Question

use identities to find values of the sine and cosine functions of the function for the angle measure.
2x, given tan x = - 2 and cos x < 0
cos 2x = - \frac{3}{5}
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression )
sin 2x = \square
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression )

Explanation:

Step1: Determine the quadrant of \(x\)

Since \(\tan x=- 2<0\) (tangent is negative) and \(\cos x < 0\) (cosine is negative), \(x\) is in the second quadrant.

Step2: Use the identity \(\tan x=\frac{\sin x}{\cos x}=-2\), so \(\sin x=- 2\cos x\)

Also, from the Pythagorean identity \(\sin^{2}x+\cos^{2}x = 1\)
Substitute \(\sin x=-2\cos x\) into \(\sin^{2}x+\cos^{2}x = 1\):
\((-2\cos x)^{2}+\cos^{2}x=1\)
\(4\cos^{2}x+\cos^{2}x = 1\)
\(5\cos^{2}x=1\)
\(\cos^{2}x=\frac{1}{5}\)
Since \(x\) is in the second quadrant, \(\cos x=-\frac{1}{\sqrt{5}}=-\frac{\sqrt{5}}{5}\)
Then \(\sin x=-2\cos x=-2\times(-\frac{\sqrt{5}}{5})=\frac{2\sqrt{5}}{5}\)

Step3: Use the double - angle formula for sine \(\sin2x = 2\sin x\cos x\)

Substitute \(\sin x=\frac{2\sqrt{5}}{5}\) and \(\cos x=-\frac{\sqrt{5}}{5}\) into \(\sin2x = 2\sin x\cos x\)
\(\sin2x=2\times\frac{2\sqrt{5}}{5}\times(-\frac{\sqrt{5}}{5})\)
\(\sin2x=2\times\frac{-2\times5}{25}\)
\(\sin2x=-\frac{4}{5}\)

Answer:

\(\sin2x =-\frac{4}{5}\)