Question

What is a formula of
$\left( 1 \right){\text{ sin 4}}\theta$
$\left( 2 \right){\text{ cos 4}}\theta$

Answer 1

Hint : For solving the first part of the question, we will assume $u = 2\theta$ and then will solve it to expand the identity by using the formula $\sin 2u = 2\sin u\cos u$ . And solving furthermore, we will get the formula for it. For the second question, in the same way for this, we will expand the identity, and the formula used will be $\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta$ , we will get the solution for it.
Formula used:
Trigonometric identities used are
$\sin 2u = 2\sin u\cos u$
$\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta$
${\cos ^2}\theta + {\sin ^2}\theta = 1$
${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$
Here,
$\theta ,u$ , will be the angle

Complete step-by-step answer :
$\left( 1 \right){\text{ sin 4}}\theta$
Let us assume that $u = 2\theta$ , then we have the identity as
$\Rightarrow \sin 4\theta = \sin 2u$
Since we know that $\sin 2u = 2\sin u\cos u$
Therefore, on substituting the value $u = 2\theta$ , we get
$\Rightarrow \sin \left( {2 \cdot 2\theta } \right) = 2\sin 2\theta \cos 2\theta$
And on solving it we get
$\Rightarrow \sin \left( {4\theta } \right) = 2\sin 2\theta \cos 2\theta$
And we also know that $\sin 2u = 2\sin u\cos u$
So the identity will be
$\Rightarrow \sin \left( {4\theta } \right) = 4\sin \theta \cos \theta \cos 2\theta$
Since to get the formula we have to remove that $\cos 2\theta$
As from the formula of double angle sum, we have
$\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta$
So by using the above formula we have the identity as
$\Rightarrow \sin \left( {4\theta } \right) = 4\sin \theta \cos \theta \left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right)$
And on solving it we get
$\Rightarrow \sin \left( {4\theta } \right) = 4\sin \theta {\cos ^3}\theta - 4{\sin ^3}\theta \cos \theta$
Hence the formula will be $\sin \left( {4\theta } \right) = 4\sin \theta {\cos ^3}\theta - 4{\sin ^3}\theta \cos \theta$
So, the correct answer is “ $\sin \left( {4\theta } \right) = 4\sin \theta {\cos ^3}\theta - 4{\sin ^3}\theta \cos \theta$ ”.

$\left( 2 \right){\text{ cos 4}}\theta$
As we know that $\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta$ and also we know ${\cos ^2}\theta + {\sin ^2}\theta = 1$
So from the above formula, and on solving it, we get
$\Rightarrow \cos 2\theta = 2{\cos ^2}\theta - 1$
Therefore, from the above
$\Rightarrow \cos 4\theta = 2{\cos ^2}2\theta - 1$
And on expanding it by substituting the value we have got just above, we get
$\Rightarrow \cos 4\theta = 2{\left( {2{{\cos }^2}\theta - 1} \right)^2} - 1$
And by using the formula, ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$ we get
$\Rightarrow \cos 4\theta = 2\left( {4{{\cos }^4}\theta - 4{{\cos }^2}\theta + 1} \right) - 1$
And on solving the above solution, we get
$\Rightarrow \cos 4\theta = 8{\cos ^4}\theta - 8{\cos ^2}\theta + 1$
Hence the formula will be $\cos 4\theta = 8{\cos ^4}\theta - 8{\cos ^2}\theta + 1$
So, the correct answer is “ $\cos 4\theta = 8{\cos ^4}\theta - 8{\cos ^2}\theta + 1$ ”.

Note : This type of question can be solved easily if we know how to morph the trigonometric identities by using the formula. So we should have to memorize the formula to solve such problems. Also, the second problem can be solved by using the concept of de Moivre’s theorem, which is ${\left( {\cos \theta + i\sin \theta } \right)^n} = \cos n\theta + i\sin n\theta$ and from this substituting it in the identity and equating the reals parts we will get the same.