Trigonometric Ratios of Some Specific Angles - Practice Questions & MCQ

Value Trigonometric Ratios of some Particular Angles (Part 1)
1. $\sin 18^{\circ}$

Let $\theta=18^{\circ}$, then $5 \theta=90^{\circ} \therefore 2 \theta+3 \theta=90^{\circ}$
or $2 \theta=90^{\circ}-3 \theta \therefore \sin 2 \theta=\sin \left(90^{\circ}-3 \theta\right)$
or $\sin 2 \theta=\cos 3 \theta$ or $2 \sin \theta \cos \theta=4 \cos ^3 \theta-3 \cos \theta$
or $2 \sin \theta=4 \cos ^2 \theta-3 \quad$ [ dividing by $\cos \theta$ ]
or $2 \sin \theta=4\left(1-\sin ^2 \theta\right)-3=1-4 \sin ^2 \theta$
or $4 \sin ^2 \theta+2 \sin \theta-1=0$
$\therefore \sin \theta=\frac{-2 \pm \sqrt{4+16}}{8}=\frac{-2 \pm 2 \sqrt{5}}{8}=\frac{-1 \pm \sqrt{5}}{4}$
Thus $\sin \theta=\frac{-1+\sqrt{5}}{4}, \frac{-1-\sqrt{5}}{4}$
$\because \theta=18^{\circ}$
$\therefore \sin \theta=\sin 18^{\circ}>0$, as $18^{\circ}$ lies in the 1 st quadrant
$\therefore \sin \theta$ i.e., $\sin 18^{\circ}=\frac{\sqrt{5}-1}{4}$

2. $\cos 18^{\circ}$

$
\begin{aligned}
\cos ^2 18^{\circ} & =1-\sin ^2 18^{\circ}=1-\left(\frac{\sqrt{5}-1}{4}\right)^2 \\
& =1-\frac{5+1-2 \sqrt{5}}{16}=1-\frac{6-2 \sqrt{5}}{16} \\
& =\frac{16-6+2 \sqrt{5}}{16}=\frac{10+2 \sqrt{5}}{16} \\
\therefore \quad \cos 18^{\circ} & =\frac{1}{4} \sqrt{10+2 \sqrt{5}} \quad\left[\because \cos 18^{\circ}>0\right]
\end{aligned}
$

3. $\sin 72^{\circ}$ and $\cos 72^{\circ}$

$
\begin{aligned}
& \sin 72^{\circ}=\sin \left(90^{\circ}-18^{\circ}\right)=\cos 18^{\circ}=\frac{1}{4} \sqrt{10+2 \sqrt{5}} \\
& \cos 72^{\circ}=\cos \left(90^{\circ}-18^{\circ}\right)=\sin 18^{\circ}=\frac{\sqrt{5}-1}{4}
\end{aligned}
$

Value of Trigonometric Ratios of some Particular Angles (Part 2)
4. $\cos 36^{\circ}$

We know that, $\cos 2 \theta=1-2 \sin ^2 \theta$
Put $\theta=18^{\circ}$
$\cos 2 \times 18^{\circ}=1-2 \sin ^2\left(18^{\circ}\right)$
$\cos 36^{\circ}=1-2 \sin ^2\left(18^{\circ}\right)$

$
\begin{aligned}
\cos 36^{\circ} & =1-2 \times\left(\frac{\sqrt{5}-1}{4}\right)^2 \\
& =1-2 \times \frac{5-2 \sqrt{5}+1}{16} \\
& =1-\frac{3-\sqrt{5}}{4} \\
& =\frac{4-3+\sqrt{5}}{4}=\frac{\sqrt{5}+1}{4}
\end{aligned}
$
Hence, $\cos 36^{\circ}=\frac{\sqrt{5}+1}{4}$

$
\begin{aligned}
& \text { 5. } \sin 36^{\circ} \\
& \sin ^2 \theta+\cos ^2 \theta=1 \\
& \sin ^2 \theta=1-\cos ^2 \theta \\
& \sin ^2\left(36^{\circ}\right)=1-\cos ^2\left(36^{\circ}\right) \\
& =1-\left(\frac{\sqrt{5}+1}{4}\right)^2 \\
& =1-\frac{6+2 \sqrt{5}}{16}=\frac{16-6-2 \sqrt{5}}{16}=\frac{10-2 \sqrt{5}}{16} \\
& \therefore \sin 36^{\circ}=\frac{\sqrt{10-2 \sqrt{5}}}{4} \quad\left[\because \sin 36^{\circ}>0\right] \\
& \text { 6. } \sin 54^{\circ} \text { and } \cos 54^{\circ} \\
& \sin 54^{\circ}=\sin \left(90^{\circ}-36^{\circ}\right)=\cos 36^{\circ} \\
& \therefore \sin 54^{\circ}=\frac{\sqrt{5}+1}{4} \\
& \cos 54^{\circ}=\cos \left(90^{\circ}-36^{\circ}\right)=\sin 36^{\circ} \\
& \therefore \cos 36^{\circ}=\frac{1}{4} \sqrt{10-2 \sqrt{5}}
\end{aligned}
$

7. $\cos 22.5^{\circ}$

Let $\theta=22.5^{\circ}$, then $2 \theta=45^{\circ}$
Use the Identity, $\cos 2 \theta=2 \cos ^2 \theta-1$

$
\begin{aligned}
\cos ^2\left(22.5^{\circ}\right) & =\frac{1+\cos \left(45^{\circ}\right)}{2} \\
& =\frac{1+\frac{1}{\sqrt{2}}}{2} \\
& =\frac{\sqrt{2}+1}{2 \sqrt{2}}=\frac{2+\sqrt{2}}{4} \\
\therefore \cos 22.5^{\circ} & =\frac{1}{2} \sqrt{2+\sqrt{2}}
\end{aligned}
$

8. $\sin 22.5^{\circ}$

Let $\theta=22.5^{\circ}$, then $2 \theta=45^{\circ}$
Use the Identity, $\cos 2 \theta=1-2 \sin ^2 \theta$

$
\begin{aligned}
\sin ^2\left(22.5^{\circ}\right) & =\frac{1-\cos \left(45^{\circ}\right)}{2} \\
& =\frac{1-\frac{1}{\sqrt{2}}}{2} \\
& =\frac{\sqrt{2}-1}{2 \sqrt{2}}=\frac{2-\sqrt{2}}{4} \\
\therefore \sin 22.5^{\circ} & =\frac{1}{2} \sqrt{2-\sqrt{2}}
\end{aligned}
$