# 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 to

A. 0

B. 2

C. 1

D. 3

Answer

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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.

Complete step-by-step answer:

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.

Complete step-by-step answer:

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.

Last updated date: 01st Jun 2023

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