Yep - you are dead right with regards to factorising the LHS.
Once you've done that use a Pythagorian Identity and you should be good to go!
Ok, our teacher has been very busy lately so he hasn't been around to help us and he isn't available after/before school at the moment. So here i am XD
This one i have no clue where to begin:
1-2cos^2(x)+cos^4(x)=sin^4(x)Prove that each side of the equation equals each other (or something like that, i don't have the book infront of me)
*I'd assume that you have to factor the cos's but i have no clue how to do that*
I also don't know where to begin with this one..
*Again i'd assume u have to factor the cos's into something?*
Thanks for ANY help anyone can give!!!
Oh, and for the second equation just use the Pythagorian identity:
cos^2(x) + sin^2(x) = 1
Start by dividing through by (1 - sin^2(x)) and putting the new denominator in terms of cos. Just one more identity after that on the RHS and a bit of algebraic manipulation and you should be there
I tried to give you a hint to get you going on the second one
To prove that they equate to one another:
(i) divide both sides by [1 - sin^2(x)]
(1 + sin^2(x)) = [(2cos^2(x) - cos^4(x)] / (1 - sin^2(x))
(ii) Using the trig identity (cos^2(x) + sin^2(x) = 1) which therefore leads to 1 - sin^2(x) = cos^2(x) (through simple manipulation) substitute the RHS denominator
1 + sin^2(x) = [(2cos^2(x) - cos^4(x)] / cos^2(x)
Do you see where I'm going? now simplify the RHS and use another identity but in sin^2(x) on the RHS
Thank you guys so much!!!
Although i now know how to do both ways (thanks!) i think my teacher was looking for just using the LHS and making it look like the RHS and not crossing sides. Or thats what he did in all the examples? It probably doesn't matter either way but Thanks again!