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
- 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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