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Question: The maximum value of \(f(x) = \sin x(1 + \cos x)\) A. \(3\dfrac{{\sqrt 3 }}{4}\) B. \(3\dfrac{{\...

The maximum value of f(x)=sin⁡x(1+cos⁡x)f(x) = \sin x(1 + \cos x)
A. 3343\dfrac{{\sqrt 3 }}{4}
B. 3323\dfrac{{\sqrt 3 }}{2}
C. 333\sqrt 3
D. 3\sqrt 3

Explanation

Solution

We can use various trigonometric identities such as cos⁡2x−sin⁡2x=cos⁡2x{\cos ^2}x - {\sin ^2}x = \cos 2x and cos⁡2x=2cos⁡2x−1\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)f(x) = \sin x(1 + \cos x)
On Taking the derivative with respect to xx on both the sides we get
f′(x)=sin⁡x(−sin⁡x)+(1+cos⁡x)cos⁡xf'(x) = \sin x( - \sin x) + (1 + \cos x)\cos x
Now On multiplying the respective terms we get
f′(x)=−sin⁡2x+cos⁡x+cos⁡2xf'(x) = - {\sin ^2}x + \cos x + {\cos ^2}x
Now Using the identity cos⁡2x−sin⁡2x=cos⁡2x{\cos ^2}x - {\sin ^2}x = \cos 2x we get
f′(x)=cos⁡2x+cos⁡xf'(x) = \cos 2x + \cos x
Now Using another value for cos⁡2x=2cos⁡2x−1\cos 2x = 2{\cos ^2}x - 1 we get
f′(x)=2cos⁡2x−1+cos⁡xf'(x) = 2{\cos ^2}x - 1 + \cos x
Now Splitting the value of cos⁡x\cos x as 2cos⁡x−cos⁡x2\cos x - \cos x we get
f′(x)=2cos⁡2x+2cos⁡x−cos⁡x−1f'(x) = 2{\cos ^2}x + 2\cos x - \cos x - 1
Now On arranging the terms we get
f′(x)=2cos⁡x(cos⁡x+1)−1(cos⁡x+1)f'(x) = 2\cos x(\cos x + 1) - 1(\cos x + 1)
Now On factorization we get
f′(x)=(2cos⁡x−1)(cos⁡x+1)f'(x) = (2\cos x - 1)(\cos x + 1)
Now, let f′(x)=0f'(x) = 0 to find all the critical points of the function
Then we get
(2cos⁡x−1)(cos⁡x+1)=0(2\cos x - 1)(\cos x + 1) = 0
Then either cos⁡x=12\cos x = \dfrac{1}{2} or cos⁡x=−1\cos x = - 1
Now we know that The maximum of these two values is cos⁡x=12\cos x = \dfrac{1}{2} which occurs at x=π3x = \dfrac{\pi }{3}.
Therefore the maximum value of f(x)f(x) is at x=π3x = \dfrac{\pi }{3}.
Hence, f(π3)=sin⁡(π3)(1+cos⁡(π3))f\left( {\dfrac{\pi }{3}} \right) = \sin \left( {\dfrac{\pi }{3}} \right)\left( {1 + \cos \left( {\dfrac{\pi }{3}} \right)} \right)
f(π3)=(32)(1+12) ∴f(π3)=334f\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)f(x) = \sin x(1 + \cos x) which can be rewritten as f(x)=sin⁡x(12sin⁡2x)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⁡2xf'(x) = \cos x + \cos 2x
Using the CD formula we get
f′(x)=2cos⁡3x2cos⁡x2f'(x) = 2\cos \dfrac{{3x}}{2}\cos \dfrac{x}{2}
Now, let f′(x)=0f'(x) = 0
Then we get 2cos⁡3x2cos⁡x2=02\cos \dfrac{{3x}}{2}\cos \dfrac{x}{2}= 0
Therefore either cos⁡x2=0\cos \dfrac{x}{2} = 0 or cos⁡3x2=0\cos \dfrac{{3x}}{2} = 0
Therefore we get x=πx = \pi or x=π3x = \dfrac{\pi }{3}
f(π3)=sin⁡(π3)(1+cos⁡(π3)) ⇒f(π3)=(32)(1+12) ⇒f(π3)=334f\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}
⇒f(π)=sin⁡π(1+cos⁡(π))=0\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)f(x) is at x=π3x = \dfrac{\pi }{3}.