Please oh Please Help me D':

[~Jacqueline~]

Please oh Please Help me D':

I supremely suck at math. I've been working on math problems all day =__= these are the last 2 :sadroblems that I have. Any help with a description of how you got it would be awesome.

84. Verify the identity algebraically. Use a graphingutility to confirm the identity graphically. Verify.

cos^4x-sin^4x = cos2x

I've tried to figure it out algebraically but I'm just not getting the right answer. All I've come up with is to factor, but that doesn't really work.



12. Use a graphing utility to approximate the equation’ssolutions in the interval [0, 2 ). If possible, find the exact solutionsalgebraically.
sin2xsinx = cos2x

It would be easy to solve if it was just sin2xsin=0 but the cos2x is throwing me off. I've substituted sin2x with 2sinxcosx, but I'm not sure where to go from there.


again thanks to anyone and everyone that trys to help, its very much appreciated

Jacqueline<3
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[soroban]

Hello, ~Jacqueline~

You're expected to know these two identitites:

. . \(\displaystyle \sin^2\!\theta + \cos^2\!\theta \:=\:1\)

. . \(\displaystyle \cos^2\!\theta - \sin^2\!\theta \:=\:\cos2\theta\)

\(\displaystyle \text{84. Verify: }\:\cos^4\!x - \sin^4\!x \:=\:\cos2x\)

\(\displaystyle \cos^4\!x - \sin^4\!x \:=\\cos^2\!x - \sin^2\!x)\underbrace{(\cos^2\!x + \sin^2\!x)}_{\text{This is 1}}\)
. . . . . . . . . . .\(\displaystyle =\;\;\cos^2\!x - \sin^2\!x\)

. . . . . . . . . . .\(\displaystyle =\;\;\;\cos2x\)

 

[~Jacqueline~]

Thanks!!!! Thats extremely helpfu, you're awesome!! :lol: