Posted by **Lucy** on Sunday, April 13, 2008 at 9:53pm.

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

- Math (trigonometry) -
**drwls**, Sunday, April 13, 2008 at 10:09pm
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.

- Math (trigonometry) -
**Lucy**, Sunday, April 13, 2008 at 10:26pm
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

- Math (trigonometry) -
**Phelelani**, Thursday, May 19, 2016 at 8:50am
SOH CAH TOA... IT'S sin=opposite over hypotenus..... Cos=ajacent over hypotenus.... Tan=opposite over ajacent

- Math (trigonometry) -
**Phelelani**, Thursday, May 19, 2016 at 8:51am
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

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