Question

How do you prove $\sin (2x) = 2\sin (x)\cos (x)$ using other trigonometric identities?

Answer 1

Hint: The given question asks us to prove the given identity using other identities. We know that above given identities is true in the sense that it is taught itself as an identity we have to just prove it by using basic identities. The identities we will use to solve this type of question should contain both the $\sin $ and $\cos $ components and should also have an addition of angles that will help you resolve the $2x$ present on the left hand side of the equation . We will use the identity written below for this purpose, $\sin (x + y) = \sin x\cos y + \cos x\sin y$
This identity above has two components $x$ and $y$ but we need $2x$ so we will take the $y$ portion equal to $x$ as well, and solve accordingly.

Complete step-by-step solution:
The given question asks us to prove the identity,
$\Rightarrow \sin (2x) = 2\sin (x)\cos (x)$
We will use the identity given below to prove this question
$\Rightarrow \sin (x + y) = \sin x\cos y + \cos x\sin y$
This identity above has two components $x$ and $y$ but we need $2x$ so we will take the $y$ portion equal to $x$ as well, and solve accordingly.
Solving the left hand side of the equation first,
$\Rightarrow LHS = \sin (2x)$
From above identity we write,
$\Rightarrow \sin (2x) = \sin x\cos x + \cos x\sin x$
Which upon solving becomes,
$\Rightarrow \sin (2x) = 2\sin x\cos x$
So $LHS = 2\sin x\cos x$
And $
\Rightarrow RHS = 2\sin x\cos x \\
    \\
$
Thus $LHS = RHS$
Hence the equation is proved.

Note: Whenever solving this type of question where angle are in double or triple form always use the above identity also in case the cosine version is asked with the compound angles (double or triple or addition of two angles) then we solve the question by the below given identity,
$\cos (x + y) = \cos x\cos y - \sin x\sin y$