Still trying to perfect these, I ain't too good at them yet.
How do I show that (1+cos#)(1-cos#)=sin^2#
I get to 1-cos^2#=sin^2#
But then I forget what to do :o
Trigonometric Identities.
Still trying to perfect these, I ain't too good at them yet.
How do I show that (1+cos#)(1-cos#)=sin^2#
I get to 1-cos^2#=sin^2#
But then I forget what to do :o
1-cos<sup>2</sup> = sin<sup>2</sup>
Seriously? I was hoping it was that easy! xD
I might be back for more help in a bit xD
don't forget about sec^2=1 + tan^2!
Okay, Boko. We get it. You're smart.
If I was smart I wouldn't be asking for help Yar-Yars!
Yeah, there are 5 that you absolutely have to memorize, but 4 of them are derivable from the firstOriginally Posted by RubyLapiz
sin<sup>2</sup> + cos<sup>2</sup> = 1
subtract sin<sup>2</sup> to get
cos<sup>2</sup> = 1 - sin<sup>2</sup>
subtract cos<sup>2</sup> to get
sin<sup>2</sup> = 1 - cos<sup>2</sup>
divide by sin<sup>2</sup> to get
1 + cot<sup>2</sup> = csc<sup>2</sup>
divide by cos<sup>2</sup> to get
1 + tan<sup>2</sup> = sec<sup>2</sup>
Just wait until you get the double angles, half angles and sum and difference equations mixed in there. You will have oh so much fun XD
can you post the double angle and half angle formulas? I always forget them.
This is why I like teachers who will let you have a cheat sheet with these identities on tests.
My teacher wants me to memorize them.
sadly, calculus teachers are not some of those teachers
Double Angle Theorem (typed up from tav's link):
cos(2x )= cos<sup>2</sup>(x) - sin<sup>2</sup>(x)
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