Question

If $A = {30^ \circ }$, verify that $\sin 2A = 2\sin A\cos A$.

Answer 1

Hint:
For solving this question we will substitute the value of $A = {30^ \circ }$ in $\sin 2A$ and $2\sin A\cos A$ , we will then verify both the sides of the identities that the left-hand side and right-hand side should be equal. On comparing both the equations we will conclude. And in this way, we will solve this question.

Complete step by step solution:
So we have the identities given as $\sin 2A = 2\sin A\cos A$
Now taking the left-hand side of it and substituting the values in it, we get
$ \Rightarrow \operatorname{Sin} 2A$
Substituting the value, $A = {30^ \circ }$ , we get
$ \Rightarrow \sin 2({30^ \circ })$
On multiplying, we get
$ \Rightarrow \sin ({60^ \circ })$
As we know that $sin({60^ \circ }) = \dfrac{{\sqrt 3 }}{2}$
Therefore, we get
$ \Rightarrow \dfrac{{\sqrt 3 }}{2}$ , and we will name it equation $1$
Let us now substitute the value of $A = {30^ \circ }$ in $2\sin A\cos A$
So we get
$ \Rightarrow 2\sin ({30^ \circ })\cos ({30^ \circ })$
As we know that $sin({30^ \circ }) = \dfrac{1}{2}\& \cos ({30^ \circ }) = \dfrac{{\sqrt 3 }}{2}$
So on putting these values in the right-hand side of the equation, we get
$ \Rightarrow 2\left( {\dfrac{1}{2}} \right)\left( {\dfrac{{\sqrt 3 }}{2}} \right)$
And on solving the above equation by doing the multiplication one by one, we get
$ \Rightarrow \dfrac{{\sqrt 3 }}{2}$ , and we will name it equation $2$
Here, we can see that the value of both the equations is the same. Hence we can see that the left-hand side is equal to the right and side.

Therefore, trigonometric identity $\sin 2A = 2\sin A\cos A$ is verified.

Note:
This type of question becomes very easy if we know the values of it. So for proving the trigonometric functions, trigonometric values must be remembered. And also for proving this type of question, we should always take both the left-hand side and the right-hand side one by one and find the values for them and then compare, it will be easier when we do like this.