Exercise 3.3 - Chapter 3 Trigonometry 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

$
\operatorname{Ex} 3.3
$
Question 1.
Find the values of
(i) $\sin \left(480^{\circ}\right)$
(ii) $\sin \left(-1110^{\circ}\right)$
(iii) $\cos \left(300^{\circ}\right)$
(iv) $\tan \left(1050^{\circ}\right)$
(v) $\cot \left(660^{\circ}\right)$
(vi) $\tan \left(\frac{19 \pi}{3}\right)$
(vii) $\sin \left(-\frac{11 \pi}{3}\right)$
Solution:
$
\begin{aligned}
& \text { (i) } \sin \left(480^{\circ}\right)=\sin \left(360^{\circ}+120^{\circ}\right)=\sin 120^{\circ} \\
& =\sin \left(90^{\circ}+30^{\circ}\right)=\cos 30^{\circ}=\sqrt{3} / 2 \\
& \text { (ii) } \sin \left(-1110^{\circ}\right)=-\sin \left(1110^{\circ}\right) \\
& =-\sin \left(360^{\circ} \times 3+30^{\circ}\right) \\
& =-\sin 30^{\circ}=-1 / 2
\end{aligned}
$
(iii) $\cos \left(300^{\circ}\right)=\cos \left(270^{\circ}+30^{\circ}\right)=\sin 30^{\circ}=1 / 2$
$
\begin{aligned}
& \text { (iv) } \tan \left(1050^{\circ}\right)=\tan \left[3\left(360^{\circ}\right)-30^{\circ}\right] \\
& =-\tan 30^{\circ}=-\frac{1}{\sqrt{3}}
\end{aligned}
$
$
\begin{aligned}
& \text { (v) } \cot \left(660^{\circ}\right)=\cot \left(360^{\circ} \times 2-60^{\circ}\right) \\
& =-\cot 60^{\circ}=-\frac{1}{\sqrt{3}}
\end{aligned}
$

(vi)
$
\begin{aligned}
\tan \left(\frac{19 \pi}{3}\right) & =\tan \frac{19}{3} \times 180=\tan \frac{19}{3} \times \frac{360}{2} \\
& =\tan \frac{19}{6}\left(360^{\circ}\right) \\
& =\tan 3 \frac{1}{6}\left(360^{\circ}\right) \\
& =\tan \left[3(360)^{\circ}+\frac{360^{\circ}}{6}\right]=\tan 60^{\circ}=\sqrt{3}
\end{aligned}
$
(vii)
$
\begin{aligned}
\sin \left(-\frac{11 \pi}{3}\right) & =-\sin \frac{11 \pi}{3}=-\sin \frac{11}{3} \times 180 \\
& =-\sin \frac{11}{3} \times \frac{360}{2}=-\sin 360^{\circ}\left(\frac{11}{6}\right) \\
& =-\sin 360^{\circ}\left(2-\frac{1}{6}\right) \\
& =-\left[\sin \left(360^{\circ} \times 2\right)-\frac{360^{\circ}}{6}\right] \\
& =-\left(-\sin 60^{\circ}\right)=\sqrt{3} / 2
\end{aligned}
$
Question 2.
$\left(\frac{5}{7}, \frac{2 \sqrt{6}}{7}\right)$ is a point on the terminal side of an angle $\theta$ is standard position. Determine the trigonometric function values of angle $\theta$.

Solution:

In the diagram $\mathrm{ON}=\frac{5}{7} ; \mathrm{PN}=\frac{2 \sqrt{6}}{7} ; \mathrm{ON}^2+\mathrm{NP}^2=\mathrm{OP}^2$
(i.e) $\frac{25}{49}+\frac{24}{49}=\frac{49}{49}=\mathrm{OP}^2 \Rightarrow \mathrm{OP}=1$
$\sin \theta=\frac{\mathrm{PN}}{\mathrm{OP}}=\frac{2 \sqrt{6} / 7}{1}=\frac{2 \sqrt{6}}{7}$
$\cos \theta=\frac{\mathrm{ON}}{\mathrm{OP}}=\frac{5 / 7}{1}=\frac{5}{7}$
$\tan \theta=\frac{P N}{O N}=\frac{2 \sqrt{6} / 7}{5 / 7}=\frac{2 \sqrt{6}}{5}$
$\operatorname{cosec} \theta=\frac{1}{\sin \theta}=\frac{7}{2 \sqrt{6}}$
$\sec \theta=\frac{1}{\cos \theta}=\frac{7}{5}$
$\cot \theta=\frac{1}{\tan \theta}=\frac{5}{2 \sqrt{6}}$

