Showing how pi/2+theta = cos using unit circle.

Printable View

January 16th 2011, 08:06 PM
Oiler

Showing how pi/2+theta = cos using unit circle.

Hey all,

Been stuck with my classwork lately, I can not visualize why pi/2+theta = cos theta. This is similar to my last post except i was able to understand it when its in the first quadrant.

Thanks.
January 16th 2011, 08:10 PM
dwsmith

$\sin{\left(\frac{\pi}{2}+\theta\right)}=\cos{\thet a}$

You mean this correct?
January 16th 2011, 08:27 PM
Oiler

dwsmith, yes.
January 16th 2011, 08:29 PM
pickslides

Quote:

Originally Posted by Oiler

I can not visualize why pi/2+theta = cos theta. Thanks.

Draw $\sin{\left(\frac{\pi}{2}+\theta\right)}$ and $\cos{\theta}$ separately on the same set of axes.

What do you notice?
January 16th 2011, 08:35 PM
Oiler

The angles are symmetrical at pi/2 ?

\ | /
\ | /
__\|/___
January 16th 2011, 09:28 PM
pickslides

Quote:

Originally Posted by Oiler

The angles are symmetrical at pi/2 ?

\ | /
\ | /
__\|/___

At $\theta = \frac{\pi}{2}$ they are the same.

Did you plot them together yet?
January 16th 2011, 09:47 PM
Oiler

This is all i can imagine the angle to be xx - Slimber.com: Drawing and Painting Online
January 16th 2011, 10:16 PM
jacs

1 Attachment(s)

If you put the $\theta$ between the vertical axis instead, then where you currently have $\theta$ will be $\pi-(\theta+\frac{\pi}2)$ and we know $sin ({\pi-(\theta+\frac{\pi}2})) = sin( \theta+\frac{\pi}2)$ and by alternate angles, the other angle in the triangle will be $\theta$ .
see diagram

$cos \theta = \frac{y}1$
$sin (\pi-(\theta+\frac{\pi}2}))=\frac{y}1$
hence $sin (\theta+\frac{\pi}2)=cos \theta$

Attachment 20466
January 16th 2011, 10:23 PM
Oiler

Thanks jacs. i can clearly see this now.
January 16th 2011, 10:29 PM
Oiler

This in theory should also hold true to cos(pi/2+theta)=sinx ?..
January 16th 2011, 10:33 PM
pickslides

Quote:

Originally Posted by Oiler

This in theory should also hold true to cos(pi/2+theta)=sinx ?..

Nope, it is $\cos \left( \theta -\frac{\pi}{2}\right) = \sin \theta$
January 16th 2011, 10:48 PM
Oiler

hmm for cos(pi/2+theta) i get cos(pi/2+theta) = sin(theta), I am guessing it should be -sin(theta)
January 17th 2011, 11:43 AM
HallsofIvy

Quote:

Originally Posted by Oiler

hmm for cos(pi/2+theta) i get cos(pi/2+theta) = sin(theta), I am guessing it should be -sin(theta)

Looking at the unit circle, adding pi/2 to an angle in the first quadrant gives and angle in the second quadrant where the "x" coordinate is negative. You should also be able to see that the vertical and horizontal sides of the right triangles formed (|x| and |y|) are swapped. That is (x, y), in the first quadrant, with both x and y positive, is rotated to (-y, x).

Since, on the unit circle, the first coordinate is [itex]cos(\theta)[/itex] and the second coordinate is [itex]sin(theta)[/itex], [itex]cos(\theta+ \pi/2)= -y= -sin(\theta)[/itex].