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.

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- January 16th 2011, 08:06 PMOilerShowing 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 PMdwsmith

You mean this correct? - January 16th 2011, 08:27 PMOiler
dwsmith, yes.

- January 16th 2011, 08:29 PMpickslides
- January 16th 2011, 08:35 PMOiler
The angles are symmetrical at pi/2 ?

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__\|/___ - January 16th 2011, 09:28 PMpickslides
- January 16th 2011, 09:47 PMOiler
This is all i can imagine the angle to be xx - Slimber.com: Drawing and Painting Online

- January 16th 2011, 10:16 PMjacs
If you put the between the vertical axis instead, then where you currently have will be and we know and by alternate angles, the other angle in the triangle will be .

see diagram

hence

Attachment 20466 - January 16th 2011, 10:23 PMOiler
Thanks jacs. i can clearly see this now.

- January 16th 2011, 10:29 PMOiler
This in theory should also hold true to cos(pi/2+theta)=sinx ?..

- January 16th 2011, 10:33 PMpickslides
- January 16th 2011, 10:48 PMOiler
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 AMHallsofIvy
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].