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

Updated On May 15, 2024

By SaraNextGen

Ex 3.5
Question 1.
Find the value of $\cos 2 \mathrm{~A}, \mathrm{~A}$ lies in the first quadrant, when
(i) $\cos A=\frac{15}{17}$ (ii) $\sin A=\frac{4}{5}$ (iii) $\tan A=\frac{16}{63}$
Solution:
(i)
$
\begin{aligned}
\cos A & =\frac{15}{17} \\
B C & =\sqrt{17^2-15^2}=\sqrt{289-225}=\sqrt{64}=8
\end{aligned}
$
Now $\cos A=\frac{15}{17} ; \sin A \quad \frac{8}{17}$
$
\begin{aligned}
\cos 2 \mathrm{~A} & =\cos ^2 \mathrm{~A}-\sin ^2 \mathrm{~A}= \\
& =\frac{225}{289}-\frac{64}{289}=\frac{161}{289}
\end{aligned}
$
(ii)

$
\begin{aligned}
\sin A & =\frac{4}{5} \\
\cos 2 A & =1-2 \sin ^2 A=1-2\left(\frac{4}{5}\right)^2 \\
& =1-2\left(\frac{16}{25}\right)=1-\frac{32}{25} \\
& =\frac{25-32}{25}=\frac{-7}{25}
\end{aligned}
$
(iii)
$
\begin{aligned}
\tan A & =16 / 63 \\
\cos 2 \mathrm{~A} & =\frac{1-\tan ^2 \mathrm{~A}}{1+\tan ^2 \mathrm{~A}}=\frac{1-(16 / 63)^2}{1+(16 / 63)^2}=\frac{3713}{4225}
\end{aligned}
$
Question 2.
If $\theta$ is an acute angle, then find

(i) $\sin \left(\frac{\pi}{4}-\frac{\theta}{2}\right)$, when $\sin \theta=\frac{1}{25}$.
(ii) $\cos \left(\frac{\pi}{4}+\frac{\theta}{2}\right)$, when $\sin \theta=\frac{8}{9}$
Solution:

(i) Given $\sin \theta=\frac{1}{25}$, To find $\sin \left(\frac{\pi}{4}-\frac{\theta}{2}\right)$ Now $\cos 2\left(\frac{\pi}{4}-\frac{\theta}{2}\right)=\cos \left(\frac{\pi}{2}-\theta\right)=\sin \theta=\frac{1}{25}$ (given)
We know $\cos 2 \theta=1-2 \sin ^2 \theta$
$
\begin{aligned}
\Rightarrow \quad 1-2 \sin ^2\left(\frac{\pi}{4}-\frac{\theta}{2}\right) & =\frac{1}{25} \\
\Rightarrow \quad 2 \sin ^2\left(\frac{\pi}{4}-\frac{\theta}{2}\right) & =1-\frac{1}{25}=\frac{24}{25} \\
\text { So } \sin ^2\left(\frac{\pi}{4}-\frac{\theta}{2}\right) & =\frac{24}{25 \times 2}=\frac{12}{25} \\
\Rightarrow \quad \sin \left(\frac{\pi}{4}-\frac{\theta}{2}\right) & =\sqrt{\frac{12}{25}}=\frac{2 \sqrt{3}}{5}
\end{aligned}
$

(ii) Given $\sin \theta=\frac{8}{9}$, To find $\cos \left(\frac{\pi}{4}+\frac{\theta}{2}\right)$ Now $\cos 2\left(\frac{\pi}{4}+\frac{\theta}{2}\right)=\cos \left(\frac{\pi}{2}+\theta\right)=-\sin \theta=-\frac{8}{9}$
[We know $\cos 2 \theta=2 \cos ^2 \theta-1$ ]
$
\Rightarrow \quad 2 \cos ^2\left(\frac{\pi}{4}+\frac{\theta}{2}\right)-1=-\frac{8}{9}
$
So, $\quad 2 \cos ^2\left(\frac{\pi}{4}+\frac{\theta}{2}\right)=1-\frac{8}{9}=\frac{1}{9}$
$
\Rightarrow \quad \cos ^2\left(\frac{\pi}{4}+\frac{\theta}{2}\right)=\frac{1}{9 \times 2}=\frac{1}{18}
$
$
\Rightarrow \quad \cos \left(\frac{\pi}{4}+\frac{\theta}{2}\right)=\sqrt{\frac{1}{18}}=\frac{1}{3 \sqrt{2}}
$

