"Use the given conditions to find the exact value of the expression."

$\sin(\alpha) = -\frac {5}{13}, \quad \tan(\alpha) > 0, \quad \sin(\alpha-\frac \pi 3)$

First off I'm guessing that "the expression" is the first one since it has an equals sign. The second formula implies that $\alpha$ is either in quadrant I or quadrant IV. The third formula I do not understand! How does it give me any relevant information? I am familiar with the addition and subtraction formulas for trig functions. I'm assuming they somehow apply here.

Also, $ \sin^-1(\frac {-5} {13}) $ yields -22.61986495 degrees which doesn't help with an exact answer. Without a good angle it's hard to find a good reference triangle.

Any insights are greatly appreciated!

  • $\begingroup$ The condition $\tan\alpha > 0$ implies that $\sin\alpha$ and $\cos\alpha$ have the same sign, so $\tan\alpha > 0$ implies that $\alpha$ is a first-quadrant or third-quadrant angle. $\endgroup$ – N. F. Taussig Oct 29 '15 at 11:58

Hint: use the angle sum formula $Sin(A-B)= SinA cosB-CosA SinB$

let's first solve for $cos(\alpha)$.

You can either draw out the triangle to find that the adjacent side is 12 ,or use the $sin^2 x + cos^2 x = 1$ identity. Either way, $cos(\alpha)=\frac{-12}{13}$

$sin(\alpha-\frac{\pi}{3}) = sin(\alpha)cos(\frac{\pi}{3})-cos(\alpha)sin( \frac{\pi}{3})= \frac{-5}{13}\frac{1}{2}-\frac{-12}{13}\frac{\sqrt3}{2}= \frac{12\sqrt{3}-5}{26}$

  • 1
    $\begingroup$ I think you meant the sin subtraction formula $\endgroup$ – gordlonious Oct 29 '15 at 3:17
  • $\begingroup$ yes thanks I edited my answer $\endgroup$ – More water plz Oct 29 '15 at 3:17

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