Azret's blog

Trigonometry proof [CLOSED FAST :D]
By Azret, history, 9 years ago, translation, In English

Hello.

I am trying to prove that sin (α - β) = sin α * cos β - cos α * sin β but getting something wrong.

OPα = (cos α, sin α)
OPβ = (cos β, sin β)
|OPα × OPβ| = |OPα| * |OPβ| * sin (γ)
|OPα| = |OPβ| = 1
γ = α - β
Hence, |OPα × OPβ| = sin (α - β)
Since, |A × B| = Ax * By - Ay * Bx
|OPα × OPβ| = cos α * sin β - sin α * cos β
Hence, sin (α - β) = cos α * sin β - sin α * cos β

But all formulas in internet say that sin (α - β) = sin α * cos β - cos α * sin β , i.e reverse of what I wrote above. Can someone find my mistake?

P.S Below is the correct proof written in white. You can find my mistake yourself first. ;)
As I understood I made mistake in computing angle. Since angles are + in counter-clockwise direction then
|OPα × OPβ| = ... * sin(360 - γ)
|OPα × OPβ| = |OPα| * |OPβ| * ( - sin (γ))
Hence, |OPα × OPβ| =  - sin (α - β)
And  - sin (α - β) = cos α * sin β - sin α * cos β
Leads to sin (α - β) = sin α * cos β - cos α * sin β

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9 years ago, # |
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You take absolute value only from part of the expression. I think it should be:

|OPα × OPβ| = |OPα| * |OPβ| * |sin (γ)|

|OPα × OPβ| = |sin (α - β)|

And as result:

|sin (α - β)| = |cos α * sin β - sin α * cos β|

And that is correct.

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    9 years ago, # ^ |
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    |A| is magnitude of vector A, not an absolute value. :D

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9 years ago, # |
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as you know, A × B =  - B × A, and you made such a mistake in order of operands in following expression: |OPα × OPβ| = sin(α - β).

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