1. ## Double Angle Forumla

The angle 2x lies in the fourth quadrant such that cos2x=8/17.

2. Determine an exact value for cosx
3. What is the measure of x in radians?

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I know that quadrant 4 has 2x in it, so quadrant _____ has to have x ?
for part 2, the exact measure of cosx would, it be 4/8.5??? I divided 8 by two and 17 by two.. I dont know if it is right. Check?

For part 3, the measure of x, would I have to take cos^-1(4/8.5) [ if part b is right] to get the measure of x? Thanks!

2. Originally Posted by skeske1234
2. Determine an exact value for cosx
for part 2, the exact measure of cosx would, it be 4/8.5??? I divided 8 by two and 17 by two.. I dont know if it is right. Check?
No, $\displaystyle \cos{2x}$ is not the same as $\displaystyle 2\cos{x}$

The double angle formula states that $\displaystyle \cos{2x} = 2\cos^2{x} - 1$.

Therefore $\displaystyle 2\cos^2{x} - 1 = \frac{8}{17}$ , can you work this one out now?

Originally Posted by skeske1234
For part 3, the measure of x, would I have to take cos^-1(4/8.5) [ if part b is right] to get the measure of x?
No I'm afraid this is incorrect.

Let me give you an example, if $\displaystyle \cos{nx} = q$ for example where $\displaystyle n$ and $\displaystyle q$ are just numbers, then $\displaystyle nx = \cos^{-1}{q}$, therefore $\displaystyle x = \frac{\cos^{-1}{q}}{n}$.

Can you work this one through?

3. As for the first part I'm afraid I'm unsure about the quadrant method.