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
sin(a-b)=sin(a)cos(b)-cos(a)sin(b)
But I am not sure what to do.
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
sin(a-b)=sin(a)cos(b)-cos(a)sin(b)
But I am not sure what to do.
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} $ $=\frac{288-(-7)5}{13\cdot25}=\frac{323}{325}$