Question

For what value of A, the following is true:
sin2A=2sinA
$
  {\text{A}}{\text{. 0}} \\
  {\text{B}}{\text{. 1 }} \\
  {\text{C }}{\text{. 4}} \\
  {\text{D}}{\text{. - 1}} \\
 $

Answer 1

Hint: To solve this first we have to put ${\text{B}}$ equal to ${\text{A}}$ in trigonometric formula of ${\text{sin(A + B) = sinAcosB + cosAsinB}}$. And then simplify the obtained equation.

Complete step-by-step answer:
From trigonometric formulas,
We know that
$
  \sin ({\text{A + B) = }}\sin {\text{A}}\cos {\text{B + }}\cos {\text{A}}\sin {\text{B}} \\
$
Now, replace B by A, then
$
   \Rightarrow \sin (2{\text{A) = }}\sin {\text{A}}\cos {\text{A + }}\sin {\text{A}}\cos {\text{A }} \\
   \Rightarrow \sin {\text{2A = 2}}\sin {\text{A}}\cos {\text{A ---- eq}}{\text{.1}} \\
$

Now it is given that
$\sin {\text{2A = 2sinA --- eq 2}}{\text{.}}$

Equate eq1. and eq.2, we get

$ 2\sin {\text{A}}\cos {\text{A = 2sinA}} \\
   \Rightarrow 2\sin {\text{A}}\cos {\text{A - 2sinA = 0}} \\
   \Rightarrow \sin {\text{A}}(\cos {\text{A}} - 1){\text{ = 0}} \\
$
From above equation it possible that either sinA = 0 or cosA = 1
On solving these two situations separately, we get
$
   \Rightarrow {\text{sinA = 0}} \\
   \Rightarrow {\text{ A = n}}\pi {\text{ \{ }}\ {\text{ n = 0,1,2,3}}...........{\text{\} }} \\
   \Rightarrow {\text{ A = 0 or }}\pi ........{\text{ eq}}{\text{.3}} \\
$
And,
$
   \Rightarrow \cos {\text{A = 1}} \\
   \Rightarrow {\text{ A = 2n}}\pi {\text{ \{ }}\ n = 0,1,2......\} \\
   \Rightarrow {\text{ A = 0 or 2}}\pi ......{\text{ eq}}{\text{.4}} \\
$
From eq. 3 and eq.4, we get,
$
 \Rightarrow {\text{ A = 0}} \\
$
Hence, option A is correct.

Note: Whenever you get this type of question the key concept of solving is you have to use the trigonometric formula of ${\text{sin(A + B) = sinAcosB + cosAsinB}}$. And then put A, B as according to the question. After that, resolve the equation into simplest forms and get their general solutions.