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

Question

$$\cos2v = - \dfrac{1}{9}$$ 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.

Answer 1

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.

Answer 2

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.

Lacking an exact answer, the best we can do is the answer NoChance has already given.