Question 3.
If $\cos \theta=\frac{1}{2}\left(a+\frac{1}{a}\right)$, show that $\cos 3 \theta=\frac{1}{2}\left(a^3+\frac{1}{a^3}\right)$.
Solution:
$
\begin{aligned}
& \text { Given, } \cos \theta=\frac{1}{2}\left(a+\frac{1}{a}\right) \\
& \cos ^3 \theta=\left[\frac{1}{2}\left(a+\frac{1}{a}\right)\right]^3=\frac{1}{8}\left(a+\frac{1}{a}\right)^3 \\
& =\frac{1}{8}\left[a^3+\frac{1}{a^3}+3(a)\left(\frac{1}{a}\right)\left(a+\frac{1}{a}\right)\right] \\
& =\frac{1}{8}\left[a^3+\frac{1}{a^3}+3\left(a+\frac{1}{a}\right)\right] \\
& \text { Now } \cos 3 \theta=4 \cos ^3 \theta-3 \cos \theta \\
& =4\left[\frac{1}{8}\left\{\left(a^3+\frac{1}{a^3}\right)+3\left(a+\frac{1}{a}\right)\right\}\right]-3\left[\frac{1}{2}\left(a+\frac{1}{a}\right)\right] \\
& =\frac{4}{8}\left(a^3+\frac{1}{a^3}\right)+\frac{12}{8}\left(a+\frac{1}{a}\right)-\frac{3}{2}\left(a+\frac{1}{a}\right) \\
& =\frac{1}{2}\left(a^3+\frac{1}{a^3}\right)+\frac{3}{2}\left(a+\frac{1}{a}\right)-\frac{3}{2}\left(a+\frac{1}{a}\right) \\
& =\frac{1}{2}\left(a^3+\frac{1}{a^3}\right)=\mathrm{RHS} \\
&
\end{aligned}
$

Question 4.
Prove that $\cos ^5 \theta=16 \cos ^5 \theta-20 \cos ^3 \theta+5 \cos \theta$.
Solution:
$
\begin{aligned}
& \cos ^5 \theta=\cos (2 \theta+3 \theta)=\cos 2 \theta \cos 3 \theta-\sin 2 \theta \sin 3 \theta \\
& =\left(2 \cos ^2 \theta-1\right)\left(4 \cos ^3 \theta-3 \cos \theta\right)-2 \sin \theta \cos \theta\left(3 \sin \theta-4 \sin ^3 \theta\right) \\
& =8 \cos ^5 \theta-6 \cos ^3 \theta-4 \cos ^3 \theta+3 \cos \theta-6 \sin ^2 \theta \cos \theta+8 \cos \theta \sin ^4 \theta \\
& =8 \cos ^5 \theta-6 \cos ^3 \theta-4 \cos ^3 \theta+3 \cos \theta-6\left(1-\cos ^2 \theta\right) \cos \theta+8 \cos \theta\left(1-\cos ^2 \theta\right)^2 \\
& =8 \cos ^5 \theta-6 \cos ^3 \theta-4 \cos ^3 \theta+3 \cos \theta-6 \cos \theta+6 \cos ^3 \theta+8 \cos 0\left(1+\cos ^4 \theta-2 \cos ^2 \theta\right)
\end{aligned}
$
$
\begin{aligned}
& =8 \cos ^5 \theta-6 \cos ^3 \theta-4 \cos ^3 \theta+3 \cos \theta-6 \cos \theta+6 \cos ^3 \theta+8 \cos \theta+8 \cos ^5 \theta-16 \cos ^3 \theta \\
& =16 \cos ^5 \theta-20 \cos ^3 \theta+5 \cos \theta=\mathrm{RHS}
\end{aligned}
$
Question 5.
Prove that $\sin 4 \alpha=4 \tan \alpha \frac{1-\tan ^2 \alpha}{\left(1+\tan ^2 \alpha\right)^2}$.
Solution:
$
\begin{aligned}
\text { RHS } & =4 \tan \alpha \frac{1-\tan ^2 \alpha}{\left(1+\tan ^2 \alpha\right)^2} \\
& =2\left(\frac{2 \tan \alpha}{1+\tan ^2 \alpha}\right) \times\left(\frac{1-\tan ^2 \alpha}{1+\tan ^2 \alpha}\right) \\
& =2 \sin 2 \alpha \cos 2 \alpha=\sin 4 \alpha=\text { LHS }
\end{aligned}
$

