Ok...
the trig identity "pathagorian identity"
ok just to clearify...when i have sin(^2)x that is sin squared....NOT sin raised to the 2x...ok?
well it states:
1 - sin(^2)x = cos(^2)
so...
is
1 - sin(^4)x = cos(^4)
???
Thanks!!
Super needed!!
~Jeff~
Trig. Identities...
- Gostridah
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Re: Trig. Identities...
Let's rewrite those as
a. sin(^2)x+cos(^2)x=1,
b. sin(^4)x+cos(^4)x=1 (?)
If in a we square both sides, we get:
(sin(^2)x+cos(^2)x)^2=1^2=1, so
sin(^4)x+2.sin(^2)x.cos(^2)x+cos(^4)x=1, so
sin(^4)x+cos(^4)x=1-2.sin(^2)x.cos(^2)x, which isn't 1 unless sinx or cosx is 0.
a. sin(^2)x+cos(^2)x=1,
b. sin(^4)x+cos(^4)x=1 (?)
If in a we square both sides, we get:
(sin(^2)x+cos(^2)x)^2=1^2=1, so
sin(^4)x+2.sin(^2)x.cos(^2)x+cos(^4)x=1, so
sin(^4)x+cos(^4)x=1-2.sin(^2)x.cos(^2)x, which isn't 1 unless sinx or cosx is 0.
- SirPostAlot
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Re: Trig. Identities...
hmm...yeah i think i see what your saying...
okies i understand...haha
thanks!
~Jeff~
okies i understand...haha
thanks!
~Jeff~