# Math Help - How do I simplify inverse functions trig?

1. ## How do I simplify inverse functions trig?

Okay, i don't really know where to start
"simplify completely. your answers should not contain trigonometric or inverse trigonometric functions
$sin(Arcsin2/3+Arccos\sqrt3/2)$
So I think I should solve for $Arcsin2/3$ and $Arcos\sqrt3/2$, then add them to get one number and take the sin of that number.
Would that be the steps?
however, I don't know how to solve for $Arcsin2/3$ and $Arccos\sqrt3/2$

2. Hello dorkymichelle
Originally Posted by dorkymichelle
Okay, i don't really know where to start
"simplify completely. your answers should not contain trigonometric or inverse trigonometric functions
$sin(Arcsin2/3+Arccos\sqrt3/2)$
So I think I should solve for $Arcsin2/3$ and $Arcos\sqrt3/2$, then add them to get one number and take the sin of that number.
Would that be the steps?
however, I don't know how to solve for $Arcsin2/3$ and $Arccos\sqrt3/2$
Get rid of inverse trig functions as quickly as you can - they're not nice! Use new letters to stand for them. So, let:
$x =\arcsin \tfrac23 \Rightarrow \sin x = \tfrac 23$
and
$y = \arccos\tfrac{\sqrt3}{2} \Rightarrow \cos y = \tfrac{\sqrt3}{2}$

Now, we want $\sin (x+y)$, so that's $\sin x \cos y + \cos x \sin y$.

We know $\sin x$ and $\cos y$. Can you work out $\cos x$ and $\sin y$, using Pythagoras? When you've done that, plug all the values in, and you're there.

I make the answer $\frac{2\sqrt3 + \sqrt5}{6}$. Do you agree?