Exercise 3.6-Additional Questions - Chapter 3 Trigonometry 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Additional Questions
Question 1.
Prove that $\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ}=\frac{1}{16}$.
Solution:
Let $\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ}=\mathrm{x}$
Multiply by $2 \sin 20^{\circ}$ on both sides.
$
\begin{aligned}
& 2 x \sin 20^{\circ}=2 \sin 20^{\circ} \cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ} \\
& =\sin 40^{\circ} \cos 40^{\circ} \cos 80^{\circ} \quad[\because \sin 2 \theta=2 \sin \theta \cos \theta] \\
& =\frac{1}{2} \sin 80^{\circ} \cos 80^{\circ}=\frac{1}{2} \cdot \frac{1}{2} \sin 160^{\circ} \\
& =\frac{1}{4} \sin \left(180^{\circ}-20^{\circ}\right)=\frac{1}{4} \sin 20^{\circ} \\
& \therefore \quad 2 x \sin 20^{\circ}=\frac{1}{4} \sin 20^{\circ} \Rightarrow x=\frac{1}{8} \\
& \therefore \cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ}=\frac{1}{8} \cdot \cos 60^{\circ}=\frac{1}{8} \cdot \frac{1}{2} \quad\left[\because \cos 60^{\circ}=\frac{1}{2}\right] \\
& =\frac{1}{16}=\mathrm{RHS} \\
&
\end{aligned}
$

Question 2.
Prove that $\sin 20^{\circ} \sin 40^{\circ} \sin 60^{\circ} \sin 80^{\circ}=\frac{3}{16}$.
Solution:
$
\begin{aligned}
\mathrm{LHS} & =\sin 20^{\circ} \sin 40^{\circ} \sin 60^{\circ} \sin 80^{\circ} \\
& =\frac{1}{2}\left(\cos 20^{\circ}-\cos 60^{\circ}\right) \frac{\sqrt{3}}{2} \sin 80^{\circ} \\
& =\frac{\sqrt{3}}{4}\left[\cos 20^{\circ} \sin 80^{\circ}-\frac{1}{2} \sin 80^{\circ}\right] \\
& =\frac{\sqrt{3}}{4}\left[\frac{1}{2}\left(\sin 100^{\circ}+\sin 60^{\circ}\right)\right]-\frac{\sqrt{3}}{8} \sin 80^{\circ} \\
& =\frac{\sqrt{3}}{8} \sin 100^{\circ}+\frac{\sqrt{3}}{8} \sin 60^{\circ}-\frac{\sqrt{3}}{8} \sin 80^{\circ} \\
& =\frac{\sqrt{3}}{8} \sin \left(180^{\circ}-80^{\circ}\right)+\frac{\sqrt{3}}{8} \cdot \frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{8} \sin 80^{\circ} \\
& =\frac{\sqrt{3}}{8} \sin 80^{\circ}+\frac{3}{16}-\frac{\sqrt{3}}{8} \sin 80^{\circ}=\frac{3}{16}=\mathrm{RHS}
\end{aligned}
$
Question 3.
Prove that $\sin 50^{\circ}-\sin 70^{\circ}+\cos 80^{\circ}=0$.
Solution:
$
\begin{aligned}
& \text { LHS }=\sin 50^{\circ}-\sin 70^{\circ}+\cos 80^{\circ} \\
& =2 \cos \left(\frac{50^{\circ}+70^{\circ}}{2}\right) \sin \left(\frac{50^{\circ}-70^{\circ}}{2}\right)+\cos 80^{\circ} \\
& =2 \cos 60^{\circ} \sin \left(-10^{\circ}\right)+\cos 80^{\circ} \\
& =-2 \frac{1}{2} \sin 10^{\circ}+\cos \left(90^{\circ}-10^{\circ}\right) \\
& =-\sin 10^{\circ}+\sin 10^{\circ}=0=\mathrm{RHS}
\end{aligned}
$
Question 4.
Prove that $(\cos \alpha+\cos \beta)^2+(\sin \alpha-\sin \beta)^2=4 \cos ^2\left(\frac{\alpha+\beta}{2}\right)$

