Trig problem

[MeDiCS]

I've been trying to prove that sin(a+b)=sin(a)cos(b)+sin(b)cos(a), and I've come across the following triangle:

View attachment 4352

The problem is that the cosine law an the pythagorean theorem disagree as to the value of H.

Using the smaller triangle and the pythagorean theorem:
H^2 = [1 - sen(a)]^2 + cos^2(a)
H^2 = 1 - 2sen(a) + sen^2(a) + cos^2(a)
H^2 = 2 - 2sen(a)
H = sqr(2[1 - sen(a)])

Using the bigger triangle and the cosine law:
H^2 = 1^2 + 1^2 - 2*1*1*cos(a)
H^2 = 2 - 2cos(a)
H = sqr(2[1 - cos(a)])

If this is right, then cos(a) = sen(a), which is not true for most values of a. What am I doing wrong here?

 

[martins]

I think in your picture, you've got the sin and cos labels the wrong way round. If you correct them, then for the smaller triangle:

[math]
H^2 = \sin^2 a + (1-\cos a)^2 = \sin^2 a + 1 - 2\cos a + \cos^2 a = 2 - 2\cos a
[/math]
which corresponds with your other result.

 

[MeDiCS]

I think in your picture, you've got the sin and cos labels the wrong way round.

:rofl:, true. I do this mistake often...

Thanks for the help :thumbup:!