Question

The value of $\cos \left(15°\right)-\sin \left(15°\right)$ is equal to

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{1}{2}$
• C. $-\frac{1}{\sqrt{2}}$
• D. $0$

Answer 1

The correct option is A

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

Explanation for correct option

Given trigonometric function is $\cos \left(15°\right)-\sin \left(15°\right)$

$$\begin{gathered} \cos \left(15°\right)=\cos \left(45°-30°\right)=\cos \left(30°\right)\cos \left(45°\right)+\sin \left(30°\right)\sin \left(45°\right)=\frac{\sqrt{3}}{2}·\frac{1}{\sqrt{2}}+\frac{1}{2}·\frac{1}{\sqrt{2}}\left[\because \cos \left(A-B\right)=\cos \left(A\right)\cos \left(B\right)+\sin \left(A\right)\sin \left(B\right)\right] \\ \sin \left(15°\right)=\sin \left(45°-30°\right)=\cos \left(30°\right)\sin \left(45°\right)-\sin \left(30°\right)\cos \left(45°\right)=\frac{\sqrt{3}}{2}·\frac{1}{\sqrt{2}}-\frac{1}{2}·\frac{1}{\sqrt{2}}\left[\because \sin \left(A-B\right)=\sin \left(A\right)\cos \left(B\right)-\cos \left(A\right)\sin \left(B\right)\right] \\ \therefore \cos \left(15°\right)-\sin \left(15°\right)=\frac{\sqrt{3}+1}{2\sqrt{2}}-\frac{\sqrt{3}-1}{2\sqrt{2}} \\ =\frac{1}{\sqrt{2}} \end{gathered}$$Hence, the correct option is $\mathrm{Option}\left(\mathrm{A}\right)$