2}\right)\sin \left(\frac{5x-3x}{2}\right)\right]\\ [\because \sin A-\sin B=2\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)]\end{array}$ $\begin{array}{c}=\frac{\cos x}{\sin x}\left[2\cos 4x\sin x\right]\\ =2\cos 4x\cdot \cos x\\ L\cdot H\cdot S.=R.H.S\end{array}$ 
Q16. Prove the following: $\frac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x}=-\frac{\sin 2x}{\cos 10x}$
Answer. It is known that $\cos A-\cos B=-2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right),\sin A-\sin B=2\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)$ $\therefore L.H.S=\frac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x}$ $=\frac{-2\sin \left(\frac{9x+5x}{2}\right)\cdot \sin \left(\frac{9x-5x}{2}\right)}{2\cos \left(\frac{17x+3x}{2}\right)\cdot \sin \left(\frac{17x-3x}{2}\right)}$ $\begin{array}{cc}& =\frac{-2\sin 7x\cdot \sin 2x}{2\cos 10x\cdot \sin 7x}\\ & =-\frac{\sin 2x}{\cos 10x}\\ & =R.H.S\end{array}$ 
Q17. Prove the following: $\frac{\sin 5x+\sin 3x}{\cos 5x+\cos 3x}=\tan 4x$

Answer. It is known that $\sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right),\cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$ $\therefore L\cdot H\cdot S.=\frac{\sin 5x+\sin 3x}{\cos 5x+\cos 3x}$ $\begin{array}{c}2\sin \left(\frac{5x+3x}{2}\right)\cdot \cos \left(\frac{5x-3x}{2}\right)\\ 2\cos \left(\frac{5x+3x}{2}\right)\cdot \cos \left(\frac{5x-3x}{2}\right)\\ =\frac{2\sin 4x\cdot \cos x}{2\cos 4x\cdot \cos x}\\ =\frac{\sin 4x}{\cos 4x}\end{array}$ = tan 4x = R.H.S

Q18. Prove the following: $\frac{\sin x-\sin y}{\cos x+\cos y}=\tan \frac{x-y}{2}$
Answer. It is known that $\sin A-\sin B=2\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right),\cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$ $\therefore$ = $=\frac{\sin x-\sin y}{\cos x+\cos y}$ $=\frac{2\cos \left(\begin{array}{c}x+y\\ 2\end{array}\right)\cdot \sin \left(\begin{array}{c}x-y\\ 2\end{array}\right)}{2\cos \left(\frac{x+y}{2}\right)\cdot \cos \left(\frac{x-y}{2}\right)}$ $\begin{array}{cc}& \sin \left(\frac{x-y}{2}\right)\\ & =\frac{\sin \left(\frac{x-y}{2}\right)}{\cos \left(\frac{x-y}{2}\right)}\\ =& \tan \left(\frac{x-y}{2}\right)=R.H.S\end{array}$
Q19. Prove the following: $\frac{\sin x+\sin 3x}{\cos x+\cos 3x}=\tan 2x$
$\mathrm{Answer.\; It\; is\; known\; that sin}A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right),\cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$ $\therefore L.H.S.=\frac{\sin x+\sin 3x}{\cos x+\cos 3x}$ $\begin{array}{cc}& =\frac{2\sin \left(\frac{x+3x}{2}\right)\cos \left(\frac{x-3x}{2}\right)}{2\cos \left(\frac{x+3x}{2}\right)\cos \left(\frac{x-3x}{2}\right)}\\ & =\frac{\sin 2x}{\cos 2x}\\ & =\tan 2x\\ & =R.HS\end{array}$
Q20. Prove the following: $\frac{\sin x-\sin 3x}{{\sin }^{2}x-{\cos }^{2}x}=2\sin$x

Answer. It is known that $\sin A-\sin B=2\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right),{\cos }^{2}A-{\sin }^{2}A=\cos 2A$ $\therefore L\cdot H\cdot S=\frac{\sin x-\sin 3x}{{\sin }^{2}x-{\cos }^{2}x}$ $\begin{array}{cc}& 2\cos \left(\frac{x+3x}{2}\right)\sin \left(\frac{x-3x}{2}\right)\\ & =\frac{2\cos 2x\sin (-x)}{-\cos 2x}\\ & =-2\times (-\sin x)\\ & =2\sin x=R.H.S\end{array}$

Q21. Prove the following: $\frac{\cos 4x+\cos 3x+\cos 2x}{\sin 4x+\sin 3x+\sin 2x}=\cot 3x$

