I have a question. How would I prove the following.

sin((2 arcsin(4/5)-arccos(12/13))=323/325

How would I solve this I have an idea I know


But I am not sure what to do.

  • 1
    $\begingroup$ Why not compute $\arcsin(4/5)$ and $\arccos(12/13)$, then substitute those values into the given expression? $\endgroup$ – JohnD Dec 14 '12 at 3:15
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    $\begingroup$ that's awfully similar to your earlier post. Does anything you learned from that post help here? math.stackexchange.com/q/258289/9003 $\endgroup$ – amWhy Dec 14 '12 at 3:17
  • $\begingroup$ There is no statement, and therefore nothing to "prove". Perhaps you mean that you want a way to rewrite it in a simpler form that no longer has trig and inverse trig functions? $\endgroup$ – Jonas Meyer Dec 14 '12 at 3:18
  • $\begingroup$ I mean how would I make it equal 323/325 I do not know how to do it. $\endgroup$ – Fernando Martinez Dec 14 '12 at 3:24
  • $\begingroup$ I know in the last problem someone offered me a formula to help solve the problem, but I am not sure what I would use here. $\endgroup$ – Fernando Martinez Dec 14 '12 at 3:25

Let $\arccos \frac{12}{13}=B$ and $\arcsin\frac45 =A$

$\displaystyle\implies 0\le B\le \pi$ and $\displaystyle-\frac\pi2\le A\le\frac\pi2$ (using the definition of principal value)

$\displaystyle\implies \sin B\ge0$ and $\displaystyle\cos A\ge0$

$\displaystyle\cos B=\frac{12}{13}\implies \sin B=+\sqrt{1-\left(\frac{12}{13}\right)^2}=+\frac5{13}$

and $\displaystyle\sin A=\frac45\implies \cos A=+\sqrt{1-\left(\frac45\right)^2}=+\frac35$

So, $$\sin(2A-B)=\sin 2A\cos B-\cos 2A\sin B$$ $$=2\sin A\cos A\cos B-(1-2\sin^2A)\sin B$$ $$=2\cdot\frac45\cdot \frac35\cdot\frac{12}{13}-\{1-2\left(\frac45 \right)^2\}\cdot \frac5{13} $$ $$=\frac{288-(-7)5}{13\cdot25}=\frac{323}{325}$$


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