Trigonometry identities are so hard...

Question

I need some help proving these identities:
*Oh, and I'm only in grade 11, so the identities we use are quotient identity and Pythagorean identity.

sinx/(sinx + cosx) = tanx/(1 + tanx)

cos^2x - sin^2x = 2cos^2x - 1

Thanks!
Lucy

Answer 1

By inverting the fractions (a perfectly legal operation), the first equation can be converted to
(sin x + cos x)/sin x = (1 + tan x)/x
1 + cot x = 1 + 1/tan x
= 1 + cot x

In the second problem, substitute 1 - cos^2 x for sin^2 x on the left side.

Answer 2

Thanks for your help!

I understand the second problem now.
Except I'm confused about what you did in the first problem. We haven't learned anything about cotx yet...

I inverted the fractions, though and ended up with:

(sin x + cos x) / sin x = (1 + tan x) / tanx
(I'm just wondering...why did you write (1 + tan x) / x on the left side?)

Then simplifies to... cos x = 1 ??

I'm confused... :S

Answer 3

*Sorry, should be:

I'm just wondering...why did you write (1 + tan x) / x on the *right* side?

Instead of (1 + tan x) / tan x?

Answer 4

SOH CAH TOA... IT'S sin=opposite over hypotenus..... Cos=ajacent over hypotenus.... Tan=opposite over ajacent

Answer 5

SOH CAH TOA... IT'S sin=opposite over hypotenus..... Cos=ajacent over hypotenus.... Tan=opposite over ajacent..... When you want to fine thiter or any angle