Question

find \\( \sin ( 2 a ) \\), if \\( \sin ( a ) = - \frac { 33 } { 65 } \\), and \\( a \\) is in quadrant 3.

Explanation:

Step1: Find $\cos(A)$

Use the identity $\sin^{2}(A)+\cos^{2}(A) = 1$.
Given $\sin(A)=-\frac{33}{65}$, then $\cos^{2}(A)=1-\sin^{2}(A)=1 - (-\frac{33}{65})^{2}=1-\frac{1089}{4225}=\frac{4225 - 1089}{4225}=\frac{3136}{4225}$.
Since $A$ is in quadrant 3, $\cos(A)<0$. So $\cos(A)=-\sqrt{\frac{3136}{4225}}=-\frac{56}{65}$.

Step2: Use the double - angle formula for sine

The double - angle formula for sine is $\sin(2A)=2\sin(A)\cos(A)$.
Substitute $\sin(A)=-\frac{33}{65}$ and $\cos(A)=-\frac{56}{65}$ into the formula:
$\sin(2A)=2\times(-\frac{33}{65})\times(-\frac{56}{65})=\frac{2\times33\times56}{65\times65}=\frac{3696}{4225}$.

Answer:

$\frac{3696}{4225}$