Question

Simplify $ \sin \left( {4\theta } \right) $ to trigonometric functions of unit $ \theta $

Answer 1

Hint: The given problem can be solved by using the double angle formulae of sine and cosine. Use of double angle formulae help us to convert the $ \sin \left( {4\theta } \right) $ to trigonometric functions of unit $ \left( {2\theta } \right) $ and then to trigonometric functions of unit $ \theta $ . Double angle formulae for sine and cosine are: $ \sin \left( {2x} \right) = 2\sin (x)\cos (x) $ and \[\cos \left( {2x} \right) = \left( {1 - 2{{\sin }^2}x} \right) = \left( {2{{\cos }^2}x - 1} \right)\]

Complete step-by-step answer:
For simplifying $ \sin \left( {4\theta } \right) $ to trigonometric functions of unit $ \theta $ , we first use double angle formulae of sine to convert $ \sin \left( {4\theta } \right) $ to trigonometric functions of unit $ \left( {2\theta } \right) $ .
Using $ \sin \left( {2x} \right) = 2\sin (x)\cos (x) $ in the given problem, we get,
 $ \sin \left( {4\theta } \right) $ $ = $ \[\sin \left\{ {2\left( {2\theta } \right)} \right\}\]
 \[ = \] $ 2 $ $ \sin \left( {2\theta } \right) $ $ \cos \left( {2\theta } \right) $ $ $
Now, we have to convert trigonometric functions of unit $ \left( {2\theta } \right) $ into trigonometric functions of unit $ \theta $ by using the double angle formulae.
Again using $ \sin \left( {2x} \right) = 2\sin (x)\cos (x) $
 \[ = \] $ 2\left( {2\sin \theta \cos \theta } \right) $ $ \cos \left( {2\theta } \right) $
Now, we have to use a double angle formula for cosine to convert $ \cos \left( {2\theta } \right) $ into trigonometric functions of unit $ \theta $ .
Using \[\cos \left( {2x} \right) = \left( {1 - 2{{\sin }^2}x} \right)\] ,
 \[ = \] $ 2\left( {2\sin \theta \cos \theta } \right)\left( {1 - 2{{\sin }^2}\theta } \right) $
 \[ = \] $ 2\left( {2\sin \theta \cos \theta - 4{{\sin }^3}\theta \cos \theta } \right) $
On simplifying further, we get,
 \[ = \] $ 4\sin \theta \cos \theta - 8{\sin ^3}\theta \cos \theta $
Hence, $ \sin \left( {4\theta } \right) $ in terms of trigonometric functions of unit $ \theta $ is $ \left( {4\sin \theta \cos \theta - 8{{\sin }^3}\theta \cos \theta } \right) $ .
So, the correct answer is “$ \left( {4\sin \theta \cos \theta - 8{{\sin }^3}\theta \cos \theta } \right) $”.

Note: The above question can also be solved by using compound angle formulae instead of double angle formulae such as $ \sin (A + B) = \left( {\sin A\cos B + \cos A\sin B} \right) $ and $ \cos (A + B) = \left( {\cos A\cos B - \sin A\sin B} \right) $ . This method can also be used to get to the answer of the given problem.