Question

Find The values of $\cos {255^ \circ } + \sin {165^ \circ } =$
A.0
B.$\dfrac{{\sqrt 3 - 1}}{{\sqrt 3 }}$
C.$\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}$
D.$\dfrac{{\sqrt 2 + 1}}{{\sqrt 2 }}$

Answer 1

In this question we can also write $255 = 120 + 135$, $165 = 120 + 45$. Then we will use $\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$ and $\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$after substituting the respective values and simplify to get the answer.

Complete step-by-step answer:
$\cos {255^ \circ } + \sin {165^ \circ } =$
We can also write
$\Rightarrow \cos \left( {120 + 135} \right) + \sin \left( {120 + 45} \right)$
We know that $\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$ and $\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$ so the equation will become
$\Rightarrow \left( {\cos {{120}^ \circ }\cos {{135}^ \circ } - \sin {{120}^ \circ }\sin {{135}^ \circ }} \right) + \left( {\sin {{120}^ \circ }\cos {{45}^ \circ } + \cos {{120}^ \circ }\sin {{45}^ \circ }} \right)$
Substituting the respective values, we get
$\Rightarrow \left( {\dfrac{{ - 1}}{2} \times \dfrac{{ - 1}}{{\sqrt 2 }} - \dfrac{{\sqrt 3 }}{2} \times \dfrac{1}{{\sqrt 2 }}} \right) + \left( {\dfrac{{\sqrt 3 }}{2} \times \dfrac{1}{{\sqrt 2 }} + \dfrac{{ - 1}}{2} \times \dfrac{1}{{\sqrt 2 }}} \right) \\\ \Rightarrow 0 \\\$
So, $\cos {255^ \circ } + \sin {165^ \circ } =$0
Answer is (A)

Note: Trigonometric formula used are $\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$ and $\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$. Values of some used here must be have to memorized
$\sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }} = \cos {45^ \circ } \\\ \sin {135^ \circ } = \dfrac{1}{{\sqrt 2 }} \\\ \cos {135^ \circ } = \dfrac{{ - 1}}{{\sqrt 2 }} \\\ \sin {120^ \circ } = \dfrac{{\sqrt 3 }}{2} \\\ \cos {120^ \circ } = \dfrac{{ - 1}}{2} \\\$