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

$\displaystyle sin(Arcsin2/3+Arccos\sqrt3/2)$

So I think I should solve for $\displaystyle Arcsin2/3$ and $\displaystyle Arcos\sqrt3/2$, 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 $\displaystyle Arcsin2/3$ and $\displaystyle Arccos\sqrt3/2$

Get rid of inverse trig functions as quickly as you can - they're not nice! Use new letters to stand for them. So, let:$\displaystyle x =\arcsin \tfrac23 \Rightarrow \sin x = \tfrac 23$

and$\displaystyle y = \arccos\tfrac{\sqrt3}{2} \Rightarrow \cos y = \tfrac{\sqrt3}{2}$

That feels better already!

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

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

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

Grandad