Answer. L.H.S $=\frac{\cos 4x+\cos 3x+\cos 2x}{\sin 4x+\sin 3x+\sin 2x}$ $\begin{array}{cc}& =\frac{(\cos 4x+\cos 2x)+\cos 3x}{(\sin 4x+\sin 2x)+\sin 3x}\\ & 2\cos \left(\frac{4x+2x}{2}\right)\cos \left(\frac{4x-2x}{2}\right)+\cos 3x\\ & =\frac{2\sin (4x+2x}{2})\cos \left(\frac{4x-2x}{2}\right)+\sin 3x\end{array}$ $[\because \cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right),\sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)]$ $\begin{array}{cc}& =\frac{2\cos 3x\cos x+\cos 3x}{2\sin 3x\cos x+\sin 3x}\\ & =\frac{\cos 3x(2\cos x+1)}{\sin 3x(2\cos x+1)}\\ & =\cot 3x=R.H.S\end{array}$

Q22. Prove the following: $\cot x\cot 2x-\cot 2x\cot 3x-\cot 3x\cot x=1$ 
Answer. $\begin{array}{c}\text{L. H.S.}=\cot x\cot 2x-\cot 2x\cot 3x-\cot 3x\cot x\\ =\cot x\cot 2x-\cot 3x(\cot 2x+\cot x)\\ =\cot x\cot 2x-\cot (2x+x)(\cot 2x+\cot x)\\ =\cot x\cot 2x-\left[\frac{\cot 2x\cot x-1}{\cot x+\cot 2x}\right](\cot 2x+\cot x)\end{array}$ $\begin{array}{c}[\because \cot (A+B)=\frac{\cot A\cot B-1}{\cot A+\cot B}]\\ =\cot x\cot 2x-(\cot 2x\cot x-1)\\ =1=R.H.S\end{array}$
Q23. Prove the following: $\tan 4x=\frac{4\tan x(1-{\tan }^{2}x)}{1-6{\tan }^{2}x+{\tan }^{4}x}$
Answer. It is known that $\tan 2A=\frac{2\tan A}{1-{\tan }^{2}A}$ $\begin{array}{c}\therefore L\text{H.S.}=\tan 4x=\tan 2\left(2x\right)\\ =\frac{2\tan 2x}{1-{\tan }^2\left(2x\right)}\\ =\frac{2\left(\frac{2\tan x}{1-{\tan }^{2}x}\right)}{1-{\left(\frac{2\tan x}{1-{\tan }^{2}x}\right)}^2}\end{array}$ $=\frac{\left(\frac{4\tan x}{1-{\tan }^{2}x}\right)}{1-\frac{4{\tan }^{2}x}{{(1-{\tan }^{2}x)}^2}]}$ $=\frac{\left(\frac{4\tan x}{1-{\tan }^{2}x}\right)}{\left[\frac{{(1-{\tan }^{2}x)}^2-4{\tan }^{2}x}{{(1-{\tan }^{2}x)}^2}\right]}$ $\begin{array}{cc}& =\frac{4\tan x(1-{\tan }^{2}x)}{1+{\tan }^{4}x-2{\tan }^{2}x-4{\tan }^{2}x}\\ & =\frac{4\tan x(1-{\tan }^{2}x)}{1-6{\tan }^{2}x+{\tan }^{4}x}=RHS\end{array}$
Q24.Prove that: cos 4x = 1-8 sin2 x cos2 x
Answer. $\begin{array}{c}\text{L.H.S.}=\cos 4x\\ =\cos 2\left(2x\right)\\ =1-2{\sin }^{2}2x[\cos 2A=1-2{\sin }^{2}A]\\ =1-2{\left(2\sin x{\cos }^{2}x\right)}^2[\sin 2A=2\sin A\cos A]\\ =1-8{\sin }^{2}x{\cos }^{2}x\\ =\text{R.H.S.}\end{array}$
Q25.Prove that: cos 6x = 32 cos6 x – 48 cos4 x + 18 cos2 x – 1
Ans:
We know that: cos 3x = 4 cos3 x – 3cos x

On replacing x by 2x, we get

cos 3(2x) = 4 cos3 (2x) – 3 cos 2x
⇒ cos 6x = 4 (2cos2 x – 1)3 – 3 (2cos2 x – 1)
[∵ cos 2x = 2cos2 x – 1]
= 4 [8 cos6 x – 12 cos4 x + 6 cos2 x – 1] – 6 cos2 x + 3
[∵ (a – b)3 = a3 – 3a2b + 3ab2 – b3]
= 32 cos6 x – 48 cos4 x + 24 cos2 x – 4 – 6 cos2 x + 3
⇒ cos 6x = 32 cos6 x – 48 cos4 x + 18 cos2 x – 1
Hence proved.

