Exercise 3.5-Additional Question - Chapter 3 Trigonometry 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

AdditionalQuestions
Question 1.
If $\tan \alpha=\frac{1}{3}$ and $\tan \beta=\frac{1}{7}$ show that $2 \alpha+\beta=\frac{\pi}{4}$.
Solution:
$
\begin{aligned}
& \tan 2 \alpha=\frac{2 \tan \alpha}{1-\tan ^2 \alpha}=\frac{2\left(\frac{1}{3}\right)}{1-\left(\frac{1}{3}\right)^2}=\frac{\frac{2}{3}}{\frac{8}{9}}=\frac{3}{4} \\
& \tan (2 \alpha+\beta)=\frac{\tan 2 \alpha+\tan \beta}{1-\tan 2 \alpha+\tan \beta}=\frac{\frac{3}{4}+\frac{1}{7}}{1-\left(\frac{3}{4}\right)\left(\frac{1}{7}\right)}=\frac{\frac{21+4}{28}}{\frac{28-3}{28}}=1 \\
& \therefore 2 \alpha+\beta=45^{\circ}=\frac{\pi}{4}
\end{aligned}
$
Question 2.
If $2 \cos \theta=x+\frac{1}{x}$ then prove that $\cos 2 \theta=\frac{1}{2}\left(x^2+\frac{1}{x^2}\right)$
Solution:
$
\begin{aligned}
2 \cos \theta & =x+\frac{1}{x} \\
\therefore 4 \cos ^2 \theta & =\left(x+\frac{1}{x}\right)^2=x^2+\frac{1}{x^2}+2 \\
4 \cos ^2 \theta-2 & =x^2+\frac{1}{x^2} \\
2\left(2 \cos ^2 \theta-1\right) & =x^2+\frac{1}{x^2} \\
\Rightarrow \quad \cos 2 \theta & =\frac{1}{2}\left(x^2+\frac{1}{x^2}\right)
\end{aligned}
$

Question 3.
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} \\
&=\cos 20^{\circ}\left[\cos \left(60^{\circ}-20^{\circ}\right) \cos \left(60^{\circ}+20^{\circ}\right)\right] \\
&=\cos 20^{\circ}\left[\cos ^2 60^{\circ}-\sin ^2 20^{\circ}\right]=\cos 20^{\circ}\left[\frac{1}{4}-\sin ^2 20^{\circ}\right] \\
&=\frac{1}{4} \cos 20^{\circ}\left\{1-4\left(1-\cos ^2 20^{\circ}\right)\right\} \\
&=\frac{1}{4}\left\{4 \cos ^3 20^{\circ}-3 \cos 20^{\circ}\right\} \\
&=\frac{1}{4}\left[\cos 3 \times 20^{\circ}\right]=\frac{1}{4} \times \cos 60^{\circ}=\frac{1}{8}=\text { RHS }
\end{aligned}
$
Question 4.
Show that $4 \sin \mathrm{A} \sin \left(60^{\circ}+\mathrm{A}\right) \cdot \sin \left(60^{\circ}-\mathrm{A}\right)=\sin 3 \mathrm{~A}$
Solution:
LHS $=4 \sin \mathrm{A} \sin \left(60^{\circ}+\mathrm{A}\right) \cdot \sin \left(60^{\circ}-\mathrm{A}\right)$
$=4 \sin \mathrm{A}\left\{\sin \left(60^{\circ}+\mathrm{A}\right) \cdot \sin \left(60^{\circ}-\mathrm{A}\right)\right\}$
$\left.=4 \sin \mathrm{A}\left\{\sin ^2 60^{\circ}-\sin ^2 \mathrm{~A}\right)\right\}$
$=4 \sin A\left\{\frac{3}{4}-\sin ^2 A\right\}=3 \sin A-4 \sin ^3 A=\sin 3 A=R H S$