Question

write the following expression as the sine, cosine, or tangent of a double - angle. then find the exact value of the expression.
\\( \frac { 2 \tan \frac { 5 \pi } { 12 } } { 1 - \tan ^ { 2 } \frac { 5 \pi } { 12 } } \\)
\\( \frac { 2 \tan \frac { 5 \pi } { 12 } } { 1 - \tan ^ { 2 } \frac { 5 \pi } { 12 } } = \tan \frac { 5 \pi } { 6 } \\)
the exact value of the expression is
(type an exact answer, using radicals as needed. rationalize the denominator.)

Explanation:

Step1: Recall the double - angle formula for tangent

The double - angle formula for tangent is \(\tan(2\alpha)=\frac{2\tan\alpha}{1 - \tan^{2}\alpha}\).
Let \(\alpha=\frac{5\pi}{12}\), then \(\frac{2\tan\frac{5\pi}{12}}{1-\tan^{2}\frac{5\pi}{12}}=\tan(2\times\frac{5\pi}{12})=\tan\frac{5\pi}{6}\).

Step2: Calculate the value of \(\tan\frac{5\pi}{6}\)

We know that \(\tan\theta=\frac{\sin\theta}{\cos\theta}\), and for \(\theta = \frac{5\pi}{6}\), \(\sin\frac{5\pi}{6}=\frac{1}{2}\), \(\cos\frac{5\pi}{6}=-\frac{\sqrt{3}}{2}\).
So \(\tan\frac{5\pi}{6}=\frac{\sin\frac{5\pi}{6}}{\cos\frac{5\pi}{6}}=\frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}}=-\frac{1}{\sqrt{3}}=-\frac{\sqrt{3}}{3}\).

Answer:

The exact value of the expression is \(-\frac{\sqrt{3}}{3}\)