Question 6.
If $A+B=45^{\circ}$, show that $(1+\tan A)(1+\tan B)=2$.
Solution:
$
\begin{aligned}
& A+B=45^{\circ} \\
& \Rightarrow \tan (A+B)=\tan 45^{\circ}=1
\end{aligned}
$
(i.e) $\frac{\tan A+\tan B}{1-\tan A \tan B}=1 \Rightarrow \tan A+\tan B=1-\tan A \tan B$
Now LHS $=(1+\tan \mathrm{A})(1+\tan B)$
$=\tan \mathrm{A}+\tan \mathrm{B}+\tan \mathrm{A} \tan \mathrm{B}+1$
$=(1-\tan A \tan B)+(\tan A \tan B+1)$ from (1)
$=2=$ RHS
Question 7.
Prove that $\left(1+\tan 1^{\circ}\right)\left(1+\tan 2^{\circ}\right)\left(1+\tan 3^{\circ}\right) \ldots\left(1+\tan 44^{\circ}\right)$ is a multiple of 4 .
Solution:
$
\begin{aligned}
& 1+\tan 44^{\circ}=1+\tan \left(45^{\circ}-1^{\circ}\right) \\
& =1+\frac{\tan 45^{\circ}-\tan 1^{\circ}}{1-\tan 45^{\circ} \tan 1^{\circ}}=1+\frac{1-\tan 1^{\circ}}{1+\tan 1^{\circ}} \\
& =\frac{1+\tan 1^{\circ}+1-\tan 1^{\circ}}{1-1 \tan 1^{\circ}}=\frac{2}{1-1 \tan 1^{\circ}} \\
& \left(1+\tan 1^{\circ}\right)\left(1+\tan 44^{\circ}\right)=2 \\
& \text { Similarly }\left(1+\tan 2^{\circ}\right)\left(1+\tan 43^{\circ}\right)=2 \\
& \left(1+\tan 3^{\circ}\right)\left(1+\tan 42^{\circ}\right)=2 \\
& \left(1+\tan 22^{\circ}\right)\left(1+\tan 23^{\circ}\right)=2 \\
& =\left(1+\tan 1^{\circ}\right)\left(1+\tan 2^{\circ}\right) \ldots\left(1+\tan 44^{\circ}\right)=2 \times 2 \times \ldots 22 \text { times }
\end{aligned}
$
It is a multiple of 4 .

Question 8 .
Prove that $\tan \left(\frac{\pi}{4}+\theta\right)-\tan \left(\frac{\pi}{4}-\theta\right)=2 \tan 2 \theta$.
Solution:
$
\begin{aligned}
\tan \left(\frac{\pi}{4}+\theta\right) & =\frac{\tan \frac{\pi}{4}+\tan \theta}{1-\tan \frac{\pi}{4} \tan \theta}=\frac{1+\tan \theta}{1-\tan \theta} \\
\tan \left(\frac{\pi}{4}-\theta\right) & =\frac{\tan \frac{\pi}{4}-\tan \theta}{1+\tan \frac{\pi}{4} \tan \theta}=\frac{1-\tan \theta}{1+\tan \theta} \\
\text { Now LHS } & =\tan \left(\frac{\pi}{4}+\theta\right)-\tan \left(\frac{\pi}{4}-\theta\right) \\
& =\frac{1+\tan \theta}{1-\tan \theta}-\frac{1-\tan \theta}{1+\tan \theta}=\frac{(1+\tan \theta)^2-(1-\tan \theta)^2}{1-\tan ^2 \theta} \\
& =\frac{1+\tan { }^2 \theta+2 \tan \theta-1-\tan ^2 \theta+2 \tan \theta}{1-\tan ^2 \theta} \\
& =\frac{4 \tan \theta}{1-\tan { }^2 \theta}=2\left(\frac{2 \tan \theta}{1-\tan ^2 \theta}\right) \\
& =2 \tan 2 \theta=\text { RHS }
\end{aligned}
$
Question 9.
Show that $\cot \left(7 \frac{1}{2}^{\circ}\right)=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}$.
Solution:

