Question

give the exact value of the expression without using a calculator.

\\( \sin \left( 2 \tan ^ { - 1 } \frac { 8 } { 15 } \
ight) \\)

\\( \sin \left( 2 \tan ^ { - 1 } \frac { 8 } { 15 } \
ight) = \square \\)
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Let \( \theta = \tan^{-1}\frac{8}{15} \), so \( \tan\theta=\frac{8}{15} \)

We can consider a right triangle where the opposite side to \( \theta \) is 8 and the adjacent side is 15.

Step2: Find \( \sin\theta \) and \( \cos\theta \)

First, find the hypotenuse \( r \) using the Pythagorean theorem: \( r = \sqrt{8^2 + 15^2}=\sqrt{64 + 225}=\sqrt{289} = 17 \)
Then, \( \sin\theta=\frac{8}{17} \) and \( \cos\theta=\frac{15}{17} \)

Step3: Use the double - angle formula for sine \( \sin(2\alpha)=2\sin\alpha\cos\alpha \)

Here, \( \alpha=\theta \), so \( \sin(2\tan^{-1}\frac{8}{15})=\sin(2\theta) \)
Substitute \( \sin\theta=\frac{8}{17} \) and \( \cos\theta=\frac{15}{17} \) into the double - angle formula:
\( \sin(2\theta)=2\times\frac{8}{17}\times\frac{15}{17}=\frac{240}{289} \)

Answer:

\( \frac{240}{289} \)