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.