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# Show on unit circle that cos (θ + 3pi/2) = sin θ

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I know how to prove this equivalent expression through an identity but how do i show this on a unit circle visually?

UpTheChels  Nov 22, 2017

#1
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First let's look at the two angles drawn on a unit circle separately:

Now....let's look at each triangle...

Look at the first triangle. Since there are  pi  radians in every triangle

∠ABC   =   pi - pi/2 - θ   =   pi/2 - θ

Look at the second triangle. Since there are  2pi  radians in every circle

∠DEF  =   2pi - (θ + 3pi/2)   =   2pi - 3pi/2 - θ   =   pi/2 - θ

And both triangles have a side of length  1 .  So....

△ABC   is  congruent to  △DEF   by  the Ange-Angle-Side rule. And

side  EF   =   side  BC

cos( θ + 3pi/2 )   =   sin θ

hectictar  Nov 23, 2017
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#1
+6558
+4

First let's look at the two angles drawn on a unit circle separately:

Now....let's look at each triangle...

Look at the first triangle. Since there are  pi  radians in every triangle

∠ABC   =   pi - pi/2 - θ   =   pi/2 - θ

Look at the second triangle. Since there are  2pi  radians in every circle

∠DEF  =   2pi - (θ + 3pi/2)   =   2pi - 3pi/2 - θ   =   pi/2 - θ

And both triangles have a side of length  1 .  So....

△ABC   is  congruent to  △DEF   by  the Ange-Angle-Side rule. And

side  EF   =   side  BC

cos( θ + 3pi/2 )   =   sin θ