Solution:
$
\begin{aligned}
\text { LHS }=(\cos \alpha+\cos \beta)^2 & +(\sin \alpha-\sin \beta)^2 \\
& =\left(2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}\right)^2+\left(2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}\right)^2 \\
& =4 \cos ^2\left(\frac{\alpha+\beta}{2}\right) \cos ^2\left(\frac{\alpha-\beta}{2}\right)+4 \cos ^2\left(\frac{\alpha+\beta}{2}\right) \sin ^2\left(\frac{\alpha-\beta}{2}\right) \\
& =4 \cos ^2\left(\frac{\alpha+\beta}{2}\right)\left(\cos ^2\left(\frac{\alpha-\beta}{2}\right)+\sin ^2\left(\frac{\alpha-\beta}{2}\right)\right) \\
& =4 \cos ^2\left(\frac{\alpha+\beta}{2}\right)=\text { RHS }
\end{aligned}
$
Question 5.
Prove that $2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}=0$
Solution:
$
\begin{aligned}
& \text { LHS }=2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13} \\
&=\cos \left(\frac{\pi}{13}+\frac{9 \pi}{13}\right)+\cos \left(\frac{\pi}{13}-\frac{9 \pi}{13}\right)+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13} \\
&=\cos \frac{10 \pi}{13}+\cos \left(-\frac{8 \pi}{13}\right)+\cos \left(\pi-\frac{10 \pi}{13}\right)+\cos \left(\pi-\frac{8 \pi}{13}\right) \\
&=\cos \frac{10 \pi}{13}+\cos \frac{8 \pi}{13}-\cos \frac{10 \pi}{13}-\cos \frac{8 \pi}{13}=0=\text { RHS }
\end{aligned}
$
Question 6.
Prove that $\frac{2 \cos 2 \theta+1}{2 \cos 2 \theta-1}=\tan \left(60^{\circ}+\theta\right)\left(\tan 60^{\circ}-\theta\right)$

Solution:
$
\text { RHS }=\tan \left(60^{\circ}+\theta\right) \tan \left(60^{\circ}-\theta\right)
$
Let $60^{\circ}+\theta=\alpha$ and $60^{\circ}-\theta=\beta$
$
\begin{aligned}
\text { RHS } & =\tan \alpha \tan \beta=\frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta} \\
& =\frac{2 \sin \alpha \sin \beta}{2 \cos \alpha \cos \beta}=\frac{\cos (\alpha-\beta)-\cos (\alpha+\beta)}{\cos (\alpha+\beta)+\cos (\alpha-\beta)} \\
& =\frac{\cos 2 \theta-\cos 120^{\circ}}{\cos 120^{\circ}+\cos 2 \theta}=\frac{\cos 2 \theta-\left(-\frac{1}{2}\right)}{-\frac{1}{2}+\cos 2 \theta} \\
& {\left[\because \cos 120^{\circ}=\cos \left(90^{\circ}+30^{\circ}\right)=-\sin 30^{\circ}=-\frac{1}{2}\right] } \\
& =\frac{2 \cos 2 \theta+1}{2 \cos 2 \theta-1}=\text { LHS }
\end{aligned}
$
Question 7.
Prove that $\cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ}=\frac{1}{8}$.
Solution:

$
\begin{aligned}
& \text { LHS }=\cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ} \\
& =\frac{1}{2} \cos 20^{\circ}\left(2 \cos 80^{\circ} \cos 40^{\circ}\right) \\
& \left.=\frac{1}{2} \cos 20^{\circ}\left[\cos 80^{\circ}+40^{\circ}\right)+\cos \left(80^{\circ}-40^{\circ}\right)\right] \\
& {[\because 2 \cos \mathrm{A} \cos \mathrm{B}=\cos (\mathrm{A}+\mathrm{B})+\cos (\mathrm{A}-\mathrm{B})]} \\
& =\frac{1}{2} \cos 20^{\circ}\left(\cos 120^{\circ}+\cos 40^{\circ}\right) \\
& =\frac{1}{2} \cos 20^{\circ}\left(-\frac{1}{2}+\cos 40^{\circ}\right) \\
& =-\frac{1}{4} \cos 20^{\circ}+\frac{1}{2} \cos 40^{\circ} \cos 20^{\circ} \\
& =-\frac{1}{4} \cos 20^{\circ}+\frac{1}{4}\left(2 \cos 40^{\circ} \cos 20^{\circ}\right) \\
& =-\frac{1}{4} \cos 20^{\circ}+\frac{1}{4}\left[\cos \left(40^{\circ}+20^{\circ}\right)+\cos \left(40^{\circ}-20^{\circ}\right)\right] \\
& =-\frac{1}{4} \cos 20^{\circ}+\frac{1}{4}\left(\cos 60^{\circ}+\cos 20^{\circ}\right) \\
& =-\frac{1}{4} \cos 20^{\circ}+\frac{1}{4} \cos 60^{\circ}+\frac{1}{4} \cos 20^{\circ} \\
& =\frac{1}{4} \cos 60^{\circ}=\frac{1}{4}\left(\frac{1}{2}\right)=\frac{1}{8}=\text { RHS } \\
&
\end{aligned}
$
Question 8.
Prove that $\tan 70^{\circ}-\tan 20^{\circ}-2 \tan 40^{\circ}=4 \tan 10^{\circ}$

Solution:

