Question

Given that $\sin 2\alpha = \frac{\sqrt{3}}{2}$, the value of $\sin 3\alpha$ is :

• A. $\frac{3\sqrt{3}}{4}$
• B. $\frac{1}{2}$
• C. $1$
• D. $\frac{\sqrt{3}}{4}$

Hint:

Always keep standard trigonometric tables memorized:
$\sin 30^\circ = \frac{1}{2}$, $\sin 45^\circ = \frac{1}{\sqrt{2}}$, $\sin 60^\circ = \frac{\sqrt{3}}{2}$, and $\sin 90^\circ = 1$.
Knowing these standard values allows you to quickly work back and forth between angles and their trigonometric ratios.

Answer 1

The Correct Option is C

Step 1: Find the angle alpha the same way, since we need its value regardless of method.
We are given $\sin 2\alpha = \frac{\sqrt{3}}{2}$. Since $\sin 60^\circ = \frac{\sqrt{3}}{2}$:
\[ 2\alpha = 60^\circ \implies \alpha = 30^\circ \]

Step 2: Instead of directly substituting into $3\alpha$ and reading off the standard value, use the triple-angle identity for sine.
Recall the triple angle formula:
\[ \sin 3\alpha = 3\sin\alpha - 4\sin^3\alpha \]

Step 3: Find sin(alpha) first.
Since $\alpha = 30^\circ$:
\[ \sin\alpha = \sin 30^\circ = \frac{1}{2} \]

Step 4: Substitute this value into the triple-angle formula and compute term by term.
\[ \sin 3\alpha = 3\left(\frac{1}{2}\right) - 4\left(\frac{1}{2}\right)^3 \]
\[ \sin 3\alpha = \frac{3}{2} - 4 \times \frac{1}{8} \]
\[ \sin 3\alpha = \frac{3}{2} - \frac{1}{2} \]
\[ \sin 3\alpha = 1 \]

Final Answer:
The value of $\sin 3\alpha$ is $1$, which matches Option (C). \[ \boxed{\sin 3\alpha = 1} \]