We have to prove that $\cot \left(7 \frac{1}{2}^{\circ}\right)=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}$
$
\text { LHS }=\cot \left(7 \frac{1}{2}^{\circ}\right)=\frac{\cos \left(7 \frac{1}{2}{ }^{\circ}\right)}{\sin \left(71 / 2^{\circ}\right)} \text {. }
$
To find $\frac{\cos \theta}{\sin \theta}$, multiply numerator \& denominator by $2 \cos \theta$
Let $\theta=71 / 2^{\circ}$
$2 \theta=15^{\circ}$
$
\begin{aligned}
& \frac{2 \cos ^2 \theta}{2 \sin \theta \cos \theta}=\frac{1+\cos 2 \theta}{\sin 2 \theta}=\frac{1+\cos 15^{\circ}}{\sin 15^{\circ}} \\
= & \frac{1+\sqrt{3}+1 / 2 \sqrt{2}}{\sqrt{3}-1 / 2 \sqrt{2}}=\frac{2 \sqrt{2}+\sqrt{3}+1}{\sqrt{3}-1}
\end{aligned}
$
Multiply numerator $\&$ denominator by $\sqrt{3}+1$
$
=\frac{(2 \sqrt{2}+\sqrt{3}+1)(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}
$
$
=\frac{2 \sqrt{2}+3+\sqrt{3}+\sqrt{3}+1}{3-1}
$
$
=\frac{2 \sqrt{3}+2 \sqrt{2}+4}{2}
$
$
\begin{aligned}
& =\frac{2(\sqrt{2}+\sqrt{3}+\sqrt{6}+2)}{2} \\
& =\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6} \\
& =\text { RHS }
\end{aligned}
$
Question 10.
Prove that $(1+\sec 2 \theta)(1+\sec 4 \theta) \ldots .\left(1+\sec 2^{\mathrm{n}} \theta\right)=\tan 2^{\mathrm{n}} \theta$

Solution:
$
\begin{aligned}
& \text { LHS }(1+\sec 2 \theta)=1+\frac{1}{\cos 2 \theta}+\frac{\cos 2 \theta+1}{\cos 2 \theta}=\frac{2 \cos ^2 \theta}{\cos 2 \theta} \\
& (1+\sec 4 \theta)=1+\frac{1}{\cos 4 \theta}=\frac{\cos 4 \theta+1}{\cos 4 \theta}=\frac{2 \cos ^2(2 \theta)}{\cos 4 \theta} \\
& \left(1+\sec 2^n \theta\right)=1+\frac{1}{2^n \theta}=\frac{\cos 2^n \theta+1}{2^n \theta}=\frac{2 \cos ^2 2^{n-1} \theta}{\cos 2^n \theta} \\
& (1+\sec 2 \theta)(1+\sec 4 \theta) \ldots .\left(1+\sec 2^n \theta\right) \\
& =\frac{2^n \cos ^2 \theta}{\cos 2 \theta} \cdot \frac{\cos ^2 2 \theta}{\cos 4 \theta} \ldots \frac{\cos ^2 2^{n-1} \theta}{\cos 2^n \theta} \\
& =\frac{2^n \cos \theta}{\cos 2^n \theta}\left\{\cos \theta \cdot \cos 2 \theta \ldots \cos 2^{n-1} \theta\right\} \\
& =\frac{2^n \cos \theta\left\{\sin 2^n \theta\right\}}{2^n \sin \theta \cos 2^n \theta}=\tan 2^n \theta \cdot \cos \theta \\
&
\end{aligned}
$

Question 11.
Prove that $32(\sqrt{3}) \sin \frac{\pi}{48} \cos \frac{\pi}{48} \cos \frac{\pi}{24} \cos \frac{\pi}{12} \cos \frac{\pi}{6}=3$
Solution:
$
32 \sqrt{3}\left[\sin \frac{\pi}{48} \times \cos \frac{\pi}{48}\right]=16 \sqrt{3}\left[2 \sin \frac{\pi}{48} \cos \frac{\pi}{48}\right]=16 \sqrt{3} \sin \frac{\pi}{24}\left(\frac{2 \pi}{48}=\frac{\pi}{24}\right)
$
Now $16 \sqrt{3}\left[\sin \frac{\pi}{24} \times \cos \frac{\pi}{24}\right]=8 \sqrt{3}\left[2 \sin \frac{\pi}{24} \cos \frac{\pi}{24}\right]=8 \sqrt{3}\left[\sin \frac{2 \pi}{24}\right]=8 \sqrt{3} \sin \frac{\pi}{12}$
Now $8 \sqrt{3}\left[\sin \frac{\pi}{12} \cos \frac{\pi}{12}\right]=4 \sqrt{3}\left[2 \sin \frac{\pi}{12} \cos \frac{\pi}{12}\right]$
$
=4 \sqrt{3}\left[\sin \frac{2 \pi}{12}\right]=4 \sqrt{3}\left(\sin \frac{\pi}{6}\right)
$
Now $4 \sqrt{3} \sin \frac{\pi}{6} \cos \frac{\pi}{6}=2 \sqrt{3}\left[2 \sin \frac{\pi}{6} \cos \frac{\pi}{6}\right]$
$
2 \sqrt{3}\left[\sin \frac{2 \pi}{6}\right]=2 \sqrt{3} \sin \frac{\pi}{3}=2 \sqrt{3} \times \frac{\sqrt{3}}{2}=3=\text { RHS }
$