Question

The value of the expression ${{\cos }^{2}}10{}^\circ -\cos 10{}^\circ \cos 50{}^\circ +{{\cos }^{2}}50{}^\circ $ is
[a] $\dfrac{3}{2}\left( 1+\cos 20{}^\circ \right)$
[b] $\dfrac{3}{4}$
[c] $\dfrac{3}{4}+\cos 20{}^\circ $
[d] $\dfrac{3}{2}$

Answer 1

Hint: Take cos 10 common from the first two terms and use the fact that ${{\cos }^{2}}A-{{\sin }^{2}}B=\cos \left( A+B \right)\cos \left( A-B \right)$ and hence prove that the given expression is equal to $1+\dfrac{\cos 40{}^\circ }{2}-\cos 10{}^\circ \cos 50{}^\circ $. Use the fact that $2\cos A\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)$ and hence prove that the given expression is equal to $1+\dfrac{\cos 40{}^\circ }{2}-\dfrac{\cos 60{}^\circ }{2}-\dfrac{\cos 40{}^\circ }{2}$. Hence find which of the options are correct.

Complete step-by-step solution:
Let $l={{\cos }^{2}}10{}^\circ -\cos 10{}^\circ \cos 50{}^\circ +{{\cos }^{2}}50{}^\circ $
We know that ${{\cos }^{2}}x=1-{{\sin }^{2}}x$
Hence, we have
$l={{\cos }^{2}}10{}^\circ -\cos 10{}^\circ \cos 50{}^\circ +1-{{\sin }^{2}}50{}^\circ $
We know that ${{\cos }^{2}}A-{{\sin }^{2}}B=\cos \left( A+B \right)\cos \left( A-B \right)$
Put $A=10{}^\circ $ and $B=50{}^\circ $, we get
${{\cos }^{2}}10{}^\circ -{{\sin }^{2}}50{}^\circ =\cos \left( 10{}^\circ +50{}^\circ \right)\cos \left( 10{}^\circ -50{}^\circ \right)=\cos 60{}^\circ \cos 40{}^\circ $
We know that $\cos 60{}^\circ =\dfrac{1}{2}$. Hence, we have
${{\cos }^{2}}10{}^\circ -{{\sin }^{2}}50{}^\circ =\dfrac{\cos 40{}^\circ }{2}$
Hence, we have
$l=\dfrac{\cos 40{}^\circ }{2}-\cos 10{}^\circ \cos 50{}^\circ $
We know that $2\cos A\cos B=\cos \left( A+B\right)+\cos \left(A-B \right)$
Put $A=10{}^\circ $ and $B=50{}^\circ $, we get
$\begin{align}
  & 2\cos 10{}^\circ \cos 50{}^\circ =\cos \left({10{}^\circ +50{}^\circ } \right)+\cos \left({10{}^\circ -50{}^\circ }\right) \\
 & =\cos 60{}^\circ +\cos 40{}^\circ \\
\end{align}$
We know that $\cos 60{}^\circ =\dfrac{1}{2}$. Hence, we have
$2\cos 10{}^\circ \cos 50{}^\circ =\dfrac{1}{2}+\cos 40{}^\circ $
Dividing both sides by 2, we get
$\cos 10{}^\circ \cos 50{}^\circ =\dfrac{1}{4}+\dfrac{\cos 40{}^\circ }{2}$
Hence, we have
$l=1+\dfrac{\cos 40{}^\circ }{2}-\dfrac{1}{4}-\dfrac{\cos 40{}^\circ }{2}=\dfrac{3}{4}$
Hence, we have
${{\cos }^{2}}10{}^\circ -\cos 10{}^\circ \cos 50{}^\circ +1-{{\sin }^{2}}50{}^\circ =\dfrac{3}{4}$
Hence option [b] is correct.

Note:[1] In these types of questions, we should try all the possible ways to simplify the expression. If by one method, does not simplify the problem or gets stuck we should another way to simplify the expression. Usually after at most three trials the problem gets simplified.
[2] One can use the identity ${{\cos }^{2}}x=\dfrac{1+\cos 2x}{2}$ followed by $\cos A+\cos B=2\cos \dfrac{A+B}{2}\cos \dfrac{A-B}{2}$ instead of ${{\cos }^{2}}A-{{\sin }^{2}}B=\cos \left( A+B \right)\cos \left( A-B \right)$ to simplify the expression.