Q2(iii):

Choose the correct option and justify your choice : (iii) $\sin 2A = 2 \sin A$ is true when $A =$

Given: The trigonometric equation $\sin 2A = 2 \sin A$.

To find: The value of $A$ for which the given equation holds true, choosing from the standard options usually provided in this context: (A) $0^\circ$, (B) $30^\circ$, (C) $45^\circ$, (D) $60^\circ$.

Visual Representation:

Step 1: Testing Option (A) where $A = 0^\circ$

Substitute $A = 0^\circ$ into the Left Hand Side (LHS) of the equation:

LHS $= \sin 2A = \sin(2 \times 0^\circ) = \sin 0^\circ$

[Since the value of $\sin 0^\circ = 0$ from trigonometric ratio tables]

LHS $= 0$

Now, substitute $A = 0^\circ$ into the Right Hand Side (RHS) of the equation:

RHS $= 2 \sin A = 2 \sin 0^\circ$

[Since $\sin 0^\circ = 0$]

RHS $= 2 \times 0 = 0$

Since LHS = RHS, the equation is true for $A = 0^\circ$.

Step 2: Testing Option (B) where $A = 30^\circ$

LHS $= \sin 2(30^\circ) = \sin 60^\circ$

[Using the standard value $\sin 60^\circ = \frac{\sqrt{3}}{2}$]

LHS $= \frac{\sqrt{3}}{2}$

RHS $= 2 \sin 30^\circ$

[Using the standard value $\sin 30^\circ = \frac{1}{2}$]

RHS $= 2 \times \frac{1}{2} = 1$

Since $\frac{\sqrt{3}}{2} \neq 1$, the equation is false for $A = 30^\circ$.

Step 3: Testing Option (C) where $A = 45^\circ$

LHS $= \sin 2(45^\circ) = \sin 90^\circ$

[Since $\sin 90^\circ = 1$]

LHS $= 1$

RHS $= 2 \sin 45^\circ = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2}$

Since $1 \neq \sqrt{2}$, the equation is false for $A = 45^\circ$.

Step 4: Testing Option (D) where $A = 60^\circ$

LHS $= \sin 2(60^\circ) = \sin 120^\circ$

[Using the identity $\sin(180^\circ - \theta) = \sin \theta$, $\sin 120^\circ = \sin 60^\circ = \frac{\sqrt{3}}{2}$]

LHS $= \frac{\sqrt{3}}{2}$

RHS $= 2 \sin 60^\circ = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}$

Since $\frac{\sqrt{3}}{2} \neq \sqrt{3}$, the equation is false for $A = 60^\circ$.

Conclusion: Comparing the results, the equation $\sin 2A = 2 \sin A$ holds true only when $A = 0^\circ$.

Final Answer: The correct option is (A) $0^\circ$.