Question

Derivation of tan(a+b). I felt like my trig was rusting from disuse, so I thought I would review the derivation of the common identities before working a few exercise sets. I got stuck on tan(a+b), so I'm hoping for some insight.

I don't mean "stuck" like I can't follow the simple instructions to write sin(a+b)/cos(a+b) as (sin(a)cos(b)+sin(b)cos(a))/(cos(a)cos(b)-sin(a)sin(b)) then divide the numerator and denominator by cos(a)cos(b). Where I'm stuck is trying to understand the intuition behind the step of dividing out the cos(a)cos(b).

Answer 1

maybe scale a little back further more, then work your way up again?

Answer 2

I'm not sure I follow you. The furthest back I can scale is tan(a+b), right? Unless I'm missing something, that is a non-starter (which was why the next step is re-writing it as sin(a+b)/cos(a+b). Could you try to explain it differently?

Answer 3

quotient rule follows

Answer 4

@bahrom7893 @modphysnoob

Answer 5

then chain rule

Answer 6

If it is any help, I've been looking at the wikipedia page on "Proofs of trigonometric identities" and they have a pretty decent graphic that makes the both sin(a+b) and cos(a+) seem intuitive, but I've been staring at it for a while and not seeing how to apply the same process to tan(a+b).

Answer 7

Those identities were reasoned though long before calculus. If possible I'd like to follow their steps without resorting to it.

Answer 8

what are you trying to do? tan(a+b)?

Answer 9

trying to get a feel for the intuition behind the step where you divide the num/den by cos(a)cos(b). Like, why did they select THAT to divide (other than "it just works")

Answer 10

I'm going to leave this open for a day or so, just in case anyone comes along who might know. But, it's not like a life or death issue. I'm just curious.