$
\begin{aligned}
& \text { LHS }=\tan 70^{\circ}-\tan 20^{\circ}-2 \tan 40^{\circ} \\
& =\left(\frac{\sin 70^{\circ}}{\cos 70^{\circ}}-\frac{\sin 20^{\circ}}{\cos 20^{\circ}}\right)-2 \tan 40^{\circ} \\
& =\left(\frac{\sin 70^{\circ} \cos 20^{\circ}-\cos 70^{\circ} \sin 20^{\circ}}{\cos 70^{\circ} \cos 20^{\circ}}\right)-2 \tan 40^{\circ} \\
& =\frac{\sin \left(70^{\circ}-20^{\circ}\right)}{\cos 70^{\circ} \cos 20^{\circ}}-2 \tan 40^{\circ} \\
& =\frac{2 \sin 50^{\circ}}{2 \cos 70^{\circ} \cos 20^{\circ}}-2 \tan 40^{\circ} \\
& =\frac{2 \sin 50^{\circ}}{\cos 90^{\circ}+\cos 50^{\circ}}-2 \tan 40^{\circ} \\
& =\frac{2 \sin 50^{\circ}}{\cos 50^{\circ}}-2 \tan 40^{\circ} \quad\left[\because \cos 90^{\circ}=0\right] \\
& =2\left(\frac{\sin 50^{\circ}}{\cos 50^{\circ}}-\frac{\sin 40^{\circ}}{\cos 40^{\circ}}\right) \\
& =2\left(\frac{\sin 50^{\circ} \cos 40^{\circ}-\cos 50^{\circ} \sin 40^{\circ}}{\cos 50^{\circ} \cos 40^{\circ}}\right) \\
& =2\left[\frac{\sin \left(50^{\circ}-40^{\circ}\right)}{\cos 50^{\circ} \cos 40^{\circ}}\right]=\frac{4 \sin 10^{\circ}}{2 \cos 50^{\circ} \cos 40^{\circ}} \\
& =\frac{4 \sin 10^{\circ}}{\cos 90^{\circ}+\cos 10^{\circ}}=\frac{4 \sin 10^{\circ}}{\cos 10^{\circ}} \\
& =4 \tan 10^{\circ}=\text { RHS } \\
&
\end{aligned}
$
Question 9
Prove that $\cos \theta+\cos \left(\frac{2 \pi}{3}-\theta\right)+\cos \left(\frac{2 \pi}{3}+\theta\right)=0$

Solution:

We have
$
\begin{aligned}
\text { LHS }= & \cos \theta+\left[\cos \left(\frac{2 \pi}{3}-\theta\right)+\cos \left(\frac{2 \pi}{3}+\theta\right)\right] \\
= & \cos \theta+2 \cos \frac{\frac{2 \pi}{3}-\theta+\frac{2 \pi}{3}+\theta}{2} \cos \frac{\frac{2 \pi}{3}-\theta-\frac{2 \pi}{3}-\theta}{2} \\
= & \cos \theta+2 \cos \frac{2 \pi}{3} \cos (-\theta)=\cos \theta+2\left(-\frac{1}{2}\right) \cos \theta \\
& {\left[\because \cos \frac{2 \pi}{3}=\cos 120^{\circ}=-\frac{1}{2} \text { and } \cos (-\theta)=\cos \theta\right] } \\
= & \cos \theta-\cos \theta=0=\text { RHS }
\end{aligned}
$
Question 10
Prove that $\frac{\sin 11 A \sin A+\sin 7 A \sin 3 A}{\cos 11 A \sin A+\cos 7 A \sin 3 A}=\tan 8 A$
Solution:
Multiplying numerator and denominator by 2 , we have
$
\begin{aligned}
\text { LHS } & =\frac{(2 \sin 11 \mathrm{~A} \sin \mathrm{A})+(2 \sin 7 \mathrm{~A} \sin 3 \mathrm{~A})}{(2 \cos 11 \mathrm{~A} \sin \mathrm{A})+(2 \cos 7 \mathrm{~A} \sin 3 \mathrm{~A})} \\
& =\frac{(\cos 10 \mathrm{~A}-\cos 12 \mathrm{~A})+(\cos 4 \mathrm{~A}-\cos 10 \mathrm{~A})}{(\sin 12 \mathrm{~A}-\sin 10 \mathrm{~A})+(\sin 10 \mathrm{~A}-\sin 4 \mathrm{~A})} \\
& =\frac{\cos 4 \mathrm{~A}-\cos 12 \mathrm{~A}}{\sin 12 \mathrm{~A}-\sin 4 \mathrm{~A}} \\
& =\frac{2 \sin 8 \mathrm{~A} \sin 4 \mathrm{~A}}{2 \cos 8 \mathrm{~A} \sin 4 \mathrm{~A}}=\tan 8 \mathrm{~A}=\text { RHS }
\end{aligned}
$