Question

Find the value of ${\cos ^2}{76^ \circ } + {\cos ^2}{16^ \circ } - \cos {76^ \circ }.\cos {16^ \circ }$
A. $\dfrac{{ - 1}}{4}$
B. $\dfrac{1}{2}$
C. 0
D. $\dfrac{3}{4}$

Answer 1

This problem is simply using all the identities that are in trigonometric functions. We just need to take care of which identity is to be used where. This is simply the problem in which we will use multiple angle formula, factorization and defactorization formula. So let’s start!
Formula used:
$2{\cos ^2}\theta = \cos 2\theta + 1$
$2\cos A\cos B = \cos \left( {A + B} \right) + \cos \left( {A - B} \right)$
$\cos C + \cos D = 2\cos \dfrac{{\left( {C + D} \right)}}{2}.\cos \dfrac{{\left( {C - D} \right)}}{2}$

Complete step by step answer:
Given the sum is,
${\cos ^2}{76^ \circ } + {\cos ^2}{16^ \circ } - \cos {76^ \circ }.\cos {16^ \circ }$
Now we will multiply and divide the given equation by 2.
$= \dfrac{1}{2}\left( {2{{\cos }^2}{{76}^ \circ } + 2{{\cos }^2}{{16}^ \circ } - 2\cos {{76}^ \circ }.\cos {{16}^ \circ }} \right)$
Now it is of the form we need. Now we will use the first and second formula listed above.
$= \dfrac{1}{2}\left( {\cos \left( {2 \times {{76}^ \circ }} \right) + 1 + \cos \left( {2 \times {{16}^ \circ }} \right) + 1 - \cos \left( {{{76}^ \circ } + {{16}^ \circ }} \right) - \cos \left( {{{76}^ \circ } - {{16}^ \circ }} \right)} \right)$
On observing carefully we get,
$= \dfrac{1}{2}\left( {2 + \cos \left( {{{152}^ \circ }} \right) + \cos \left( {{{32}^ \circ }} \right) - \cos \left( {{{92}^ \circ }} \right) - \cos \left( {{{60}^ \circ }} \right)} \right)$
Now we will use the third formula,
$= \dfrac{1}{2}\left( {2 + 2\cos \dfrac{{{{152}^ \circ } + {{32}^ \circ }}}{2}.\cos \dfrac{{{{152}^ \circ } - {{32}^ \circ }}}{2} - \cos \left( {{{92}^ \circ }} \right) - \cos \left( {{{60}^ \circ }} \right)} \right)$
Add the respective angles,
$= \dfrac{1}{2}\left( {2 + 2\cos \dfrac{{{{184}^ \circ }}}{2}.\cos \dfrac{{{{120}^ \circ }}}{2} - \cos \left( {{{92}^ \circ }} \right) - \cos \left( {{{60}^ \circ }} \right)} \right)$
On dividing by 2 we get,
$= \dfrac{1}{2}\left( {2 + 2\cos {{92}^ \circ }.\cos {{60}^ \circ } - \cos \left( {{{92}^ \circ }} \right) - \cos \left( {{{60}^ \circ }} \right)} \right)$
Now as we know the value of $\cos \left( {{{60}^ \circ }} \right) = \dfrac{1}{2}$ putting this in the above equation we get,
$= \dfrac{1}{2}\left( {2 + 2\cos {{92}^ \circ } \times \dfrac{1}{2} - \cos \left( {{{92}^ \circ }} \right) - \dfrac{1}{2}} \right)$
On solving we get,
$= \dfrac{1}{2}\left( {2 + \cos {{92}^ \circ } - \cos \left( {{{92}^ \circ }} \right) - \dfrac{1}{2}} \right)$
We can cancel the same angle with different sign,
$= \dfrac{1}{2}\left( {2 - \dfrac{1}{2}} \right)$
On solving the bracket we get,
$= \left( {\dfrac{3}{4}} \right)$
This is the answer.
${\cos ^2}{76^ \circ } + {\cos ^2}{16^ \circ } - \cos {76^ \circ }.\cos {16^ \circ } = \dfrac{3}{4}$

So, the correct answer is “Option D”.

Note: Note that we all know each and every formula of trigonometry but it is vain if we can’t use them at proper places. Thus it is advised to practice these types of questions. Also note that, the factorization formula is the hint that the angles will be eliminated simply. When we look at the question, we definitely jump to the most common identity ${\sin ^2}\theta + {\cos ^2}\theta = 1$ but we have not used it a single time. Thus note these things.