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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? 

 Nov 22, 2017

Best Answer 

 #1
avatar+9460 
+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

 

I hope this made sense!  smiley

 Nov 23, 2017
 #1
avatar+9460 
+4
Best Answer

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

 

I hope this made sense!  smiley

hectictar Nov 23, 2017
 #2
avatar+79 
+2

Thanks hectictar, I appreciate the help, you're a lifesaver! :)

UpTheChels  Nov 23, 2017

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