Question

The trigonometric function $ \sin 4\theta $ can be written as:
A. $ 4\sin \theta \left( 1-2{{\sin }^{2}}\theta \right)\sqrt{1-{{\sin }^{2}}\theta } $
B. $ 2\sin \theta \cos \theta {{\sin }^{2}}\theta $
C. $ 4\sin \theta -6{{\sin }^{3}}\theta $
D. $ 2\sin 2\theta $

Answer 1

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

Complete step-by-step answer:
We can write $ \sin 4\theta =\sin (2\theta +2\theta ) $ .
Using the identity $ \sin (A+B)=\sin A\cos B+\sin B\cos A $ , we get:
= $ \sin 2\theta \cos 2\theta +\sin 2\theta \cos 2\theta $
= $ 2\sin 2\theta \cos 2\theta $
Writing $ 2\theta $ as $ \theta + \theta $ ; we get
= $ 2\sin (\theta +\theta )\cos (\theta +\theta ) $
Using the $ \sin (A+B)=\sin A\cos B+\sin B\cos A $ and $ \cos (A+B)=\cos A\cos B-\sin A\sin B $ , we get:
= $ 2\left[ 2\sin \theta \cos \theta \left( {{\cos }^{2}}\theta -{{\sin }^{2}}\theta \right) \right] $
Which can be written as:
= $ 4\sin \theta \sqrt{{{\cos }^{2}}\theta }\left( {{\cos }^{2}}\theta -{{\sin }^{2}}\theta \right) $
Using the identity $ {{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta $ , we get:
= $ 4\sin \theta \sqrt{1-{{\sin }^{2}}\theta }\left( 1-{{\sin }^{2}}\theta -{{\sin }^{2}}\theta \right) $
= $ 4\sin \theta \left( 1-2{{\sin }^{2}}\theta \right)\sqrt{1-{{\sin }^{2}}\theta } $
Therefore, the correct answer option is A. $ 4\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 \theta =\dfrac{P}{H} $ , $ \cos \theta =\dfrac{B}{H} $ , $ \tan \theta =\dfrac{P}{B} $
  $ \tan \theta =\dfrac{\sin \theta }{\cos \theta } $ , $ \cot \theta =\dfrac{\cos \theta }{\sin \theta } $
  $ \csc \theta =\dfrac{1}{\sin \theta } $ , $ \sec \theta =\dfrac{1}{\cos \theta } $ , $ \tan \theta =\dfrac{1}{\cot \theta } $
Using the Pythagoras' theorem:
  $ {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 $
  $ {{\tan }^{2}}\theta +1={{\sec }^{2}}\theta $
  $ 1+{{\cot }^{2}}\theta ={{\csc }^{2}}\theta $
Sum-Product formula:
  $ \sin 2A+\sin 2B=2\sin (A+B)\cos (A-B) $
  $ \sin 2A-\sin 2B=2\cos (A+B)\sin (A-B) $
  $ \cos 2A+\cos 2B=2\cos (A+B)\cos (A-B) $
  $ \cos 2A-\cos 2B=-2\sin (A+B)\sin (A-B) $