Thread: [SOLVED] Identities Help? Thanks!

1. [SOLVED] Identities Help? Thanks!

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:

Prove that each side of the equation equals each other (or something like that, i don't have the book infront of me)
1-2cos^2(x)+cos^4(x)=sin^4(x)

*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..

(1-sin^2(x))(1+sin^2(x))=2cos^2(x)-cos^4(x)
*Again i'd assume u have to factor the cos's into something?*

Thanks for ANY help anyone can give!!!

2. Hi there,

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!

Good luck

3. 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

D

4. Re:

Ok, maybe im mistaken? but i think u misunderstood me.

the second equation i also have to prove that one side of the equation equals the other.

Thank you so much though!! Figured out the 1st one! thanks again!

5. 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

6. 2nd is super easy

again working only on the LHS, just use Pythagorian Identity again.

Starting with:
$\displaystyle (1-sin^2x)(1+sin^2x)$
sub the sines with $\displaystyle sin^2x = 1-cos^2x$ and it all falls out.

7. yes, this is a quicker way of doing it.

8. Thank You!!

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!