Exercise 3.4

Question 1:
Find the principal and general solutions of the equation, tan x = √3
Answer. $\tan x=\sqrt{3}$ It is known that $\tan \frac{\pi }{3}=\sqrt{3}\text{and tan}\left(\frac{4\pi }{3}\right)=\tan (\pi +\frac{\pi }{3})=\tan \frac{\pi }{3}=\sqrt{3}$ Therefore, the principal solutions are x = $\frac{\pi }{3}\text{and}\frac{4\pi }{3}$ Now, tan x = $\tan \frac{\pi }{3}$ $\Rightarrow x=n\pi +\frac{\pi }{3},\text{where}n\in Z$ Therefore, the general solution is $x=n\pi +\frac{\pi }{3},\text{where}n\in Z$.
Question 2:
Find the principal and general solutions of the equation: sec x = 2
Answer. Sec x = 2 It is known that $\sec \frac{\pi }{3}=2\text{and}\sec \frac{5\pi }{3}=\sec (2\pi -\frac{\pi }{3})=\sec \frac{\pi }{3}=2$ Therefore, the principal solutions are x = $\frac{\pi }{3}\text{and}\frac{5\pi }{3}$. $\begin{array}{c}{\text{Now}}_7\sec x=\sec \frac{\pi }{3}\\ \Rightarrow \cos x=\cos \frac{\pi }{3}\\ \Rightarrow x=2n\pi \pm \frac{\pi }{3},\text{where}n\in Z\end{array}$ Therefore, the general solution is $x=2n\pi \pm \frac{\pi }{3},\text{where}n\in Z$ 
Question 3:Find the principal and general solutions of the equation cot cot x = – √3.
Answer. Cot x = $-\sqrt{3}$ It is known that $\cot \frac{\pi }{6}=\sqrt{3}$ $\begin{array}{c}\therefore \cot (\pi -\frac{\pi }{6})=-\cot \frac{\pi }{6}=-\sqrt{3}\text{and}\cot (2\pi -\frac{\pi }{6})=-\cot \frac{\pi }{6}=-\sqrt{3}\\ \text{i.e.,}\cot \frac{5\pi }{6}=-\sqrt{3}\text{and}\cot \frac{11\pi }{6}=-\sqrt{3}\end{array}$ Therefore, the principal solutions are x = $\frac{5\pi }{6}\text{and}\frac{11\pi }{6}$ Now, $\cot x=\cot \frac{5\pi }{6}$ $\Rightarrow \tan x=\tan \frac{5\pi }{6}[\cot x=\frac{1}{\tan x}]$ $\Rightarrow x=n\pi +\frac{5\pi }{6},\text{where}n\in Z$ Therefore, the general solution is $x=n\pi +\frac{5\pi }{6},\text{where}n\in Z$.
Question 4:
Find the principal and general solutions of cosec x = – 2
Answer. Cosec x = -2 It is known that $\begin{array}{c}cosec\frac{\pi }{6}=2\\ \therefore \cos \mathrm{cc}(\pi +\frac{\pi }{6})=-\cos \mathrm{cc}\frac{\pi }{6}=-2\text{and cosec}(2\pi -\frac{\pi }{6})=-\cos \mathrm{cc}\frac{\pi }{6}=-2\end{array}$ $cosec\frac{7\pi }{6}=-2\text{and}cosec\frac{11\pi }{6}=-2$ Therefore, the principal solutions are x = $\frac{7\pi }{6}\text{and}\frac{11\pi }{6}$ Now, cosec x = $\cos \mathrm{ec}\frac{7\pi }{6}$ $\Rightarrow \sin x=\sin \frac{7\pi }{6}[cosecx=\frac{1}{\sin x}]$ $\Rightarrow x=n\pi +(-1{)}^n\frac{7\pi }{6},\text{where}n\in Z$ Therefore, the general solution is $x=n\pi +(-1{)}^n\frac{7\pi }{6},\text{where}n\in Z$.
Question 5:
Find the general solution of the equation: cos 4x = cos 2x
Answer. $\begin{array}{c}\cos 4x=\cos 2x\\ \Rightarrow \cos 4x-\cos 2x=0\\ \Rightarrow -2\sin \left(\frac{4x+2x}{2}\right)\sin \left(\frac{4x-2x}{2}\right)=0\end{array}$ $\begin{array}{c}[\because \cos A-\cos B=-2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)]\\ \Rightarrow \sin 3x\sin x=0\\ \Rightarrow \sin 3x=0\text{or}\sin x=0\\ \therefore 3x=n\pi \\ \Rightarrow x=\frac{n\pi }{3}\text{or}x=n\pi ,\text{where}n\in Z\end{array}$

