Re: Deriving a trig formula

Quote:

Originally Posted by

**Lepzed** I'm practicing trig and so I'm working through some exercises. I'm having some trouble figuring this one out:

$\displaystyle tan(\alpha + \beta) = \frac{tan\, \alpha + tan\, \beta}{1 - tan\, \alpha\, tan\, \beta}$

First I've rewritten $\displaystyle tan(\alpha + \beta)$ as $\displaystyle \frac{sin(\alpha + \beta) }{cos(\alpha + \beta)}$ which equates to:

$\displaystyle \frac{sin\, \alpha\, cos\, \beta + \cos\,\alpha\,sin\beta}{cos\,\alpha\,cos\beta - sin\,\alpha\,sin\,\beta}$.

Divide every term by $\displaystyle \cos(\alpha)\cos(\beta)$.

Re: Deriving a trig formula

Aah, got it now yeah! Thanks!

Right, so I know HOW to solve it, but how could I get to sin a cos b? Just try?

Re: Deriving a trig formula

Do you mean 'How could I get to cos(a)cos(b)'? Or? ...

Re: Deriving a trig formula

Yeah, well, you know how to solve this, so for you it's obvious to divide by cos(a)cos(b). For me it's not, now I can just go about trying some stuff till I find something that eventually would work, but that doesn't seem really productive or educated.

Re: Deriving a trig formula

You can just make an educated guess; you have a $\displaystyle \sin\alpha\cos\beta$ as the first term in the numerator, and you need to have a $\displaystyle \tan\alpha$ which implies the division, and you can then verify it by checking with each other term.