- Jul 30, 2009
- 4,849
- 10
PM me if you need help. Ill do them for $$. I'm in Calculus BC btw.
Follow along with the video below to see how to install our site as a web app on your home screen.
Note: This feature may not be available in some browsers.
Originally Posted by cbrooks3
Originally Posted by bkdan1
That stuff isn't terribly hard, but it is time consuming. Just manipulate the identities. I.E. sin^2x + cos^2x = 1, divide both sides by cos^2x to get sin^2x/cos^2x + 1 = 1/cos^2x, that's the same as tan^2x + 1 = sec^2x.
my teacher says that when we do it we are only supposed to work on one side at a time so it isnt like solving an algebraec problem...thanks for all of the feedback though. much more than i expected![]()
Originally Posted by cbrooks3
hahaha i did laugh, inside my head. there are 6 problems i have to answer 4 of them...these are the problems:
2cotX=csc(squared)XsinX
2+2cot(squared)X=2cotXsecXcscX
cotX/2sinX=1+cosX
2sinX-cscX=sinX-(cotX/secX)
1=cot(squared)X=2cotXcsc2X
sinXcosX=(cosX-cos(cubed)X/sinX)
Originally Posted by DontStepOnMyShoes
![]()