$x=\arcsin(-3/5), \; \sin x = -3/5$
**Drew a triangle to find $\cos x$
$\cos x = 4/5$
Now, I don't know what to do from here. I know I have to use a double angle formula, but when I evaluate the "$\frac{1}{2}\arcsin(-3/5)$," I get:
$\sin 2x = -3/5$
How do I continue using the double angle formula. $\sin2x = 2\sin x\cos x$ ?
Thanks for the help.
$\endgroup$2 Answers
$\begingroup$If $\arcsin\left(-\dfrac35\right)=y,-\dfrac\pi2<y<0\implies \cos y>0$ and $\cos\left[\dfrac12\arcsin\left(-\dfrac35\right)\right]=\cos\dfrac y2>0$
and $\sin y=-\dfrac35\implies\cos y=+\sqrt{1-\sin^2y}$
Now use $\cos y=2\cos^2\dfrac y2-1$
and we know $\cos\dfrac y2>0$
$\endgroup$ 2 $\begingroup$From Pythagorus theorem
$$ \cos 2 \theta = 4/5 $$
$$ 2 \cos^2 \theta = (1 + \cos 2\theta ) $$
$$ \cos\theta = \frac{3}{\sqrt{10}}$$
$\endgroup$
