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

  • 1
    $\begingroup$ Sure, $\frac12\arccos(-\frac19)\approx 0.8410686705679302557765250318264307467$ $\endgroup$ – Hagen von Eitzen Sep 3 at 16:38
  • $\begingroup$ Is $v = \arcsin x$ or something? $\endgroup$ – AgentS Sep 3 at 16:42
  • $\begingroup$ I've used $$ cos2v = cos^2v - sin^2v $$ but I get that $$ cos^2v = \frac{8}{18} $$ which I cant simplify.. $\endgroup$ – Synchrowave Sep 3 at 16:52
  • $\begingroup$ @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 $\cos v=2/3.$ $\endgroup$ – user376343 Sep 3 at 17:02
  • $\begingroup$ Also, fix your question - are you asking for $v$ or for $\cos v$? $\endgroup$ – 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.

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



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.


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