Exercise 3.6-Additional Questions - Chapter 3 - Trigonometry - 11th Maths Guide Samacheer Kalvi Solutions

You can Download the Exercise 3.6-Additional Questions - Chapter 3 - Trigonometry - 11th Maths Guide Samacheer Kalvi Solutions with expert answers for all chapters. Perfect for Tamil & English Medium students to revise the syllabus and score more marks in board exams. Download and share it with your friends

Share this to Friend on WhatsApp

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}
$