Question 6:
Find the general solution of the equation cos 3x + cosx – cos 2x = 0
Answer. $\begin{array}{c}\cos 3x+\cos x-\cos 2x=0\\ \Rightarrow 2\cos \left(\frac{3x+x}{2}\right)\cos \left(\frac{3x-x}{2}\right)-\cos 2x=0[\cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)]\end{array}$ $\begin{array}{c}\Rightarrow 2\cos 2x\cos x-\cos 2x=0\\ \Rightarrow \cos 2x(2\cos x-1)=0\end{array}$ $\begin{array}{cc}\Rightarrow \cos 2x=0& \text{or}2\cos x-1=0\\ \Rightarrow \cos 2x=0& \text{or}\cos x=\frac{1}{2}\end{array}$ $\begin{array}{cc}\therefore 2x=(2n+1)\frac{\pi }{2}& \text{or}\cos x=\cos \frac{\pi }{3},\text{where}n\in Z\\ \Rightarrow x=(2n+1)\frac{\pi }{4}& \text{or}x=2n\pi \pm \frac{\pi }{3},\text{where}n\in Z\end{array}$

Question 7:
Find the general solution of the equation sin 2x + cos x = 0
Answer. $\begin{array}{c}\sin 2x+\cos x=0\\ \Rightarrow 2\sin x\cos x+\cos x=0\\ \Rightarrow \cos x(2\sin x+1)=0\\ \Rightarrow \cos x=0\text{or}2\sin x+1=0\end{array}$ $\begin{array}{c}\text{Now,}\cos x=0\Rightarrow \cos x=(2n+1)\frac{\pi }{2},\text{where}n\in Z\\ 2\sin x+1=0\\ \Rightarrow \sin x=\frac{-1}{2}=-\sin \frac{\pi }{6}=\sin (\pi +\frac{\pi }{6})=\sin (\pi +\frac{\pi }{6})=\sin \frac{7\pi }{6}\end{array}$ $\Rightarrow x=n\pi +(-1{)}^n\frac{7\pi }{6},\text{where}n\in Z$ Therefore, the general solution is $(2n+1)\frac{\pi }{2}\text{or}n\pi +(-1{)}^n\frac{7\pi }{6},n\in Z$.

Question 8:
Find the general solution of the equation sec2 2x = 1 – tan 2x.

Answer. $\begin{array}{c}{\sec }^{2}2x=1-\tan 2x\\ \Rightarrow 1+{\tan }^{2}2x=1-\tan 2x\\ \Rightarrow {\tan }^{2}2x+\tan 2x=0\\ \Rightarrow \tan 2x(\tan 2x+1)=0\\ \Rightarrow \tan 2x=0\text{or}\tan 2x+1=0\end{array}$ $\begin{array}{c}\text{Now,}\tan 2x=0\\ \Rightarrow \tan 2x=\tan 0\\ \Rightarrow 2x=n\pi +0,\text{where}n\in Z\\ \Rightarrow x=\frac{n\pi }{2},\text{where}n\in Z\\ \text{tan}2x+1=0\end{array}$ $\begin{array}{c}\Rightarrow \tan 2x=-1=-\tan \frac{\pi }{4}=\tan (\pi -\frac{\pi }{4})=\tan \frac{3\pi }{4}\\ \Rightarrow 2x=n\pi +\frac{3\pi }{4},\text{where}n\in Z\\ \Rightarrow x=\frac{n\pi }{2}+\frac{3\pi }{8},\text{where}n\in Z\end{array}$ Therefore, the general solution is $\frac{n\pi }{2}\text{or}\frac{n\pi }{2}+\frac{3\pi }{8},n\in Z$

.

Question 9:
Find the general solution of the equation sin x + sin 3x + sin 5x = 0
Answer. sin x + sin 3x + sin 5x = 0 $(\sin x+\sin 5x)+\sin 3x=0$ $\Rightarrow \left[2\sin \left(\frac{x+5x}{2}\right)\cos \left(\frac{x-5x}{2}\right)\right]+\sin 3x=0[\sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)]$ $\begin{array}{c}\Rightarrow 2\sin 3x\cos (-2x)+\sin 3x=0\\ \Rightarrow 2\sin 3x\cos 2x+\sin 3x=0\\ \Rightarrow \sin 3x(2\cos 2x+1)=0\\ \Rightarrow \sin 3x=0\text{or}2\cos 2x+1=0\\ \text{Now,}\sin 3x=0\Rightarrow 3x=n\pi ,\text{where}n\in Z\end{array}$ $\begin{array}{c}\text{i.c.,}x=\frac{n\pi }{3},\text{where}n\in Z\\ 2\cos 2x+1=0\\ \Rightarrow \cos 2x=\frac{-1}{2}=-\cos \frac{\pi }{3}=\cos (\pi -\frac{\pi }{3})\end{array}$ $\begin{array}{c}\Rightarrow \cos 2x=\cos \frac{2\pi }{3}\\ \Rightarrow 2x=2n\pi \pm \frac{2\pi }{3},\text{where}n\in Z\\ \Rightarrow x=n\pi \pm \frac{\pi }{3},\text{where}n\in Z\end{array}$ Therefore, the general solution is $\frac{n\pi }{3}\text{or}n\pi \pm \frac{\pi }{3},n\in Z$