Question

given that \\( \cos \theta = \frac { 60 } { 61 } \\) and \\( \sin \theta < 0 \\), determine the values of the sine and cosine functions for \\( 2 \theta \\) \\( \sin 2 \theta = \square \\) (type an integer or a simplified fraction.)

Explanation:

Step1: Find the value of \(\sin\theta\)

Use the identity \(\sin^{2}\theta+\cos^{2}\theta = 1\).
Given \(\cos\theta=\frac{60}{61}\), then \(\sin^{2}\theta=1-\cos^{2}\theta=1 - (\frac{60}{61})^{2}=\frac{61^{2}-60^{2}}{61^{2}}=\frac{(61 - 60)(61 + 60)}{61^{2}}=\frac{1\times121}{61^{2}}\).
Since \(\sin\theta<0\), \(\sin\theta=-\frac{11}{61}\).

Step2: Use the double - angle formula for sine

The double - angle formula for sine is \(\sin2\theta = 2\sin\theta\cos\theta\).
Substitute \(\sin\theta=-\frac{11}{61}\) and \(\cos\theta=\frac{60}{61}\) into the formula:
\(\sin2\theta=2\times(-\frac{11}{61})\times\frac{60}{61}=-\frac{1320}{3721}\).

Answer:

\(-\frac{1320}{3721}\)