Question

How do I find the value of $\text{cosec}\left( \dfrac{\pi }{12} \right)$ ?

Answer 1

Hint: First convert the angle into degrees. Then use the relation between sin(x) and cosec(x) to write the given trigonometric ratio in terms of sine. Later, use the half angle formula to find the value of the sine of the given angle.

Formula used:
$\text{cosec(}x)=\dfrac{1}{\sin (x)}$
${{\sin }^{2}}\left( \dfrac{x}{2} \right)=\dfrac{1-\cos x}{2}$, where x is an angle.

Complete step by step solution:
Let us first convert the given angle into degrees. We know that an angle of $\pi $ radians is equal to 180 degrees.Therefore, an angle of $\dfrac{\pi }{12}$ radians is equal to
$\dfrac{180}{12}=15$ degrees.
This means that we have to calculate the value of $\text{cosec(1}{{\text{5}}^{\circ }}\text{)}$.
We know that $\text{cosec(}x)=\dfrac{1}{\sin (x)}$. Therefore,
$\text{cosec(1}{{\text{5}}^{\circ }})=\dfrac{1}{\sin (\text{1}{{\text{5}}^{\circ }})}$ …. (i)
Now, we shall find the value of $\sin (\text{1}{{\text{5}}^{\circ }})$ with the help of the half angle formula. The half angle formula says that
${{\sin }^{2}}\left( \dfrac{x}{2} \right)=\dfrac{1-\cos x}{2}$ …. (ii).

In this case, $\dfrac{x}{2}={{15}^{\circ }}$.Then, this means that $x={{30}^{\circ }}$. Substitute the value x and half of x in equation (ii).Then, we get that
${{\sin }^{2}}\left( {{15}^{\circ }} \right)=\dfrac{1-\cos ({{30}^{\circ }})}{2}$ …. (iii)
Now, we know that $\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}$.
Now, substitute this value in equation (iii).With this we get that,
${{\sin }^{2}}\left( {{15}^{\circ }} \right)=\dfrac{1-\dfrac{\sqrt{3}}{2}}{2}$
If we further solve the above equation, then we get that
${{\sin }^{2}}\left( {{15}^{\circ }} \right)=\dfrac{\dfrac{2-\sqrt{3}}{2}}{2}\\
\Rightarrow{{\sin }^{2}}\left( {{15}^{\circ }} \right)=\dfrac{2-\sqrt{3}}{4}$

Now, let us take the square root on both the sides of the equation. Therefore, we get that
$\sin \left( {{15}^{\circ }} \right)=\pm \sqrt{\dfrac{2-\sqrt{3}}{4}}\\
\Rightarrow\sin \left( {{15}^{\circ }} \right) =\pm \dfrac{\sqrt{2-\sqrt{3}}}{2}$
But here 15 degrees is an acute angle and we know that sine of an acute angle is a positive number. Therefore, we discard the negative value, i.e. $\sin \left( {{15}^{\circ }} \right)=-\dfrac{\sqrt{2-\sqrt{3}}}{2}$ and consider $\sin \left( {{15}^{\circ }} \right)=\dfrac{\sqrt{2-\sqrt{3}}}{2}$.
Now, substitute this value in equation (i).
$\text{cosec(1}{{\text{5}}^{\circ }})=\dfrac{1}{\dfrac{\sqrt{2-\sqrt{3}}}{2}}\\
\therefore\text{cosec(1}{{\text{5}}^{\circ }})=\dfrac{2}{\sqrt{2-\sqrt{3}}}$

Hence, we found that $\text{cosec}\left( \dfrac{\pi }{12} \right)=\text{cosec}\left( {{15}^{\circ }} \right)=\dfrac{2}{\sqrt{2-\sqrt{3}}}$.

Note: Sometimes in some questions, the formulae may help to solve the questions.
$\sin (2\pi +\theta )=\sin (\theta )$
$\Rightarrow\cos (2\pi +\theta )=\cos (\theta )$
$\Rightarrow\sin (\pi +\theta )=-\sin (\theta )$
$\Rightarrow\cos (\pi +\theta )=-\cos (\theta )$
$\Rightarrow\sin (-\theta )=-\sin (\theta )$
$\Rightarrow\cos (-\theta )=\cos (\theta )$
With the help of these formulae you can find the formulae for the other trigonometric ratios as all the other trigonometric ratios depend on sine and cosine.