# The maximum value of $f(x) = \sin x(1 + \cos x)$

A. $3\dfrac{{\sqrt 3 }}{4}$

B. $3\dfrac{{\sqrt 3 }}{2}$

C. $3\sqrt 3 $

D. $\sqrt 3 $

Answer

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**Hint:**We can use various trigonometric identities such as ${\cos ^2}x - {\sin ^2}x = \cos 2x$ and $\cos 2x = 2{\cos ^2}x - 1$ . Try to factorize for the function as much as possible to get all the possibilities. Check all the possible values of x in the given trigonometry functions.

**Complete step by step answer:**

$f(x) = \sin x(1 + \cos x)$

On Taking the derivative with respect to $x$ on both the sides we get

$f'(x) = \sin x( - \sin x) + (1 + \cos x)\cos x$

Now On multiplying the respective terms we get

$f'(x) = - {\sin ^2}x + \cos x + {\cos ^2}x$

Now Using the identity ${\cos ^2}x - {\sin ^2}x = \cos 2x$ we get

$f'(x) = \cos 2x + \cos x$

Now Using another value for $\cos 2x = 2{\cos ^2}x - 1$ we get

$f'(x) = 2{\cos ^2}x - 1 + \cos x$

Now Splitting the value of $\cos x$ as $2\cos x - \cos x$ we get

$f'(x) = 2{\cos ^2}x + 2\cos x - \cos x - 1$

Now On arranging the terms we get

$f'(x) = 2\cos x(\cos x + 1) - 1(\cos x + 1)$

Now On factorization we get

$f'(x) = (2\cos x - 1)(\cos x + 1)$

Now, let $f'(x) = 0$ to find all the critical points of the function

Then we get

$(2\cos x - 1)(\cos x + 1) = 0$

Then either $\cos x = \dfrac{1}{2}$ or $\cos x = - 1$

Now we know that The maximum of these two values is $\cos x = \dfrac{1}{2}$ which occurs at $x = \dfrac{\pi }{3}$.

Therefore the maximum value of $f(x)$ is at $x = \dfrac{\pi }{3}$.

Hence, $f\left( {\dfrac{\pi }{3}} \right) = \sin \left( {\dfrac{\pi }{3}} \right)\left( {1 + \cos \left( {\dfrac{\pi }{3}} \right)} \right)$

$f\left( {\dfrac{\pi }{3}} \right)= \left( {\dfrac{{\sqrt 3 }}{2}} \right)\left( {1 + \dfrac{1}{2}} \right) \\

\therefore f\left( {\dfrac{\pi }{3}} \right)= 3\dfrac{{\sqrt 3 }}{4}$

If you find the above solution to be a difficult one you can follow the alternate answer which is as follows.

**Therefore option A is the correct answer.**

**Note:**Alternate solution of the given question :

$f(x) = \sin x(1 + \cos x)$ which can be rewritten as $f(x) = \sin x\left( {\dfrac{1}{2}\sin 2x} \right)$

On Taking the derivative with respect to x on both the sides we get

$f'(x) = \cos x + \cos 2x$

Using the CD formula we get

$f'(x) = 2\cos \dfrac{{3x}}{2}\cos \dfrac{x}{2}$

Now, let $f'(x) = 0$

Then we get $2\cos \dfrac{{3x}}{2}\cos \dfrac{x}{2}= 0$

Therefore either $\cos \dfrac{x}{2} = 0$ or $\cos \dfrac{{3x}}{2} = 0$

Therefore we get $x = \pi $ or $x = \dfrac{\pi }{3}$

$f\left( {\dfrac{\pi }{3}} \right) = \sin \left( {\dfrac{\pi }{3}} \right)\left( {1 + \cos \left( {\dfrac{\pi }{3}} \right)} \right) \\

\Rightarrow f\left( {\dfrac{\pi }{3}} \right)= \left( {\dfrac{{\sqrt 3 }}{2}} \right)\left( {1 + \dfrac{1}{2}} \right) \\

\Rightarrow f\left( {\dfrac{\pi }{3}} \right)= 3\dfrac{{\sqrt 3 }}{4}$

$\Rightarrow f\left( \pi \right) = \sin \pi \left( {1 + \cos \left( \pi \right)} \right) = 0$

Therefore it is clearly visible that the maximum value of $f(x)$ is at $x = \dfrac{\pi }{3}$.

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