Question 3.
Find the values of other five trigonometric functions for the following:
(i) $\cos \theta=-\frac{1}{2} ; \theta$ lies in the III quadrant.
Solution:
Taking the Numerical values
$\cos \theta=-\frac{1}{2} ; \theta$ from $\triangle \mathrm{ABC}, \mathrm{AB}=\sqrt{4-1}=\sqrt{3}$
$\theta$ is in III quadrant. $\therefore \tan \theta, \cot \theta$ are + ve

$
\begin{array}{rlrl}
\text { Now } \sin \theta & =-\frac{\sqrt{3}}{2} & ; & \operatorname{cosec} \theta=-\frac{2}{\sqrt{3}} \\
\cos \theta & =-\frac{1}{2} & ; & \sec \theta=-\frac{2}{1}=-2 \\
\tan \theta & =+\frac{\sqrt{3}}{1}=\sqrt{3} ; & \cot \theta=\frac{1}{\sqrt{3}}
\end{array}
$

(ii) $\cos \theta=\frac{2}{3} ; \theta$ lies in the I quadrant Solution:

$\cos \theta=\frac{2}{3} ; \theta$ from $\triangle \mathrm{ABC}, \mathrm{AB}=\sqrt{9-4}=\sqrt{5}$ $\theta$ is in I quadrant
$\therefore$ All trigonometric values are $+\mathrm{ve}$
$
\begin{array}{ll}
\sin \theta=\frac{\sqrt{5}}{3} & ; \operatorname{cosec} \theta=\frac{3}{\sqrt{5}} \\
\cos \theta=\frac{2}{3} & ; \sec \theta=\frac{3}{2} \\
\tan \theta=\frac{\sqrt{5}}{2} & ; \cot \theta=\frac{2}{\sqrt{5}}
\end{array}
$
(iii) $\sin \theta=-\frac{2}{3} ; \theta$ lies in the IV quadrant Solution:

$\theta$ from $\triangle \mathrm{ABC}, \mathrm{BC}=\sqrt{9-4}=\sqrt{5}$
$\theta$ is in IV quadrant $\therefore \cos \theta, \sec \theta$ are +ve $\begin{array}{ll}\sin \theta=-\frac{2}{3} & ; \operatorname{cosec} \theta=\frac{-3}{2} \\ \cos \theta=\frac{\sqrt{5}}{3} & ; \sec \theta=\frac{3}{\sqrt{5}} \\ \tan \theta=-\frac{2}{\sqrt{5}} & ; \cot \theta=-\frac{\sqrt{5}}{2}\end{array}$
(iv) $\tan \theta=-2 ; \theta$ lies in the II quadrant Solution:

$
\tan \theta=-2
$
From $\triangle \mathrm{ABC}, \mathrm{BC}=\sqrt{4+1}=\sqrt{5}$
$\theta$ is in III quadrant
$
\begin{aligned}
& \therefore \sin \theta, \operatorname{cosec} \theta \text { are + ve } \\
& \sin \theta=\frac{2}{\sqrt{5}} \quad ; \operatorname{cosec} \theta=\frac{\sqrt{5}}{2} \\
& \cos \theta=-\frac{1}{\sqrt{5}} \quad ; \sec \theta=-\frac{\sqrt{5}}{1}=-\sqrt{5} \\
& \tan \theta=-\frac{2}{1}=-2 \quad ; \cot \theta=-\frac{1}{2} \\
&
\end{aligned}
$
(v) $\sec \theta=\frac{13}{5} ; \theta$ lies in the IV quadrant
Solution:
$
\sec \theta=13 / 5
$
From $\triangle \mathrm{ABC}, \mathrm{BC}=\sqrt{13^2-5^2}=\sqrt{169-25}=\sqrt{144}=12$
$\theta$ is in IV quadrant

