Question

If $\theta $ lies in the first quadrant and $\cos \theta = \dfrac{8}{{17}},$ then the value of \[\cos \left( {{{30}^ \circ } + \theta } \right) + \cos \left( {{{45}^ \circ } - \theta } \right) + \cos \left( {{{120}^ \circ } - \theta } \right)\] is
A. $\left( {\dfrac{{\sqrt 3 - 1}}{2} + \dfrac{1}{{\sqrt 2 }}} \right)\dfrac{{23}}{{17}}$
B. $\left( {\dfrac{{\sqrt 3 + 1}}{2} + \dfrac{1}{{\sqrt 2 }}} \right)\dfrac{{23}}{{17}}$
C. $\left( {\dfrac{{\sqrt 3 - 1}}{2} - \dfrac{1}{{\sqrt 2 }}} \right)\dfrac{{23}}{{17}}$
D. $\left( {\dfrac{{\sqrt 3 + 1}}{2} - \dfrac{1}{{\sqrt 2 }}} \right)\dfrac{{23}}{{17}}$

Answer 1

Hint: In order to solve this type of question, first we will consider the given equation and simplify it by applying trigonometric identities. Then, we will substitute the given value in it and simplify it further to get the desired correct answer.

Formula used:
$\left[ {\because \cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B} \right]$
$\left[ {\because \cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B} \right]$

Complete step by step solution:
We are given that,
$\cos \theta = \dfrac{8}{{17}}$………………..equation$\left( 1 \right)$
$\Rightarrow \sin \theta = \sqrt{1- \cos^{2} \theta}$
$\Rightarrow \sin \theta = \sqrt{1- {\left(\dfrac{8}{17}\right)}^{2} }$
$ \Rightarrow \sin \theta = \dfrac{{15}}{{17}}$ ………………..equation$\left( 2 \right)$
Consider,
\[\cos \left( {{{30}^ \circ } + \theta } \right) + \cos \left( {{{45}^ \circ } - \theta } \right) + \cos \left( {{{120}^ \circ } - \theta } \right)\]
$ = \left( {\cos {{30}^ \circ }\cos \theta - \sin {{30}^ \circ }\sin \theta } \right) + \left( {\cos {{45}^ \circ }\cos \theta + \sin {{45}^ \circ }\sin \theta } \right) + \left( {\cos {{120}^ \circ }\cos \theta + \sin {{120}^ \circ }\sin \theta } \right)$ $\left[ {\because \cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B} \right],\left[ {\because \cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B} \right]$
Solving it,
$ = \left( {\dfrac{{\sqrt 3 }}{2}\cos \theta - \dfrac{1}{2}\sin \theta } \right) + \left( {\dfrac{1}{{\sqrt 2 }}\cos \theta + \dfrac{1}{{\sqrt 2 }}\sin \theta } \right) + \left( {\dfrac{{ - 1}}{2}\cos \theta + \dfrac{{\sqrt 3 }}{2}\sin \theta } \right)$ $\left[ {\because \cos {{30}^ \circ } = \dfrac{{\sqrt 3 }}{2}} \right],\left[ {\because \sin {{30}^ \circ } = \dfrac{1}{2}} \right],\left[ {\because \cos {{45}^ \circ } = \dfrac{1}{{\sqrt 2 }}} \right],\left[ {\because \sin {{45}^ \circ } = \dfrac{1}{{\sqrt 2 }}} \right],$
$\left[ {\because \cos {{120}^ \circ } = \cos \left( {{{180}^ \circ } - {{60}^ \circ }} \right) = \dfrac{{ - 1}}{2}} \right],\left[ {\because \sin {{120}^ \circ } = \sin \left( {{{180}^ \circ } - {{60}^ \circ }} \right) = \sin {{60}^ \circ } = \dfrac{{\sqrt 3 }}{2}} \right]$
Simplifying it,
$ = \cos \theta \left( {\dfrac{{\sqrt 3 }}{2} + \dfrac{1}{{\sqrt 2 }} - \dfrac{1}{2}} \right) + \sin \theta \left( {\dfrac{{\sqrt 3 }}{2} + \dfrac{1}{{\sqrt 2 }} - \dfrac{1}{2}} \right)$
$ = \left( {\dfrac{{\sqrt 3 }}{2} + \dfrac{1}{{\sqrt 2 }} - \dfrac{1}{2}} \right)\left( {\cos \theta + \sin \theta } \right)$
Substituting the values from equation $\left( 1 \right)$ and $\left( 2 \right),$
$ = \left( {\dfrac{{\sqrt 3 - 1}}{2} + \dfrac{1}{{\sqrt 2 }}} \right)\left( {\dfrac{8}{{17}} + \dfrac{{15}}{{17}}} \right)$
$ = \left( {\dfrac{{\sqrt 3 - 1}}{2} + \dfrac{1}{{\sqrt 2 }}} \right)\dfrac{{23}}{{17}}$
$\therefore $ The correct option is A.

Note: Choose the suitable trigonometric identities and be very sure while simplifying them. This type of question requires the use of correct application of trigonometric rules to get the correct answer.
Sometimes students get confused with the formulas of \[\cos \left( {A + B} \right)\] and \[\cos \left( {A + B} \right)\].