Q35 / 90 · JEE Main 2012 (26 May Online) · Mathematics

The value of $\cos 255^{\circ}+\sin 195^{\circ}$ is

  • A
    $\frac{\sqrt{3}-1}{2 \sqrt{2}}$
  • B
    $\frac{\sqrt{3}-1}{\sqrt{2}}$
  • Correct answer
    C
    $-\frac{\sqrt{3}-1}{\sqrt{2}}$
  • D
    $\frac{\sqrt{3}+1}{\sqrt{2}}$

Solution

Consider $\cos 255^{\circ}+\sin 195^{\circ}$ $$ \begin{aligned} & =\cos \left(270^{\circ}-15^{\circ}\right)+\sin \left(180^{\circ}+15^{\circ}\right) \\ & =-\sin 15^{\circ}-\sin 15^{\circ} \\ & =-2 \sin 15^{\circ}=-2\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)=-\left(\frac{\sqrt{3}-1}{\sqrt{2}}\right) \end{aligned} $$

Hence, the correct answer is option: 'C'.

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