# Thread: Deriving a trig formula

1. ## Deriving a trig formula

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

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

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

Then I'm kind of stuck, so I tried going the other way around:

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

$\frac{\frac{sin\,\alpha}{cos\,\alpha}+ \frac{sin\alpha\,\beta }{cos\,\beta}}{1 - \frac{sin\, \alpha}{cos\, \alpha}\frac{sin\, \beta}{cos\, \beta}}$

Then I multiply the 1/1 in the denominator by cos a cos b, to make them equal, to get rid of the 1, then I rewrite it according to $\frac{\frac{a}{b}}{\frac{c}{d}}$ = $\frac{ad}{bc}$ which leaves me with a lot of sin and cos, but stuck as well. Some help would be nice

2. ## Re: Deriving a trig formula

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:

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

First I've rewritten $tan(\alpha + \beta)$ as $\frac{sin(\alpha + \beta) }{cos(\alpha + \beta)}$ which equates to:
$\frac{sin\, \alpha\, cos\, \beta + \cos\,\alpha\,sin\beta}{cos\,\alpha\,cos\beta - sin\,\alpha\,sin\,\beta}$.
Divide every term by $\cos(\alpha)\cos(\beta)$.

3. ## 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?

4. ## Re: Deriving a trig formula

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

5. ## 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.

6. ## Re: Deriving a trig formula

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