Question

What is the value of $\sin {105}^0+\cos {105}^0$?
$$\begin{gathered} \text{A}\text{. }\sin {50}^0 \\ \text{B}\text{. cos}{50}^0 \\ \text{C}\text{. }{\displaystyle \frac{1}{\sqrt{2}}} \\ \text{D}\text{. 0} \end{gathered}$$

Answer 1

Hint- Here, we will be splitting the angle ${105}^0$ into the sum of ${60}^0$ and ${45}^0$ because from the trigonometric table we know the values of the trigonometric functions corresponding to ${60}^0$ and ${45}^0$.

“Complete step-by-step answer:”
As we know that $\sin \left(A+B\right)=\left(\sin A\right)\left(\cos B\right)+\left(\cos A\right)\left(\sin B\right)$
$$\begin{gathered} \sin {105}^0=\sin \left({60}^0+{45}^0\right) \\ \Rightarrow \sin {105}^0=\left(\sin {60}^0\right)\left(\cos {45}^0\right)+\left(\cos {60}^0\right)\left(\sin {45}^0\right)\to \text{(1)} \end{gathered}$$
According to trigonometric table, $\sin {60}^0={\displaystyle \frac{\sqrt{3}}{2}}$, $\cos {60}^0={\displaystyle \frac{1}{2}}$ and $sin{45}^0=\cos {45}^0={\displaystyle \frac{1}{\sqrt{2}}}$
$$\begin{gathered} \Rightarrow \sin {105}^0=\left({\displaystyle \frac{\sqrt{3}}{2}}\right)\left({\displaystyle \frac{1}{\sqrt{2}}}\right)+\left({\displaystyle \frac{1}{2}}\right)\left({\displaystyle \frac{1}{\sqrt{2}}}\right)={\displaystyle \frac{\sqrt{3}}{2\sqrt{2}}}+{\displaystyle \frac{1}{2\sqrt{2}}} \\ \Rightarrow \sin {105}^0={\displaystyle \frac{\sqrt{3}+1}{2\sqrt{2}}}\to \text{(2)} \end{gathered}$$
Also we know that $\cos \left(A+B\right)=\left(\cos A\right)\left(\cos B\right)-\left(\sin A\right)\left(\sin B\right)$
$$\begin{gathered} \cos {105}^0=\cos \left({60}^0+{45}^0\right) \\ \Rightarrow \cos {105}^0=\left(\cos {60}^0\right)\left(\cos {45}^0\right)-\left(\sin {60}^0\right)\left(\sin {45}^0\right)\to \text{(3)} \end{gathered}$$
According to trigonometric table, $\sin {60}^0={\displaystyle \frac{\sqrt{3}}{2}}$, $\cos {60}^0={\displaystyle \frac{1}{2}}$ and $sin{45}^0=\cos {45}^0={\displaystyle \frac{1}{\sqrt{2}}}$
Putting the above values in equation (3), we get
$$\begin{gathered} \Rightarrow \cos {105}^0=\left({\displaystyle \frac{1}{2}}\right)\left({\displaystyle \frac{1}{\sqrt{2}}}\right)-\left({\displaystyle \frac{\sqrt{3}}{2}}\right)\left({\displaystyle \frac{1}{\sqrt{2}}}\right)={\displaystyle \frac{1}{2\sqrt{2}}}-{\displaystyle \frac{\sqrt{3}}{2\sqrt{2}}} \\ \Rightarrow \cos {105}^0={\displaystyle \frac{1-\sqrt{3}}{2\sqrt{2}}}\to \text{(4)} \end{gathered}$$
The value of expression $\sin {105}^0+\cos {105}^0$ can be obtained by using equations (2) and (4), we get
$\sin {105}^0+\cos {105}^0={\displaystyle \frac{\sqrt{3}+1}{2\sqrt{2}}}+{\displaystyle \frac{1-\sqrt{3}}{2\sqrt{2}}}={\displaystyle \frac{\sqrt{3}+1+1-\sqrt{3}}{2\sqrt{2}}}={\displaystyle \frac{2}{2\sqrt{2}}}={\displaystyle \frac{1}{\sqrt{2}}}$
Hence, option C is correct.

Note- In this particular problem, we doesn’t know the value of trigonometric functions corresponding to ${105}^0$ directly so in order to obtain that we split this angle and then use the formulas $\sin \left(A+B\right)=\left(\sin A\right)\left(\cos B\right)+\left(\cos A\right)\left(\sin B\right)$ and $\cos \left(A+B\right)=\left(\cos A\right)\left(\cos B\right)-\left(\sin A\right)\left(\sin B\right)$ to obtain the values of $\sin {105}^0$ and $\cos {105}^0$.