Composite argument property

**wiseguy** · July 26th 2010, 07:46 AM

del

**Also sprach Zarathustra** · July 26th 2010, 07:56 AM

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.

**wiseguy** · July 26th 2010, 09:28 AM

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

**Unknown008** · July 26th 2010, 09:43 AM

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?

**Unknown008** · July 26th 2010, 09:43 AM

Deleted.

**undefined** · July 26th 2010, 09:46 AM

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..

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