trig identity

[phys1618]

1. The problem statement, all variables and given/known data

cos(α − β)cos(α + β) = cos2α - sin2 β
2. Relevant equations

cos(α + β) = cos α cos β − sin α sin β

cos(α − β) = cos α cos β + sin α sin β

3. The attempt at a solution
I worked out the LHS which makes it
cos2α cos2β - sin2α sin2β=RHS

Then, I'm stuck, however, i tried to do what I think can work...But I don't think it's right.
From here I think it's wrong.
(cos2α - sin2 β) (cos2β + sin2α)=RHS
(cos2α - sin2 β) (1) =RHSS

My problem is can we use identity for items like cos2 a + sin2 B = 1?
the variable is not the same :/

[lanedance]

can you use complex exponentials?

[sylas]

I worked out the LHS which makes it
cos2α cos2β - sin2α sin2β=RHS


Good work.


Then, I'm stuck, however, i tried to do what I think can work...But I don't think it's right.
From here I think it's wrong.
(cos2α - sin2 β) (cos2β + sin2α)=RHS
(cos2α - sin2 β) (1) =RHSS

My problem is can we use identity for items like cos2 a + sin2 B = 1?
the variable is not the same :/

The identity only works with one variable, just as you suggest. So sin2α+cos2β cannot be reduced to 1. Also, your initial step in reducing the RHS has too many powers of 2 in it.

You will do better not to use a difference of squares at this point. Basically, you have equations that involve four terms:
\sin\alpha, \sin\beta, \cos\alpha, \cos\beta You can easily get that down to two terms using the identity
\sin^2\alpha + \cos^2\alpha = 1
and the same again for β.

You will find this identity easier to use while you have the squares present in the equation. Since the RHS uses cos α and sin β, you can try eliminating the cos β and the sin α from the LHS.

Cheers -- sylas