Need help on precalc.
2 questions:
A. Prove that (sec^4x)(tan^2)=(tan^2x+tan^4x)(sec^2x)
B. Simply sin^4x to in terms of cos to the 1st power. This means that you can only use:
#'s and cos2x, cos6x, etc, etc.
January 21, 2011
10 Comments • Newest first
haha. double angle abuse.^^^ X3
i did that at first but i couldnt think of anything. however, cos4x = 1- 8cos^2x + 8cos^4x so that's how i got my answer
[quote=arrowwe]kk i think i got it.
do wut rushuko said: distribute the whole thing. Separate one of the -cos^2x. the 1-cos^2x + cos^4x becomes ((cos4x-1)/8)+1 and the -cos^2x becomes -(cos2x + 1)/2
combine the two and u got ur answer
...i hope......[/quote]
What you just said confused me.
EDIT: Worked about the 2nd problem a bit.
(sin^2x)=>1-2cos^2x+cos^4x=>
-cos2x+cos^4x
Anyone wanna tell me how to simplify the cos^4x?
kk i think i got it.
do wut rushuko said: distribute the whole thing. Separate one of the -cos^2x. the 1-cos^2x + cos^4x becomes ((cos4x-1)/8)+1 and the -cos^2x becomes -(cos2x + 1)/2
combine the two and u got ur answer
...i hope......
[quote=arrowwe]so does that mean my thing doesnt work either?[/quote]
Pretty much.
so does that mean my thing doesnt work either?
[quote=Rushuko]sec^4x=(sec^2x)(sec^2x)
sec^2x=tan^2x+1
substitute. distribute. and cirlce your answer
as for the second one..?
sin^4x= (1-cos^2x)(1-cos^2x)
Multiply the two polynomials and you get an answer with only cos's and numbers idk if thats what you are looking for though[/quote]
For the second one, it has to expressed in cos to the 1st power. So you can't have cos^2x, cos^3x, etc.
for B: i don't know if this is cheating but...
if u just change the sin^4x using trig identities like rushuko said, i guess just keep factoring to ((1+cosx)(1-cosx))^2 ?
[quote=ime1729]Have you even tried these questions yourself?
Question A does not make sense, and use double angle identities on the second one.
If you're not going to spend some time thinking about it, I'm not going to spend my time solving problems for you. Unless if you give me NX =P[/quote]
I spend a good 10 minutes working stuff out. Then I hit a roadblock. For the 1st problem this is what I did:
-Changed it to sin^2x/cos^6x
-Expanded cos^6x into a cubic sin equation
-Tried to simply the sin equation into a cubic cos equation
-Realized I was going in circles
-Raged
-Repeat
@above Eh, thanks. Didn't realize it was that simple.
A. Working with the left side
(sec^2x)(sec^2x)(tan^2x)
(sec^2x)(1+tan^2x)(tan^2x)
(sec^2x)(tan^2x + tan^4x)