Question

Exact value of $\cos 20^\circ + 2{\sin ^2}55^\circ - \sqrt 2 \sin 65^\circ$ is:
(A) $1$
(B) $\dfrac{1}{{\sqrt 2 }}$
(C) $\sqrt 2$
(D) zero

Answer 1

Firstly, find the value of $2{\sin ^2}55^\circ$ using the formula $2{\sin ^2}\theta = 1 - \cos 2\theta$ and then break the $\sin 65^\circ$ into $\sin \left( {45 + 20} \right)$ to expand it using the formula $\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$.

Complete step-by-step answer:
Given, $\cos 20^\circ + 2{\sin ^2}55^\circ - \sqrt 2 \sin 65^\circ$
Using formula $2{\sin ^2}\theta = 1 - \cos 2\theta$ to find the value of $2{\sin ^2}55^\circ$,
$\Rightarrow$ $\cos 20^\circ + \left[ {1 - \cos \left( {2 \times 55^\circ } \right)} \right] - \sqrt 2 \sin 65^\circ$
$\Rightarrow$ $\cos 20^\circ + 1 - \cos 110^\circ - \sqrt 2 \sin 65^\circ$
$\Rightarrow$ $\cos 20^\circ + 1 - \cos \left( {90 + 20} \right) - \sqrt 2 \sin 65^\circ$
$\Rightarrow$ $\cos 20^\circ + 1 - \left( { - \sin 20^\circ } \right) - \sqrt 2 \sin 65^\circ$ $\left[ {\because \cos \left( {90 + \theta } \right) = - \sin \theta } \right]$
$\Rightarrow$ $\cos 20^\circ + 1 + \sin 20^\circ - \sqrt 2 \sin 65^\circ$
$\Rightarrow$ $\cos 20^\circ + 1 + \sin 20^\circ - \sqrt 2 \sin \left( {45 + 20} \right)$ $\left( {\because 65 = 45 + 20} \right)$
Using $\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$ to find the value of $\sin \left( {45 + 20} \right)$,
$\Rightarrow$ $\cos 20^\circ + 1 + \sin 20^\circ - \sqrt 2 \left( {\sin 45^\circ \cos 20^\circ + \cos 45^\circ \sin 20^\circ } \right)$
$\Rightarrow$ $\cos 20^\circ + 1 + \sin 20^\circ - \sqrt 2 \sin 45^\circ \cos 20^\circ - \sqrt 2 \cos 45^\circ \sin 20^\circ$
$\Rightarrow$ $\cos 20^\circ + 1 + \sin 20^\circ - \sqrt 2 \times \dfrac{1}{{\sqrt 2 }} \times \cos 20^\circ - \sqrt 2 \times \dfrac{1}{{\sqrt 2 }} \times \sin 20^\circ$ $\left( {\because \sin 45^\circ = \cos 45^\circ = \dfrac{1}{{\sqrt 2 }}} \right)$
$\Rightarrow$ $\cos 20^\circ + 1 + \sin 20^\circ - \cos 20^\circ - \sin 20^\circ$
$\Rightarrow$ $1$
Therefore, $\cos 20^\circ + 2{\sin ^2}55^\circ - \sqrt 2 \sin 65^\circ$=$1$

Hence, option (A) is the correct answer.

Note: An another approach to solve this question is described below:
Given, $\cos 20^\circ + 2{\sin ^2}55^\circ - \sqrt 2 \sin 65^\circ$
Using formula $2{\sin ^2}\theta = 1 - \cos 2\theta$ to find the value of $2{\sin ^2}55^\circ$,
$\Rightarrow$ $\cos 20^\circ + \left[ {1 - \cos \left( {2 \times 55^\circ } \right)} \right] - \sqrt 2 \sin 65^\circ$
$\Rightarrow$ $\cos 20^\circ + 1 - \cos 110^\circ - \sqrt 2 \sin 65^\circ$
$\Rightarrow$ $\cos 20^\circ - \cos 110^\circ + 1 - \sqrt 2 \sin 65^\circ$
Using formula $\cos C - \cos D = 2\sin \left( {\dfrac{{C + D}}{2}} \right)\sin \left( {\dfrac{{D - C}}{2}} \right)$ to find the value of $\cos 20^\circ - \cos 110^\circ$,
$\Rightarrow$ $\left[ {2\sin \left( {\dfrac{{20 + 110}}{2}} \right)\sin \left( {\dfrac{{110 - 20}}{2}} \right)} \right] + 1 - \sqrt 2 \sin 65^\circ$
$\Rightarrow$ $2\sin 65^\circ \sin 45^\circ + 1 - \sqrt 2 \sin 65^\circ$
$\Rightarrow$ $2\sin 65^\circ \times \dfrac{1}{{\sqrt 2 }} + 1 - \sqrt 2 \sin 65^\circ$ $\left( {\because \sin 45^\circ = \dfrac{1}{{\sqrt 2 }}} \right)$
$\Rightarrow$ $\sqrt 2 \sin 65^\circ + 1 - \sqrt 2 \sin 65^\circ$
$\Rightarrow$ $1$
Therefore, $\cos 20^\circ + 2{\sin ^2}55^\circ - \sqrt 2 \sin 65^\circ$=$1$
Hence, option (A) is the correct answer.