Question

# If $f(x)={{\sin }^{2}}x+{{\sin }^{2}}\left( x+\dfrac{\pi }{3} \right)+\cos x\cos \left( x+\dfrac{\pi }{3} \right)$ and$g\left( \dfrac{5}{4} \right)=1$, then $\left( gof \right)\left( x \right)$ is equal toA. 0B. 2C. 1D. 3

Hint: In the question use the identities $\sin (A+B)=\sin A\cos B+\cos A\sin B$and $\cos (A+B)=\cos A\cos B-\sin A\sin B$and get the desired result.

In the question we are given that,
$f(x)={{\sin }^{2}}x+{{\sin }^{2}}\left( x+\dfrac{\pi }{3} \right)+\cos x\cos \left( x+\dfrac{\pi }{3} \right)$

We will now consider the identities,
$\sin (A+B)=\sin A\cos B+\cos A\sin B$
And
$\cos (A+B)=\cos A\cos B-\sin A\sin B$

Using the above mentioned identities to expand f(x), we get,
$f(x)={{\sin }^{2}}x+{{\left\{ \sin \left( x+\dfrac{\pi }{3} \right) \right\}}^{2}}+\cos x\left\{ \cos \left( x+\dfrac{\pi }{3} \right) \right\}$
$f(x)={{\sin }^{2}}x+{{\left\{ \sin x\cos \dfrac{\pi }{3}+\cos x\sin \dfrac{\pi }{3} \right\}}^{2}}+\cos x\left\{ \cos x\cos \dfrac{\pi }{3}-\sin x\sin \dfrac{\pi }{3} \right\}$
We know $\cos \dfrac{\pi }{3}=\dfrac{1}{2}$ and $\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}$,

substituting these values in above equation, we get
$f(x)={{\sin }^{2}}x+{{\left\{ \dfrac{\sin x}{2}+\dfrac{\sqrt{3}\cos x}{2} \right\}}^{2}}+\cos x\left\{ \dfrac{\cos x}{2}-\dfrac{\sqrt{3}\sin x}{2} \right\}$
Now we will expand f(x) and use the formula
${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
So,
\begin{align} & f(x)={{\sin }^{2}}x+{{\left( \dfrac{\sin x}{2} \right)}^{2}}+2\left( \dfrac{\sin x}{2} \right)\left( \dfrac{\sqrt{3}\cos x}{2} \right)+{{\left( \dfrac{\sqrt{3}\cos x}{2} \right)}^{2}}+\dfrac{{{\cos }^{2}}x}{2}-\dfrac{\sqrt{3}\cos x\sin x}{2} \\ & f(x)={{\sin }^{2}}x+\dfrac{{{\sin }^{2}}x}{4}+\dfrac{3{{\cos }^{2}}x}{4}+\dfrac{\sqrt{3}\sin x\cos x}{2}+\dfrac{{{\cos }^{2}}x}{2}-\dfrac{\sqrt{3}\cos x\sin x}{2} \\ \end{align}

By cancelling the like terms, we get
$f(x)={{\sin }^{2}}x+\dfrac{{{\sin }^{2}}x}{4}+\dfrac{3{{\cos }^{2}}x}{4}+\dfrac{{{\cos }^{2}}x}{2}$
Taking out the common terms, we get
$f(x)={{\sin }^{2}}x\left( 1+\dfrac{1}{4} \right)+{{\cos }^{2}}x\left( \dfrac{3}{4}+\dfrac{1}{2} \right)$

Taking the LCM and solving, we get
\begin{align} & f(x)={{\sin }^{2}}x\left( \dfrac{4+1}{4} \right)+{{\cos }^{2}}x\left( \dfrac{3+2}{4} \right) \\ & \Rightarrow f(x)={{\sin }^{2}}x\left( \dfrac{5}{4} \right)+{{\cos }^{2}}x\left( \dfrac{5}{4} \right) \\ \end{align}

Now we take out the common term, and write it as,
$f(x)=\dfrac{5}{4}\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)$
Now we will use the identity
${{\sin }^{2}}x+{{\cos }^{2}}x=1$
We get,
$f(x)=\dfrac{5}{4}\times 1=\dfrac{5}{4}$
Now in the equation where we were asked to find out the value of $\left( gof \right)\left( x \right)$, i.e., $g(f(x))$

Here in the above operations we got$f(x)=\dfrac{5}{4}$.
So, we get
$g(f(x))=g\left( \dfrac{5}{4} \right)$
In the question it is already given that $g\left( \dfrac{5}{4} \right)=1$
So now,
$g(f(x))$=$g\left( \dfrac{5}{4} \right)=1$
Therefore, $\left( gof \right)\left( x \right)$ is equal to 1.
Hence the correct answer is option ‘C’.

Note: Generally in these types of questions, students are always in a dilemma which identity they should use.
Another approach is substituting ${{\sin }^{2}}x=1-{{\cos }^{2}}x$ in the given equation, we get
$f(x)={{\sin }^{2}}x+1-{{\cos }^{2}}\left( x+\dfrac{\pi }{3} \right)+\cos x\cos \left( x+\dfrac{\pi }{3} \right)$
Now taking out the common term, we get
$f(x)={{\sin }^{2}}x+1+\cos \left( x+\dfrac{\pi }{3} \right)\left( \cos x-\cos \left( x+\dfrac{\pi }{3} \right) \right)$
But this becomes a tedious one.