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

How do I find the value of cosec(π12)\text{cosec}\left( \dfrac{\pi }{12} \right) ?

Explanation

Solution

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:
cosec(x)=1sin⁡(x)\text{cosec(}x)=\dfrac{1}{\sin (x)}
sin⁡2(x2)=1−cos⁡x2{{\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 π\pi radians is equal to 180 degrees.Therefore, an angle of π12\dfrac{\pi }{12} radians is equal to
18012=15\dfrac{180}{12}=15 degrees.
This means that we have to calculate the value of cosec(15∘)\text{cosec(1}{{\text{5}}^{\circ }}\text{)}.
We know that cosec(x)=1sin⁡(x)\text{cosec(}x)=\dfrac{1}{\sin (x)}. Therefore,
cosec(15∘)=1sin⁡(15∘)\text{cosec(1}{{\text{5}}^{\circ }})=\dfrac{1}{\sin (\text{1}{{\text{5}}^{\circ }})} …. (i)
Now, we shall find the value of sin⁡(15∘)\sin (\text{1}{{\text{5}}^{\circ }}) with the help of the half angle formula. The half angle formula says that
sin⁡2(x2)=1−cos⁡x2{{\sin }^{2}}\left( \dfrac{x}{2} \right)=\dfrac{1-\cos x}{2} …. (ii).

In this case, x2=15∘\dfrac{x}{2}={{15}^{\circ }}.Then, this means that x=30∘x={{30}^{\circ }}. Substitute the value x and half of x in equation (ii).Then, we get that
sin⁡2(15∘)=1−cos⁡(30∘)2{{\sin }^{2}}\left( {{15}^{\circ }} \right)=\dfrac{1-\cos ({{30}^{\circ }})}{2} …. (iii)
Now, we know that cos⁡30∘=32\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}.
Now, substitute this value in equation (iii).With this we get that,
sin⁡2(15∘)=1−322{{\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(15∘)=2−322 ⇒sin⁡2(15∘)=2−34{{\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⁡(15∘)=±2−34 ⇒sin⁡(15∘)=±2−32\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⁡(15∘)=−2−32\sin \left( {{15}^{\circ }} \right)=-\dfrac{\sqrt{2-\sqrt{3}}}{2} and consider sin⁡(15∘)=2−32\sin \left( {{15}^{\circ }} \right)=\dfrac{\sqrt{2-\sqrt{3}}}{2}.
Now, substitute this value in equation (i).
cosec(15∘)=12−32 ∴cosec(15∘)=22−3\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 cosec(π12)=cosec(15∘)=22−3\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⁡(2π+θ)=sin⁡(θ)\sin (2\pi +\theta )=\sin (\theta )
⇒cos⁡(2π+θ)=cos⁡(θ)\Rightarrow\cos (2\pi +\theta )=\cos (\theta )
⇒sin⁡(π+θ)=−sin⁡(θ)\Rightarrow\sin (\pi +\theta )=-\sin (\theta )
⇒cos⁡(π+θ)=−cos⁡(θ)\Rightarrow\cos (\pi +\theta )=-\cos (\theta )
⇒sin⁡(−θ)=−sin⁡(θ)\Rightarrow\sin (-\theta )=-\sin (\theta )
⇒cos⁡(−θ)=cos⁡(θ)\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.