Math (trigonometry)
posted by Lucy on .
Trigonometry identities are so hard...
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

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

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