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Question: If the trigonometric equation \(\sin 2A = 2\sin A\) is true then find the value of \(A\). (A) \({0...

If the trigonometric equation sin⁡2A=2sin⁡A\sin 2A = 2\sin A is true then find the value of AA.
(A) 0∘{0^ \circ }
(B) 30∘{30^ \circ }
(C) 45∘{45^ \circ }
(D) 60∘{60^ \circ }

Explanation

Solution

We know that from a trigonometric we have sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin \theta \cos \theta . Use this formula in the equation given in the question. Simplify it further and solve it to find the value of angle AA.

Complete step-by-step solution:
According to the question, we have been given a trigonometric equation and we have to find the value of an unknown angle.
The given trigonometric equation is:
⇒sin⁡2A=2sin⁡A\Rightarrow \sin 2A = 2\sin A.
For solving this equation we’ll apply the trigonometric formula sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin \theta \cos \theta on the left hand side of the equation. Doing this, we’ll get:
⇒2sin⁡Acos⁡A=2sin⁡A\Rightarrow 2\sin A\cos A = 2\sin A
We can cancel out 2 from both sides.
⇒sin⁡Acos⁡A=sin⁡A\Rightarrow \sin A\cos A = \sin A
Transferring all the terms on one side of the equation, we’ll get:
⇒sin⁡Acos⁡A−sin⁡A=0\Rightarrow \sin A\cos A - \sin A = 0
Now simplifying it further by factoring, we’ll get:
⇒sin⁡A(cos⁡A−1)=0\Rightarrow \sin A\left( {\cos A - 1} \right) = 0
For this equation to be true, both of its factors can be zero:

⇒sin⁡A=0 or cos⁡A−1=0 ⇒sin⁡A=0 or cos⁡A=1 ⇒A=0∘ or A=0∘ \Rightarrow \sin A = 0{\text{ or }}\cos A - 1 = 0 \\\ \Rightarrow \sin A = 0{\text{ or }}\cos A = 1 \\\ \Rightarrow A = {0^ \circ }{\text{ or }}A = {0^ \circ }

Thus the value of angle A is 0∘{0^ \circ } in both the cases.

(A) is the correct option.

Note: In the above problem, we have used a trigonometric formula for double angle which is:
⇒sin⁡2θ=2sin⁡θcos⁡θ\Rightarrow \sin 2\theta = 2\sin \theta \cos \theta
Another form of the formula of sin⁡2θ\sin 2\theta is:
⇒sin⁡2θ=2tan⁡θ1+tan⁡2θ\Rightarrow \sin 2\theta = \dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}
Some of the other widely used double angle trigonometric formulas are:
⇒cos⁡2θ=2cos⁡2θ−1 ⇒cos⁡2θ=1−2sin⁡2θ ⇒cos⁡2θ=cos⁡2θ−sin⁡2θ \Rightarrow \cos 2\theta = 2{\cos ^2}\theta - 1 \\\ \Rightarrow \cos 2\theta = 1 - 2{\sin ^2}\theta \\\ \Rightarrow \cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta
So we have the formula of cos⁡2θ\cos 2\theta in three different forms. We can use any of them as per the requirement of the question.
Similarly the formula for tan⁡2θ\tan 2\theta is:
⇒tan⁡2θ=2tan⁡θ1−tan⁡2θ\Rightarrow \tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}