0
$\begingroup$

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

$\endgroup$
  • 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
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.