Answer
Verified
479.4k+ views
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE