Question

Without using a calculator find the value of $ \sin \left( {{105}^{\circ }} \right) $ .

Answer 1

Hint: In this question, we need to find the value of $ \sin \left( {{105}^{\circ }} \right) $ without using the calculator. For this, we will use the trigonometric identities and formulas. We will use the trigonometric identity of sin(A+B) according to which $ \sin \left( A+B \right)=\sin A\cos B+\cos A\sin B $ . After that, we will put in the known values of sine and cosine function from the trigonometric ratio table and finally evaluate our answer.

Complete step by step answer:
Here we are given the function as $ \sin \left( {{105}^{\circ }} \right) $ . We need to evaluate it without a calculator.
From the trigonometric ratio table, we only know the value of trigonometric ratios with angles $ {{0}^{\circ }},{{30}^{\circ }},{{45}^{\circ }},{{60}^{\circ }}\text{ and }{{90}^{\circ }} $ . So we can write $ {{105}^{\circ }} $ as $ {{60}^{\circ }}+{{45}^{\circ }} $ .
Therefore our expression becomes $ \sin \left( {{105}^{\circ }} \right)=\sin \left( {{60}^{\circ }}+{{45}^{\circ }} \right) $ .
We know that the formula of the sum of angles in a sine function is given as, $ \sin \left( A+B \right)=\sin A\cos B+\cos A\sin B $ . So we can apply it on $ \sin \left( {{60}^{\circ }}+{{45}^{\circ }} \right) $ where A will be equal to $ {{60}^{\circ }} $ and B will be equal to $ {{45}^{\circ }} $ .
So our expression becomes $ \sin \left( {{105}^{\circ }} \right)=\sin {{60}^{\circ }}\cos {{45}^{\circ }}+\cos {{60}^{\circ }}\sin {{45}^{\circ }}\cdots \cdots \cdots \left( 1 \right) $ .
Let us draw trigonometric ratio table for sine and cosine function to evaluate the value of $ \sin {{60}^{\circ }},\cos {{60}^{\circ }},\sin {{45}^{\circ }},\cos {{45}^{\circ }} $ we get,

$ \theta\to $	$ {{0}^{\circ }} $	$ {{30}^{\circ }} $	$ {{45}^{\circ }} $	$ {{60}^{\circ }} $	$ {{90}^{\circ }} $
sine	0	$ \dfrac{1}{2} $	$ \dfrac{1}{\sqrt{2}} $	$ \dfrac{\sqrt{3}}{2} $	1
cosine	1	$ \dfrac{\sqrt{3}}{2} $	$ \dfrac{1}{\sqrt{2}} $	$ \dfrac{1}{2} $	0

As we can see, $ \sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2},\cos {{60}^{\circ }}=\dfrac{1}{2},\sin {{45}^{\circ }}=\dfrac{1}{\sqrt{2}} $ . So putting these values in (1) we get,
 $ \sin \left( {{105}^{\circ }} \right)=\left( \dfrac{\sqrt{3}}{2} \right)\left( \dfrac{1}{\sqrt{2}} \right)+\left( \dfrac{1}{2} \right)\left( \dfrac{1}{\sqrt{2}} \right) $ .
Simplifying it we get,
 $ \begin{align}
  & \sin \left( {{105}^{\circ }} \right)=\dfrac{\sqrt{3}}{2\sqrt{2}}+\dfrac{1}{2\sqrt{2}} \\
 & \Rightarrow \sin \left( {{105}^{\circ }} \right)=\dfrac{\sqrt{3}+1}{2\sqrt{2}} \\
\end{align} $ .
Which is our required answer.
Hence, $ \sin \left( {{105}^{\circ }} \right)=\dfrac{\sqrt{3}+1}{2\sqrt{2}} $ .

Note:
 Students should keep in mind the formula of the trigonometric function and also the trigonometric ratio table. Take care of signs while solving. Students can also rationalize the final answer to get a more simplified answer in the following way, we have $ \dfrac{\sqrt{3}+1}{2\sqrt{2}} $ .
Multiplying the numerator and denominator by $ \sqrt{2} $ we get,
 $ \dfrac{\sqrt{3}+1}{2\sqrt{2}}\times \dfrac{\sqrt{2}}{\sqrt{2}} $ .
Using $ \sqrt{a}\times \sqrt{b}=\sqrt{ab} $ and $ \sqrt{a}\sqrt{a}=a $ we get,
\[\dfrac{\sqrt{3}\sqrt{2}+\sqrt{2}}{2\sqrt{2}\sqrt{2}}=\dfrac{\sqrt{6}+\sqrt{2}}{4}\].
So we can also say that \[\sin \left( {{105}^{\circ }} \right)=\dfrac{\sqrt{6}+\sqrt{2}}{4}\].