1. ## inverse trig?

can someone help me simplify these? this is soo frusrating, im about to cry.

cos(arctan(-1))

arccos(sin(-pie/2))

cot(arcsin 3/5)

sin (arccos (t))

thanks soo much!

2. Originally Posted by akilele
arccos(sin(-pie/2))
Mmmmm pie...

Anyway, $\displaystyle \sin(-\frac{\pi}{2}) = -1$.

$\displaystyle \arccos(-1) = \pi$.

Is this enough explanation, or would you like more?

3. I get how you got -1 from sin(-pi/2) but i'm confused how you got pi. arccos(-1)=pi. But I thought cos(-1) got you pi. So what does Arc do to the equation?

4. Originally Posted by akilele
I get how you got -1 from sin(-pi/2) but i'm confused how you got pi. arccos(-1)=pi. But I thought cos(-1) got you pi. So what does Arc do to the equation?
arccos is the inverse of cos.

That means cos(x)=y implies arccos(y)=x

so cos(pi)=-1 implies arccos(-1)=pi

5. from what you gave me, i was able to do the first problem but im having trouble with the last 2 because i can't find 3/5 on the unit circle. and i don't know what t is supposed to be.

6. Originally Posted by akilele
from what you gave me, i was able to do the first problem but im having trouble with the last 2 because i can't find 3/5 on the unit circle. and i don't know what t is supposed to be.
Remember that $\displaystyle \sin^2(t)+\cos^2(t)=1$.

Let $\displaystyle y=\arccos(t) \implies t=\cos(y)$.

Therefore $\displaystyle \sin^2(y)+\cos^2(y)=\sin^2(y)+t^2=1$.

Thus $\displaystyle \sin^2(y)=1-t^2 \implies \sin(y)=\sqrt{1-t^2}$.

Now substitute $\displaystyle y=\arccos(t)$ back in to get $\displaystyle \sin(\arccos(t))=\sqrt{1-t^2}$.