# If $\cos2v = - \frac{1}{9}$ and $v$ is acute, then determine the value of $v$

$$\cos2v = - \dfrac{1}{9}$$ the angle $$v$$ is acute.

Determine the value of $$v$$.

I have tried using the double angle identity $$\cos2v = \cos^2v - \sin^2v$$ but I get a very complicated answer. I am almost certain that there must be a simpler way to solve this, since this question is labeled as "easy" in my textbook.

• Sure, $\frac12\arccos(-\frac19)\approx 0.8410686705679302557765250318264307467$ – Hagen von Eitzen Sep 3 at 16:38
• Is $v = \arcsin x$ or something? – AgentS Sep 3 at 16:42
• I've used $$cos2v = cos^2v - sin^2v$$ but I get that $$cos^2v = \frac{8}{18}$$ which I cant simplify.. – Synchrowave Sep 3 at 16:52
• @Synchrowave put the content of your last comment into your question, as your attempt. Otherwise will be the question closed. As for the cosine, 8/18=4/9 and you know that the angle is acute. Hence $\cos v=2/3.$ – user376343 Sep 3 at 17:02
• Also, fix your question - are you asking for $v$ or for $\cos v$? – user376343 Sep 3 at 17:04

As you say, we find $$\cos^2 v = \frac{8}{18} = \frac{4}{9}$$ Hence $$\cos v = \pm \frac{2}{3}$$ To my (read: Google's) knowledge, no exact answer for $$v$$ that doesn't make use of $$\cos^{-1}$$ is known. You can see here for a fairly extensive list of known exact trig and inverse trig values.

Hint

Solving: $$\cos(2x) = - \dfrac{1}{9}$$

For the general case, let the R.H.S constant be $$k$$. We'll use Radians.

You could apply $$arccos$$ function.

The definition of $$arccos(x)$$ is: $$arccos(x)=\pm cos^{-1}(x)+2\pi n$$

and solve 2 equations for $$x$$ as follows:

$$2x=\cos^{-1} \left(k\right)+2\pi n \Rightarrow x=\frac{\cos^{-1}\left(k\right)+2\pi n}{2} \tag1$$ $$2x=-\cos^{-1} \left(k\right)+2\pi n \Rightarrow x=\frac{-cos^{-1}\left(k\right)+2\pi n}{2} \tag2$$

You may now use the fact the desired angle is acute (An Acute Angle is less than 90°).

Ref: Precalculs with Trig.