To solve this type of questions, a practical way is to think of the 1.quadrant situation, that is, model the situation in right triangle trigonometry.
Check the picture.
For the moment let [tex]\displaystyle{ \cos\theta= \frac{8}{17} [/tex] (we make it positive since we are modeling using a triangle.)
Using the Pythagorean theorem, the length of the side opposite to angle [tex]\theta[/tex] is found to be 15 units.
We can see that the sine of theta is [tex]\displaystyle{ \sin\theta= \frac{opposite \ side}{hypotenuse}= \frac{15}{17} [/tex].
In the third quadrant, both sine and cosine of an angle are negative, so the actual values of the sine and cosine of theta are, eventually:
[tex]\displaystyle{ \sin\theta=- \frac{15}{17} [/tex], [tex]\displaystyle{ \cos\theta=- \frac{8}{17} [/tex].
Recall the double-angle identities (formulas):
[tex]\sin 2\theta=2\sin\theta\cos\theta[/tex], [tex]\cos2\theta=cos^2(\theta)-sin^2(\theta)[/tex].
Using these identities and the ratios found above, we have:
[tex]\displaystyle{ \cos2\theta=cos^2(\theta)-sin^2(\theta)=(- \frac{8}{17})^2-(- \frac{15}{17})^2=\frac{64}{289}-\frac{225}{289}=\frac{-161}{289}[/tex]
Similarly, applying the double angle formula for the sine we have:
[tex]\displaystyle{ \sin 2\theta=2\sin\theta\cos\theta=2\cdot(- \frac{8}{17})(- \frac{15}{17})=\frac{240}{289}[/tex]
[tex]\displaystyle{ \tan2\theta= \frac{\sin 2\theta}{\cos 2\theta}= \frac{\frac{240}{289}}{\frac{-161}{289}}= -\frac{240}{161} [/tex]
Answer:
[tex]\displaystyle{\cos2\theta=\frac{-161}{289}[/tex]
[tex]\displaystyle{ \tan2\theta= -\frac{240}{161}[/tex]
