Question

The value of $\cos \left(105°\right)+\sin \left(105°\right)$ is

• A. $\frac{1}{2}$
• B. $1$
• C. $\sqrt{2}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

The correct option is D

$\frac{1}{\sqrt{2}}$

Explanation for correct option

Given trigonometric function is $\cos \left(105°\right)+\sin \left(105°\right)$

$$\begin{gathered} \cos \left(105°\right)=\cos \left(60°+45°\right)=\cos \left(60°\right)\cos \left(45°\right)-\sin \left(60°\right)\sin \left(45°\right)=\frac{1}{2}·\frac{1}{\sqrt{2}}-\frac{\sqrt{3}}{2}·\frac{1}{\sqrt{2}}\left[\because \cos \left(A+B\right)=\cos \left(A\right)\cos \left(B\right)-\sin \left(A\right)\sin \left(B\right)\right] \\ \sin \left(105°\right)=\sin \left(60°+45°\right)=\sin \left(60°\right)\cos \left(45°\right)+\cos \left(60°\right)\sin \left(45°\right)=\frac{\sqrt{3}}{2}·\frac{1}{\sqrt{2}}+\frac{1}{2}·\frac{1}{\sqrt{2}}\left[\because \sin \left(A+B\right)=\sin \left(A\right)\cos \left(B\right)+\cos \left(A\right)\sin \left(B\right)\right] \\ \therefore \cos \left(105°\right)+\sin \left(105°\right)=\frac{1-\sqrt{3}}{2\sqrt{2}}+\frac{1+\sqrt{3}}{2\sqrt{2}} \\ =\frac{1}{\sqrt{2}} \end{gathered}$$Hence, the correct option is $\mathrm{Option}\left(\mathrm{D}\right)$