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Question: The trigonometric function \( \sin 4\theta \) can be written as: A. \( 4\sin \theta \left( 1-2{{\s...

The trigonometric function sin⁡4θ\sin 4\theta can be written as:
A. 4sin⁡θ(1−2sin⁡2θ)1−sin⁡2θ4\sin \theta \left( 1-2{{\sin }^{2}}\theta \right)\sqrt{1-{{\sin }^{2}}\theta }
B. 2sin⁡θcos⁡θsin⁡2θ2\sin \theta \cos \theta {{\sin }^{2}}\theta
C. 4sin⁡θ−6sin⁡3θ4\sin \theta -6{{\sin }^{3}}\theta
D. 2sin⁡2θ2\sin 2\theta

Explanation

Solution

Hint : Use the following identities to convert from terms involving 4θ4\theta to 2θ2\theta and to θ\theta .
sin⁡(A±B)=sin⁡Acos⁡B±sin⁡Bcos⁡A\sin (A\pm B)=\sin A\cos B\pm \sin B\cos A
cos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡B\cos (A\pm B)=\cos A\cos B\mp \sin A\sin B
Recall that sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 . Use this fact to convert the terms involving cos⁡θ\cos \theta into sin⁡θ\sin \theta .
Simplify until we are left with terms containing only sin⁡θ\sin \theta .

Complete step-by-step answer :
We can write sin⁡4θ=sin⁡(2θ+2θ)\sin 4\theta =\sin (2\theta +2\theta ) .
Using the identity sin⁡(A+B)=sin⁡Acos⁡B+sin⁡Bcos⁡A\sin (A+B)=\sin A\cos B+\sin B\cos A , we get:
= sin⁡2θcos⁡2θ+sin⁡2θcos⁡2θ\sin 2\theta \cos 2\theta +\sin 2\theta \cos 2\theta
= 2sin⁡2θcos⁡2θ2\sin 2\theta \cos 2\theta
Writing 2θ2\theta as θ+θ\theta + \theta ; we get
= 2sin⁡(θ+θ)cos⁡(θ+θ)2\sin (\theta +\theta )\cos (\theta +\theta )
Using the sin⁡(A+B)=sin⁡Acos⁡B+sin⁡Bcos⁡A\sin (A+B)=\sin A\cos B+\sin B\cos A and cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos (A+B)=\cos A\cos B-\sin A\sin B , we get:
= 2[2sin⁡θcos⁡θ(cos⁡2θ−sin⁡2θ)]2\left[ 2\sin \theta \cos \theta \left( {{\cos }^{2}}\theta -{{\sin }^{2}}\theta \right) \right]
Which can be written as:
= 4sin⁡θcos⁡2θ(cos⁡2θ−sin⁡2θ)4\sin \theta \sqrt{{{\cos }^{2}}\theta }\left( {{\cos }^{2}}\theta -{{\sin }^{2}}\theta \right)
Using the identity cos⁡2θ=1−sin⁡2θ{{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta , we get:
= 4sin⁡θ1−sin⁡2θ(1−sin⁡2θ−sin⁡2θ)4\sin \theta \sqrt{1-{{\sin }^{2}}\theta }\left( 1-{{\sin }^{2}}\theta -{{\sin }^{2}}\theta \right)
= 4sin⁡θ(1−2sin⁡2θ)1−sin⁡2θ4\sin \theta \left( 1-2{{\sin }^{2}}\theta \right)\sqrt{1-{{\sin }^{2}}\theta }
Therefore, the correct answer option is A. 4sin⁡θ(1−2sin⁡2θ)1−sin⁡2θ4\sin \theta \left( 1-2{{\sin }^{2}}\theta \right)\sqrt{1-{{\sin }^{2}}\theta } .
So, the correct answer is “Option A”.

Note : In a right-angled triangle with length of the side opposite to angle θ as perpendicular (P), base (B) and hypotenuse (H):
sin⁡θ=PH\sin \theta =\dfrac{P}{H} , cos⁡θ=BH\cos \theta =\dfrac{B}{H} , tan⁡θ=PB\tan \theta =\dfrac{P}{B}
tan⁡θ=sin⁡θcos⁡θ\tan \theta =\dfrac{\sin \theta }{\cos \theta } , cot⁡θ=cos⁡θsin⁡θ\cot \theta =\dfrac{\cos \theta }{\sin \theta }
csc⁡θ=1sin⁡θ\csc \theta =\dfrac{1}{\sin \theta } , sec⁡θ=1cos⁡θ\sec \theta =\dfrac{1}{\cos \theta } , tan⁡θ=1cot⁡θ\tan \theta =\dfrac{1}{\cot \theta }
Using the Pythagoras' theorem:
sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1
tan⁡2θ+1=sec⁡2θ{{\tan }^{2}}\theta +1={{\sec }^{2}}\theta
1+cot⁡2θ=csc⁡2θ1+{{\cot }^{2}}\theta ={{\csc }^{2}}\theta
Sum-Product formula:
sin⁡2A+sin⁡2B=2sin⁡(A+B)cos⁡(A−B)\sin 2A+\sin 2B=2\sin (A+B)\cos (A-B)
sin⁡2A−sin⁡2B=2cos⁡(A+B)sin⁡(A−B)\sin 2A-\sin 2B=2\cos (A+B)\sin (A-B)
cos⁡2A+cos⁡2B=2cos⁡(A+B)cos⁡(A−B)\cos 2A+\cos 2B=2\cos (A+B)\cos (A-B)
cos⁡2A−cos⁡2B=−2sin⁡(A+B)sin⁡(A−B)\cos 2A-\cos 2B=-2\sin (A+B)\sin (A-B)