# Evaluate cos[(1/2)[arcsin(-3/5)]]. I'm not sure what i'm doing wrong.

$x=\arcsin(-3/5), \; \sin x = -3/5$

**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.

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 $\cos\dfrac y2>0$

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From Pythagorus theorem

$$\cos 2 \theta = 4/5$$

$$2 \cos^2 \theta = (1 + \cos 2\theta )$$

$$\cos\theta = \frac{3}{\sqrt{10}}$$