4

I'm having an exam tomorrow and stumbled over an old exam question that says:

$$\sin v = \frac{12}{13},\qquad \pi/2 < v <\pi, $$

What is $\sin 2v$?

Answer exactly!

I've been sitting with this for a few hours, also asking a friend, but we are both stuck. I understand that $\sin v$ is located in the second quadrant of the unity circle, fairly close to the y-axis, and $\sin 2v$ would then be located in the third quadrant. That is about as far as I get. I've tried by calculating $\cos v$ but it doesn't really help anything.

How do I get $v$ from knowing the that $y$ is $\frac{12}{13}$? I've been looking over the known part of internet trying to find a formula for this, but no luck. I'm not sure what to search for.

Remember that it must be an exact answer. I don't expect you to give me the full solution, but help in the right direction is highly appreciated!

Start wearing purple
  • 53,234
  • 13
  • 164
  • 223
Anders
  • 199
  • 3

4 Answers4

6

Do you know that $\sin(2v)=2\sin(v)\cos(v)$?

preferred_anon
  • 17,121
6

HINT: $\sin 2v=2\sin v\cos v$, and $\cos^2v=1-\sin^2v$.

Brian M. Scott
  • 616,228
3

Two useful identities for this:

$\sin(2x) = 2\sin(x)\cos(x)$.

$\sin^2(x) + \cos^2(x) = 1$.

Begin by using the second identity. Since $\sin^2(v) = \frac{144}{169}$, it must be the case that $\cos^2(v) = \frac{25}{169}$, so, since $v$ is in the second quadrant, where $\cos$ is negative, $\cos(v) = \frac{-5}{13}$. Now, using the first identity, $\sin(2v) = 2\frac{12}{13}\frac{-5}{13} = \frac{-120}{169}$.

qaphla
  • 3,890
3

Three points:

  • $\cos^2 v+\sin^2v=1$
  • $\cos v<0$
  • $\sin(2v)=2\sin (v)\cos(v)$