trignometry proof

dagr8est

Junior Member
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Nov 2, 2004
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128
Anyone know the proof for sinӨ = cos(90−Ө)? My teacher just said it works but didn't explain why. I don't just mean plugging in a number either. :p
 
Draw a triangle on a graph with the one side starting at the orgin and going along the x axis a length "x". At the end of this line draw a verticle line going up"y" - this is the other side of the triangle.

Now finish the triangle with the hypotenuse "h"(from orgin to the top of the y line)


the angle from the x axis to the hypotenuse is theta.
the x axis line and the y axis line is a 90 degree
The remaining angle is (90-theta)

now look at sin theta = opp/hyp
sin theta = y/h

cos theta = x/h

sin (90-theta) = x/h

cos (90-theta) = y/h

and you can now see the relationship.
 
Hello, dagr8est!

Anyone know the proof for sin(θ) = cos(90 − θ)?
Are you familiar with the "compound angle" formulas?
. . . sin(A±B)  =  sin(A)cos(B)±sin(B)cos(A)\displaystyle \sin(A\,\pm\,B)\;=\;\sin(A)\cdot\cos(B)\,\pm\,\sin(B)\cdot\cos(A)
. . . cos(A±B)  =  cos(A)cos(B)sin(A)sin(B)\displaystyle \cos(A\,\pm\,B)\;=\;\cos(A)\cdot\cos(B)\,\mp\,\sin(A)\cdot\sin(B)

The right side is: .cos(90oθ)  =  cos(90o)cos(θ)+sin(90o)sin(θ)\displaystyle \cos(90^o\,-\,\theta)\;=\;\cos(90^o)\cdot\cos(\theta)\:+\:\sin(90^o)\cdot\sin(\theta)

Since cos(90o)=0\displaystyle \cos(90^o)=0 and sin(90o)=1\displaystyle \sin(90^o)=1, we have:

. . . cos(90oθ)  =  0cos(θ)+1sin(θ)  =  sin(θ)\displaystyle \cos(90^o\,-\,\theta)\;=\;0\cdot\cos(\theta)\,+\,1\cdot\sin(\theta) \;=\;\sin(\theta)
 
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