Question

Find the value of the expression $\cos \left( 60{}^\circ +x \right)+\cos \left( 60{}^\circ -x \right)$ is
[a] $\sqrt{2}\sin x$
[b] $\sqrt{2}\cos x$
[c] $\sin x$
[d] $\cos x$

Answer 1

Hint: Use the identity $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$ and $\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$. Hence determine the value of $\cos \left( 60{}^\circ -x \right)$ and of $\cos \left( 60{}^\circ +x \right)$. Add the two expressions to get the value of $\cos \left( 60{}^\circ -x \right)+\cos \left( 60{}^\circ +x \right)$. Alternatively use the fact that $\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$. Hence determine the value of the expression $\cos \left( 60{}^\circ -x \right)+\cos \left( 60{}^\circ +x \right)$.

Complete step-by-step answer:
We know that $\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$.
Put $A=60{}^\circ$ and $B=x$, we get
$\cos \left( 60{}^\circ -x \right)=\cos 60{}^\circ \cos x+\sin \left( 60{}^\circ \right)\sin x$
We know that $\cos 60{}^\circ =\dfrac{1}{2}$ and $\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$
Hence, we have
$\cos \left( 60{}^\circ -x \right)=\dfrac{\cos x}{2}+\dfrac{\sqrt{3}}{2}\sin x\text{ }\left( i \right)$
Also, we know that
$\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$
Put $A=60{}^\circ$ and $B=x$, we get
$\cos \left( 60{}^\circ +x \right)=\cos 60{}^\circ \cos x-\sin \left( 60{}^\circ \right)\sin x$
We know that $\cos 60{}^\circ =\dfrac{1}{2}$ and $\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$
Hence, we have
$\cos \left( 60{}^\circ +x \right)=\dfrac{\cos x}{2}-\dfrac{\sqrt{3}}{2}\sin x\text{ }\left( ii \right)$
Adding equation(i) and equation (ii), we get
$\cos \left( 60{}^\circ +x \right)+\cos \left( 60{}^\circ -x \right)=\dfrac{1}{2}\cos x-\dfrac{\sqrt{3}}{2}\sin x+\dfrac{1}{2}\cos x+\dfrac{\sqrt{3}}{2}\sin x=\cos x$
Hence, we have
$\cos \left( 60{}^\circ +x \right)+\cos \left( 60{}^\circ -x \right)=\cos x$ $$$$
Hence option [d] is correct.

Note: Alternative Solution:
We know that
$\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$
Put A = 60+x and B = 60-x, we get
$\cos \left( 60{}^\circ +x \right)+\cos \left( 60{}^\circ -x \right)=2\cos \left( \dfrac{60{}^\circ +x+60{}^\circ -x}{2} \right)\cos \left( \dfrac{60{}^\circ +x-60{}^\circ +x}{2} \right)$
Hence, we have
$\cos \left( 60{}^\circ +x \right)+\cos \left( 60{}^\circ -x \right)=2\cos x\cos 60{}^\circ$
We know that $\cos 60{}^\circ =\dfrac{1}{2}$
Hence, we have
$\cos \left( 60{}^\circ +x \right)+\cos \left( 60{}^\circ -x \right)=2\times \dfrac{1}{2}\cos x=\cos x$, which is the same as obtained above.
Hence option [d] is correct.