$
\begin{aligned}
& \cos \theta, \sec \theta \text { are +ve. } \\
& \sin \theta=-\frac{12}{13} ; \operatorname{cosec} \theta=-\frac{13}{12} \\
& \cos \theta=\frac{5}{13} ; \sec \theta=\frac{13}{5} \\
& \tan \theta=-\frac{12}{5} ; \cot \theta=-\frac{5}{12}
\end{aligned}
$
Question 4.
Prove that $\frac{\cot \left(180^{\circ}+\theta\right) \sin \left(90^{\circ}-\theta\right) \cos (-\theta)}{\sin \left(270^{\circ}+\theta\right) \tan (-\theta) \operatorname{cosec}\left(360^{\circ}+\theta\right)}=\cos ^2 \theta \cot \theta$.

Solution:
$
\begin{aligned}
& \cot \left(180^{\circ}+\theta\right)=\cot \theta \\
& \sin \left(90^{\circ}-\theta\right)=\cos \theta \\
& \cos (-\theta)=\cos \theta \\
& \sin (270+\theta)=-\cos \theta \\
& \tan (-\theta)=-\tan \theta \\
& \operatorname{cosec}\left(360^{\circ}+\theta\right)=\operatorname{cosec} \theta \\
& \therefore \quad \mathrm{LHS}=\frac{\cot \theta \cdot \cos \theta \cdot \cos \theta}{(-\cos \theta)(-\tan \theta)(\operatorname{cosec} \theta)} \\
& =\frac{\frac{\cos \theta}{\sin \theta} \cdot \cos \theta \cdot \cos \theta}{\cos \theta \cdot \frac{\sin \theta}{\cos \theta} \cdot \frac{1}{\sin \theta}}=\cot \theta \cos ^2 \theta=\mathrm{RHS} \\
&
\end{aligned}
$
Question 5.
Find all the angles between $0^{\circ}$ and $360^{\circ}$ which satisfy the equation $\sin ^2 \theta=\frac{3}{4}$ Solution:
$
\begin{aligned}
& \sin ^2 \theta=\frac{3}{4} \\
& \Rightarrow \sin \theta= \pm \sqrt{\frac{3}{4}}= \pm \frac{\sqrt{3}}{2} \\
& \sin \theta=+\frac{\sqrt{3}}{2} \Rightarrow \theta=60^{\circ} \text { (or) } 120^{\circ} \\
& \sin \theta=-\frac{\sqrt{3}}{2} \Rightarrow \theta=-60^{\circ} \text { (or) } 240^{\circ} \text { (or) } 300^{\circ}
\end{aligned}
$
Question 6.
Show that $\sin ^2 \frac{\pi}{18}+\sin ^2 \frac{\pi}{9}+\sin \frac{7 \pi}{18}+\sin ^2 \frac{4 \pi}{9}=2$.
Solution:
$
\begin{aligned}
& \text { LHS }=\sin ^2 10^{\circ}+\sin ^2 20^{\circ}+\sin ^2 70^{\circ}+\sin ^2 80^{\circ} \\
& =\sin ^2 10^{\circ}+\sin ^2\left(90^{\circ}-10^{\circ}\right)+\sin ^2 20^{\circ}+\sin ^2\left(90^{\circ}-20^{\circ}\right) \\
& =\sin ^2 10^{\circ}+\left(\cos 10^{\circ}\right)^2+\sin ^2 20^{\circ}+\left(\cos 20^{\circ}\right)^2 \\
& =\left(\sin ^2 10+\cos ^2 10\right)+\sin ^2 20^{\circ}+\cos ^2 20^{\circ} \\
& =1+1=2=\text { RHS }
\end{aligned}
$