.

I'm just working through the proof now, I will edit this post and write it up when I have done.

................

PROOF:

'O' = Opposite
'H' = Hypotenuse
'A' = Adjacent

We know that

\sin = \frac{\mathrm{O}}{\mathrm{H}}

Now, in a triangle defined by

\sin(t) = x

, we have defined

\mathrm{O} = x

and

\mathrm{H} = 1

.

From Pythagoras theorem, we know that

O^2 + A^2 = H^2

. So that means that

x^2 + A^2 = 1^2

,

which can be re-written as

\mathrm{A}^2 = 1 - x^2 \implies \mathrm{A} = \sqrt{1 - x^2}

.

Now we also know that

\cos = \frac{\mathrm{A}}{\mathrm{H}}

. We have already said that

\mathrm{H} = 1, \mathrm{A} = \sqrt{1 - x^2}

.

So in this triangle, we now have

\cos(t) = \frac{\sqrt{1 - x^2}}{1} = \sqrt{1 - x^2}

.

From before, we said that

\sin(t) = x \implies t = \arcsin(x)

.

So we get

\cos(\arcsin(x)) = \sqrt{1 - x^2}

(edited 11 years ago)

Reply 5

Indeterminate

Original post by SDavis123

But if I do that I get

Sin^(-1)x.sin.sin^(-1) + cos.sin^(-1) + k

How would you simplify that?

Posted from TSR Mobile

It's quite simple.

Given

x=\sin u

surely you can simplify

u\sin u

by noting that

u=\arcsin(x)

Also

\sin^2 u + \cos^2 u = 1

(edited 11 years ago)

Reply 6

SDavis123

Ok I think I get it but I definitely wouldn't have thought of this myself

And thank you so much for the help guys, I really appreciate the effort you put in, especially with that proof :smile: