Exercise 3.5-Additional Question - Chapter 3 - Trigonometry - 11th Maths Guide Samacheer Kalvi Solutions

You can Download the Exercise 3.5-Additional Question - 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

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 $=4 \sin \mathrm{A} \sin \left(60^{\circ}+\mathrm{A}\right) \cdot \sin \left(60^{\circ}-\mathrm{A}\right)$
$=4 \sin \mathrm{A}\left\{\sin \left(60^{\circ}+\mathrm{A}\right) \cdot \sin \left(60^{\circ}-\mathrm{A}\right)\right\}$
$\left.=4 \sin \mathrm{A}\left\{\sin ^2 60^{\circ}-\sin ^2 \mathrm{~A}\right)\right\}$
$=4 \sin A\left\{\frac{3}{4}-\sin ^2 A\right\}=3 \sin A-4 \sin ^3 A=\sin 3 A=R H S$