# Stumped... cosθ=sin(90-θ)

• Jul 24th 2010, 11:33 AM
wiseguy
Stumped... cosθ=sin(90-θ)
"Why is cosθ=sin(90-θ)? Use the figure to help explain."

I attached the image.

In this instance, I am at a complete loss...

??
• Jul 24th 2010, 11:51 AM
Ackbeet
Drop a perpendicular from the point (u,v) down to the u axis, and another perpendicular from (u,v) horizontally over to the v axis. Label the line from the origin to (u,v) - call it h. Now, you tell me: from the definitions, what is the cosine of theta? What is the sin of (90 - theta)?
• Jul 24th 2010, 11:52 AM
Mathelogician
Make right triangles from the point (u,v) and use the definitions of sine and cosine. Or use a unit circle for another geometric proof.
But it's simpler to use the sine Difference Formula!
• Jul 24th 2010, 12:17 PM
Quote:

Originally Posted by wiseguy
"Why is cosθ=sin(90-θ)? Use the figure to help explain."

I attached the image.

In this instance, I am at a complete loss...

??

For simplicity, if you draw a single right-angled triangle.

One acute angle is $\displaystyle \theta$ and the other is $\displaystyle 90^o-\theta$ since the acute angles sum to 90 degrees in a right-angled triangle.

Place $\displaystyle \theta$ at the acute angle coming away from the base.

Label that side at $\displaystyle \theta$ "base"

Then $\displaystyle cos\theta=\frac{base}{hypotenuse}$ as the base is adjacent $\displaystyle \theta$

$\displaystyle sin\left(90^o-\theta\right)=\frac{base}{hypotenuse}$ as the base is opposite the angle $\displaystyle 90^o-\theta$