Question

How do you half angle formula to find the value of $ \sin {{165}^{\circ }} $ ?

Answer 1

Hint:
 We start solving the problem by recalling the half angle formula for sine function as $ \sin \dfrac{x}{2}=\sqrt{\dfrac{1-\cos x}{2}} $ . We then find the value of x so that $ \dfrac{x}{2} $ is equal to the value $ {{165}^{\circ }} $ . We then make use of the results $ \cos \left( {{360}^{\circ }}-\theta \right)=\cos \theta $ , $ \cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2} $ to proceed through the problem. We then make the necessary arrangements inside the square root and make use of the fact that \[{{165}^{\circ }}\] lies in the second quadrant and sine function is positive in second quadrant to get the required answer.

Complete step by step answer:
According to the problem, we are asked to find the value of $ \sin {{165}^{\circ }} $ using the half-angle formula.
Let us recall the half-angle formula for the sine function.
We know that the half angle formula for sine function is defined as $ \sin \dfrac{x}{2}=\sqrt{\dfrac{1-\cos x}{2}} $ ---(1).
Now, we need to find the value of $ \sin {{165}^{\circ }} $ . So, we have $ \dfrac{x}{2}={{165}^{\circ }} $ ---(2).
 $ \Rightarrow x={{330}^{\circ }} $ ---(3).
Let us substitute equations (2) and (3) in equation (1).
 $ \Rightarrow \sin {{165}^{\circ }}=\sqrt{\dfrac{1-\cos {{330}^{\circ }}}{2}} $ .
\[\Rightarrow \sin {{165}^{\circ }}=\sqrt{\dfrac{1-\cos \left( {{360}^{\circ }}-{{30}^{\circ }} \right)}{2}}\] ---(4).
We know that $ \cos \left( {{360}^{\circ }}-\theta \right)=\cos \theta $ . Let us use this result in equation (4).
\[\Rightarrow \sin {{165}^{\circ }}=\sqrt{\dfrac{1-\cos {{30}^{\circ }}}{2}}\] ---(5).
We know that $ \cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2} $ . Let us use this result in equation (5).
\[\Rightarrow \sin {{165}^{\circ }}=\sqrt{\dfrac{1-\dfrac{\sqrt{3}}{2}}{2}}\].
\[\Rightarrow \sin {{165}^{\circ }}=\sqrt{\dfrac{\dfrac{2-\sqrt{3}}{2}}{2}}\].
\[\Rightarrow \sin {{165}^{\circ }}=\sqrt{\dfrac{2-\sqrt{3}}{4}}\].
\[\Rightarrow \sin {{165}^{\circ }}=\sqrt{\dfrac{2-2\left( \dfrac{\sqrt{3}}{\sqrt{2}} \right)\left( \dfrac{1}{\sqrt{2}} \right)}{4}}\].
\[\Rightarrow \sin {{165}^{\circ }}=\sqrt{\dfrac{{{\left( \dfrac{\sqrt{3}}{\sqrt{2}} \right)}^{2}}+{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}-2\left( \dfrac{\sqrt{3}}{\sqrt{2}} \right)\left( \dfrac{1}{\sqrt{2}} \right)}{4}}\] ---(6).
We know that $ {{a}^{2}}+{{b}^{2}}-2ab={{\left( a-b \right)}^{2}} $ . Let us use this result in equation (6).
\[\Rightarrow \sin {{165}^{\circ }}=\sqrt{\dfrac{{{\left( \dfrac{\sqrt{3}}{\sqrt{2}}-\dfrac{1}{\sqrt{2}} \right)}^{2}}}{{{2}^{2}}}}\].
We know that \[{{165}^{\circ }}\] lies in the second quadrant and sine function is positive in second quadrant. So, we need to consider positive square root value.
\[\Rightarrow \sin {{165}^{\circ }}=\dfrac{\dfrac{\sqrt{3}}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}}{2}\].
\[\Rightarrow \sin {{165}^{\circ }}=\dfrac{\dfrac{\sqrt{3}-1}{\sqrt{2}}}{2}\].
\[\Rightarrow \sin {{165}^{\circ }}=\dfrac{\sqrt{3}-1}{2\sqrt{2}}\].
So, we have found the value of \[\sin {{165}^{\circ }}\] as \[\dfrac{\sqrt{3}-1}{2\sqrt{2}}\].

Note:
We should perform each step carefully in order to avoid calculation mistakes and confusion. We should keep in mind about the nature of values of trigonometric functions in different quadrants while solving this type of problem. Similarly, we can expect the formulas to find the value of \[\sin {{165}^{\circ }}\] and \[\cos {{165}^{\circ }}\] using the formula of \[\cos 2\theta \].