Question

Which of the following is the exact value of ${{\cos }^{2}}57{}^\circ +{{\cos }^{2}}63{}^\circ +\cos 57{}^\circ \cos 63{}^\circ $ ?
(a) 1
(b) $\dfrac{1}{4}$
(c) $\dfrac{1}{2}$
(d) $\dfrac{3}{4}$

Answer 1

Hint: Firstly, we have to add and subtract $\cos 57{}^\circ \cos 63{}^\circ $ to the given expression. Then, we have to apply the algebraic identity and simplify the expression. We have to apply the trigonometric identities $\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ and $\cos 2x=2{{\cos }^{2}}x-1$ to simplify the expression further. Then, we have to apply the standard results of the values of trigonometric functions and simplify further.

Complete step by step answer:
We have to find the exact value of ${{\cos }^{2}}57{}^\circ +{{\cos }^{2}}63{}^\circ +\cos 57{}^\circ \cos 63{}^\circ $ . We have to add and subtract $\cos 57{}^\circ \cos 63{}^\circ $ with the given expression.
$\begin{align}
  & \Rightarrow {{\cos }^{2}}57{}^\circ +{{\cos }^{2}}63{}^\circ +\cos 57{}^\circ \cos 63{}^\circ +\cos 57{}^\circ \cos 63{}^\circ -\cos 57{}^\circ \cos 63{}^\circ \\
 & \Rightarrow \left( {{\cos }^{2}}57{}^\circ +{{\cos }^{2}}63{}^\circ +2\cos 57{}^\circ \cos 63{}^\circ \right)-\cos 57{}^\circ \cos 63{}^\circ \\
\end{align}$
We can see that the terms inside the bracket is of the form of the identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ . Therefore, the above expression can be written as
$\Rightarrow {{\left( \cos 57{}^\circ +\cos 63{}^\circ \right)}^{2}}-\cos 57{}^\circ \cos 63{}^\circ $
Now, let us multiply and divide the second term in the above expression by 2.
$\Rightarrow {{\left( \cos 57{}^\circ +\cos 63{}^\circ \right)}^{2}}-\dfrac{2\cos 57{}^\circ \cos 63{}^\circ }{2}$
Let us take the LCM and simplify.
$\Rightarrow {{\left( \cos 57{}^\circ +\cos 63{}^\circ \right)}^{2}}-\dfrac{2\cos 57{}^\circ \cos 63{}^\circ }{2}$
We know that $\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ . Therefore, we can write the above expression as
\[\begin{align}
  & \Rightarrow {{\left( 2\cos \left( \dfrac{57{}^\circ +63{}^\circ }{2} \right)\cos \left( \dfrac{57{}^\circ -63{}^\circ }{2} \right) \right)}^{2}}-\dfrac{\cos 120{}^\circ +\cos 6{}^\circ }{2} \\
 & \Rightarrow {{\left( 2\cos \left( \dfrac{120{}^\circ }{2} \right)\cos \left( \dfrac{6{}^\circ }{2} \right) \right)}^{2}}-\dfrac{\cos 120{}^\circ +\cos 6{}^\circ }{2} \\
 & \Rightarrow {{\left( 2\cos 60{}^\circ \cos 3{}^\circ \right)}^{2}}-\dfrac{\cos 120{}^\circ +\cos 6{}^\circ }{2} \\
\end{align}\]
We know that $\cos 60{}^\circ =\dfrac{1}{2}$ .
\[\begin{align}
  & \Rightarrow {{\left( 2\times \dfrac{1}{2}\cos 3{}^\circ \right)}^{2}}-\dfrac{\cos 120{}^\circ +\cos 6{}^\circ }{2} \\
 & \Rightarrow {{\cos }^{2}}3{}^\circ -\dfrac{\cos 120{}^\circ +\cos 6{}^\circ }{2} \\
\end{align}\]
We can write 120 as the sum of 90 and 30.
\[\Rightarrow {{\cos }^{2}}3{}^\circ -\dfrac{\cos \left( 90{}^\circ +30{}^\circ \right)+\cos 6{}^\circ }{2}\]
We know that $\cos \left( 90{}^\circ +\theta \right)=-\sin \theta $ . Therefore, the above expression becomes
\[\Rightarrow {{\cos }^{2}}3{}^\circ -\dfrac{-\sin 30{}^\circ +\cos 6{}^\circ }{2}\]
We know that $\sin 30{}^\circ =\dfrac{1}{2}$ .
\[\begin{align}
  & \Rightarrow {{\cos }^{2}}3{}^\circ -\dfrac{-\dfrac{1}{2}+\cos 6{}^\circ }{2} \\
 & \Rightarrow {{\cos }^{2}}3{}^\circ -\dfrac{1}{2}\left( -\dfrac{1}{2}+\cos 6{}^\circ \right) \\
\end{align}\]
We can write 6 as the product of 2 and 3.
\[\Rightarrow {{\cos }^{2}}3{}^\circ -\dfrac{1}{2}\left( -\dfrac{1}{2}+\cos \left( 2\times 3{}^\circ \right) \right)\]
We know that $\cos 2x=2{{\cos }^{2}}x-1$ . Therefore, the above expression becomes
\[\begin{align}
  & \Rightarrow {{\cos }^{2}}3{}^\circ -\dfrac{1}{2}\left( -\dfrac{1}{2}+2{{\cos }^{2}}3{}^\circ -1 \right) \\
 & \Rightarrow {{\cos }^{2}}3{}^\circ -\dfrac{1}{2}\left( -\dfrac{3}{2}+2{{\cos }^{2}}3{}^\circ \right) \\
\end{align}\]
Let us apply distributive property on the second term.
\[\begin{align}
  & \Rightarrow {{\cos }^{2}}3{}^\circ -\dfrac{1}{2}\times \left( -\dfrac{3}{2} \right)-\dfrac{1}{2}\times 2{{\cos }^{2}}3{}^\circ \\
 & \Rightarrow \require{cancel}\cancel{{{\cos }^{2}}3{}^\circ }+\dfrac{3}{4}\require{cancel}\cancel{-{{\cos }^{2}}3{}^\circ } \\
 & \Rightarrow \dfrac{3}{4} \\
\end{align}\]
Therefore, the exact value of ${{\cos }^{2}}57{}^\circ +{{\cos }^{2}}63{}^\circ +\cos 57{}^\circ \cos 63{}^\circ $ is $\dfrac{3}{4}$ .

So, the correct answer is “Option d”.

Note: Students must thoroughly learn the algebraic and trigonometric identities. They must know to find the values of basic trigonometric angles. Students must know the relation between the trigonometric functions and how they are converted using angles.