+0  
 
0
67
2
avatar+79 

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

Best Answer 

 #1
avatar+5555 
+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

hectictar  Nov 23, 2017
Sort: 

2+0 Answers

 #1
avatar+5555 
+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

3 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details