The value of $\cos 255^{\circ}+\sin 195^{\circ}$ is
The value of $\cos 255^{\circ}+\sin 195^{\circ}$ is
-
$\frac{\sqrt{3}-1}{2 \sqrt{2}}$
-
$\frac{\sqrt{3}-1}{\sqrt{2}}$
-
$-\frac{\sqrt{3}-1}{\sqrt{2}}$
-
$\frac{\sqrt{3}+1}{\sqrt{2}}$
Solution
Consider $\cos 255^{\circ}+\sin 195^{\circ}$
$
\begin{aligned}
& =\cos \left(270^{\circ}-15^{\circ}\right)+\sin \left(180^{\circ}+15^{\circ}\right) \\
& =-\sin 15^{\circ}-\sin 15^{\circ} \\
& =-2 \sin 15^{\circ}=-2\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)=-\left(\frac{\sqrt{3}-1}{\sqrt{2}}\right)
\end{aligned}
$
Asked in: JEE Main 2012 (26 May Online)
Practice more Trigonometric Ratios & Identities questions on Aicharya