Question

Pre-Cal Help!! (Image Attached)

Answer 1

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 $\displaystyle{ \cos\theta= \frac{8}{17}$ (we make it positive since we are modeling using a triangle.)

Using the Pythagorean theorem, the length of the side opposite to angle $\theta$ is found to be 15 units.

We can see that the sine of theta is $\displaystyle{ \sin\theta= \frac{opposite \ side}{hypotenuse}= \frac{15}{17}$.

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:

$\displaystyle{ \sin\theta=- \frac{15}{17}$, $\displaystyle{ \cos\theta=- \frac{8}{17}$.

Recall the double-angle identities (formulas):

$\sin 2\theta=2\sin\theta\cos\theta$, $\cos2\theta=cos^2(\theta)-sin^2(\theta)$.

Using these identities and the ratios found above, we have:

$\displaystyle{ \cos2\theta=cos^2(\theta)-sin^2(\theta)=(- \frac{8}{17})^2-(- \frac{15}{17})^2=\frac{64}{289}-\frac{225}{289}=\frac{-161}{289}$

Similarly, applying the double angle formula for the sine we have:

$\displaystyle{ \sin 2\theta=2\sin\theta\cos\theta=2\cdot(- \frac{8}{17})(- \frac{15}{17})=\frac{240}{289}$

$\displaystyle{ \tan2\theta= \frac{\sin 2\theta}{\cos 2\theta}= \frac{\frac{240}{289}}{\frac{-161}{289}}= -\frac{240}{161}$


Answer:

$\displaystyle{\cos2\theta=\frac{-161}{289}$


$\displaystyle{ \tan2\theta= -\frac{240}{161}$