1. ## Composite argument property

del

2. In the "little prince" there is a line: "Draw me a sheep!"

Now, instead of using composite argument property for cosine and instead drawing me something.... draw to yourself a unit circle.

3. I understand that cos(90-x)=sin, I just have no idea how to interwind it with the composite argument property

4. Well, let's try it.

$cos(90 - x) = cos(90)cos(x) + sin(90)sin(x)$

But cos(90) = 0 and sin(90) = 1.

So?

5. Deleted.

6. Originally Posted by wiseguy
"Use the composite argument property for cosine to prove that $cos (90° -) = sin$"

I'm aware the composite argument property of cosine is $cos(A − B) = cos A cos B + sin A sin B$ - how would you use that to show $cos (90° -) = sin$?
You do realise that your formatting is messed up, right?

From context, you mean

Given that $\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B$

prove that $\cos(90^\circ - \theta) = \sin\theta$.

So let $A = 90^\circ$ and $B = \theta$ and write it out..