Find $\sin 2 x, \cos 2 x$, and $\tan 2 x$ if $\cos x=-\frac{8}{17}$ and $x$ terminates in quadrant III.
Solution
Step 1 ：We are given that \(\cos x = -\frac{8}{17}\) and that \(x\) terminates in quadrant III. In the third quadrant, both sine and cosine are negative. We also know that \(\cos^2 x + \sin^2 x = 1\). We can use this to find \(\sin x\).
Step 2 ：Using the Pythagorean identity \(\cos^2 x + \sin^2 x = 1\), we can solve for \(\sin x\) to get \(\sin x = -\sqrt{1 - \cos^2 x} = -\sqrt{1 - \left(-\frac{8}{17}\right)^2} = -\frac{15}{17}\).
Step 3 ：We can use the double angle formulas to find \(\sin 2x\), \(\cos 2x\), and \(\tan 2x\). The double angle formulas are: \(\sin 2x = 2 \sin x \cos x\), \(\cos 2x = \cos^2 x - \sin^2 x\), and \(\tan 2x = \frac{\sin 2x}{\cos 2x}\).
Step 4 ：Substituting \(\sin x = -\frac{15}{17}\) and \(\cos x = -\frac{8}{17}\) into the double angle formulas, we get \(\sin 2x = 2 \cdot -\frac{15}{17} \cdot -\frac{8}{17} = \frac{240}{289}\), \(\cos 2x = \left(-\frac{8}{17}\right)^2 - \left(-\frac{15}{17}\right)^2 = -\frac{161}{289}\), and \(\tan 2x = \frac{\sin 2x}{\cos 2x} = -\frac{240}{161}\).
Step 5 ：Final Answer: \(\sin 2x = \boxed{\frac{240}{289}}\), \(\cos 2x = \boxed{-\frac{161}{289}}\), and \(\tan 2x = \boxed{-\frac{240}{161}}\).
