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Question

Mathematics Question on Trigonometric Functions

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

A

3−122\frac{\sqrt{3}-1}{2\sqrt{2}}

B

3+12\frac{\sqrt{3}+1}{\sqrt{2}}

C

3−12\frac{\sqrt{3}-1}{\sqrt{2}}

D

32\frac{\sqrt{3}}{\sqrt{2}}

Answer

3−12\frac{\sqrt{3}-1}{\sqrt{2}}

Explanation

Solution

Consider cos 255∘255^{\circ} + sin 195∘195^{\circ} =cos(270∘−15∘)+sin(180∘+15∘)= cos \left(270^{\circ} -15^{\circ}\right) + sin \left(180^{\circ} +15^{\circ}\right) =−sin15∘−sin15∘= -sin 15^{\circ}-sin 15^{\circ} =−2sin15∘=−2(3−122)=−(3−12)= -2 sin 15^{\circ} = -2 \left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right) = -\left(\frac{\sqrt{3}-1}{\sqrt{2}}\right)