Trigonometric Ratio of Some Angle

The trigonometric ratios of a given angle are the ratios of a right-angled triangle's sides with regard to any one of its acute angles. The six trigonometric ratios are sine (sin) , cosine (cos) , tangent(tan), cotangent(cot) , secant(sec) , cosecant(cosec). Various trigonometric formulas can be used to find the exact value of trigonometric ratios of $18^{\circ}, 72^{\circ}, 36^{\circ}$and so on.

In this article, we will cover the concept of Value Trigonometric Ratios of some Particular Angles. This category falls under the broader category of Trigonometry, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more.

Trigonometric Ratios

The trigonometric ratios of a given angle are the ratios of a right-angled triangle's sides with regard to any one of its acute angles. The six trigonometric ratios are sine (sin) , cosine (cos) , tangent(tan), cotangent(cot) , secant(sec) , cosecant(cosec).

Value of Trigonometric Ratios of Some Particular Angles

We can find the value of trigonometric ratios of some particular angles with the help of different trigonometric identities.

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}
$

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, $\cos 36^{\circ}=\frac{\sqrt{5}+1}{4}$

5. $\sin 36^{\circ}$

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

6. $\sin 54^{\circ}$ and $\cos 54^{\circ}$

$\begin{aligned} & \sin 54^{\circ}=\sin \left(90^{\circ}-36^{\circ}\right)=\cos 36^{\circ} \\ & \therefore \sin 54^{\circ}=\frac{\sqrt{5}+1}{4} \\ & \cos 54^{\circ}=\cos \left(90^{\circ}-36^{\circ}\right)=\sin 36^{\circ} \\ & \therefore \cos 36^{\circ}=\frac{1}{4} \sqrt{10-2 \sqrt{5}}\end{aligned}$

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}
$

Summary

Trigonometric identities enhance the understanding and application of trigonometry. It makes complex calculations more accessible and contributes significantly to advancements in mathematics and science. With the help of trigonometric formulas, we can find the exact value of trigonometric angles.

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Solved Examples Based on Trigonometric Ratios of Some Specific Angles

Example 1: The value of $2 \sin \left(12^{\circ}\right)-\sin \left(72^{\circ}\right)$ is : [JEE MAINS 2022]
Solution:
We need to find,

$
\begin{aligned}
& \Rightarrow 2 \sin 12^{\circ}-\sin 72^{\circ} \\
& \Rightarrow \sin 12^{\circ}+\sin 12^{\circ}-\sin 72^{\circ} \\
& \Rightarrow \sin 12^{\circ}+2 \cos 42^{\circ} \sin \left(-30^{\circ}\right) \Rightarrow \sin 12^{\circ}-\cos 42^{\circ} \\
& \Rightarrow \sin 12^{\circ}-\sin 48^{\circ} \\
& \Rightarrow 2 \cos \left(30^{\circ}\right) \sin \left(-18^{\circ}\right) \\
& \Rightarrow-\sqrt{3} \sin 18^{\circ} \\
& \Rightarrow-\sqrt{3}\left(\frac{\sqrt{5}-1}{4}\right)=\sqrt{3}\left(\frac{1-\sqrt{5}}{4}\right) \\
& \text { Hence, the required answer is } \frac{\sqrt{3}(1-\sqrt{5})}{4}
\end{aligned}
$
Hence, the required answer is $x^2-2 x-4=0$

Example 2: $\operatorname{cosec} 18^{\circ}$ is a root of the equation [JEE MAINS 2021]
Solution:
$\operatorname{cosec} 18^{\circ}=\frac{1}{\sin 18^{\circ}}=\frac{4}{\sqrt{5}-1}=\sqrt{5}+1=\alpha$
Let $\beta=1-\sqrt{5}$

$
\begin{aligned}
& \alpha+\beta=2 ; \alpha \beta=-4 \\
& \Rightarrow x^2-2 x-4=0
\end{aligned}
$

Example 3: The value of $\frac{1}{\cos 72^{\circ}}+\frac{1}{\sin 54^{\circ}}$ is
Solution:

$
\begin{aligned}
& \sin 18^{\circ}=\frac{\sqrt{5}-1}{4} \\
& \cos 36^{\circ}=\frac{\sqrt{5}+1}{4}
\end{aligned}
$
Now, the given expression can be written as

$
\begin{aligned}
& \frac{1}{\sin 18^{\circ}}+\frac{1}{\cos 36^{\circ}} \\
& =\frac{4}{\sqrt{5}-1}+\frac{4}{\sqrt{5}+1} \\
& =4\left(\frac{2 \sqrt{5}}{5-1}\right)=2 \sqrt{5}
\end{aligned}
$
Hence, the required answer is $2 \sqrt{5}$

Example 4: If $\sin A=\frac{\sqrt{5}-1}{4}$, then the value of $\frac{\sin ^2 A}{1-\cos A}$ is, (A lies in the first quadrant)

Solution:
Clearly A $=18^{\circ}$
Now,

$
\begin{aligned}
& \frac{\sin ^2 A}{1-\cos A} \\
& =\frac{1-\cos ^2 A}{1-\cos A} \\
& =\frac{(1-\cos A)(1+\cos A)}{1-\cos A} \\
& =1+\cos A \\
& =1+\frac{\sqrt{10+2 \sqrt{5}}}{4}
\end{aligned}
$
Hence, the required answer is $
1+\frac{\sqrt{10+2 \sqrt{5}}}{4}
$

Example 5: The value of $\sin 30^{\circ} \cos 60^{\circ}+\cos 30^{\circ} \sin 60^{\circ}$ is?
Solution:

$
\sin 30^{\circ} \cos 60^{\circ}+\cos 30^{\circ} \sin 60^{\circ}
$
On substituting the respective values, we get

$
\sin 30^{\circ} \cos 60^{\circ}+\cos 30^{\circ} \sin 60^{\circ}=\frac{1}{2} \times \frac{1}{2}+\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}=1
$
Hence, the required answer is 1.