trignometry proof

[dagr8est]

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.

 

[Guest]

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.

 

[soroban]

Hello, dagr8est!

Anyone know the proof for sin(θ) = cos(90 − θ)?

Are you familiar with the "compound angle" formulas?
. . . \(\displaystyle \sin(A\,\pm\,B)\;=\;\sin(A)\cdot\cos(B)\,\pm\,\sin(B)\cdot\cos(A)\)
. . . \(\displaystyle \cos(A\,\pm\,B)\;=\;\cos(A)\cdot\cos(B)\,\mp\,\sin(A)\cdot\sin(B)\)

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

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

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