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Math Help - Solving Trig Identities! HELP!

  1. #1
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    Exclamation Solving Trig Identities! HELP!

    I have these questions in my Trig Identities homework but I don't understand how to do them.

    1) 4(Cos^6X + Sin^6X) = 1 + 3 Cos^(2)2X

    I don't know what to do when the Cos or Sin has an exponent more than 2.
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  2. #2
    MHF Contributor red_dog's Avatar
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    4(\cos^6x+\sin^6x)=4[(\sin^2x)^3+(\cos^2x)^3]=

    =4[(\sin^2x+\cos^2x)^3-3\sin^2x\cos^2x(\sin^2x+\cos^2x)]=

    =4(1-3\sin^2x\cos^2x)=4-12\sin^2x\cos^2x=4-3\sin^22x=

    =4-3(1-\cos^22x)=1+3\sin^22x

    I used a^3+b^3=(a+b)^3-3ab(a+b)
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  3. #3
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    Thanks, But I kinda don't understand. Call me dense if you will :P Are you only solving the left side?
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  4. #4
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    Hello, Mineko245!

    A slight variation on red-dog's solution . . . done in baby-steps.


    1)\;\;4(\cos^6\!x + \sin^6\!x) \:= \:1 + 3\cos^2\!2x
    We have: . 4\bigg[(\cos^2\!x)^3 + (\sin^2\!x)^3\bigg] . . . sum of cubes

    . . = \;4\bigg[\underbrace{\left(\cos^2\!x + \sin^2\!x\right)}_{\text{This is 1}}\left(\cos^4\!x - \cos^2\!x\sin^2\!x + \sin^4\!x\right)\bigg]

    . . = \;4\bigg[\cos^4\!~x - \cos^2\!x\sin^2\!x + \sin^4\!x\bigg]


    Add and subtract 3\cos^2\!x\sin^2\!x

    . . 4\bigg[\cos^4\!x - \cos^2\!x\sin^2\!x \:{\color{blue}+\: 3\cos^2\!x\sin^2\!x} + 1 \:{\color{blue}- \:3\cos^2\!x\sin^2\!x}\bigg]

    . . = \;\;4\bigg[\left(\cos^4\!x + 2\cos^2\!x\sin^2\!x\right) + \sin^4\!x - 3\cos\!x\sin^2\!x\bigg]

    . . =\;\;4\bigg[\underbrace{(\cos^2\!x + \sin^2\!x)^2}_{\text{This is 1}} - 3\cos^2\!x\sin^2\!x\bigg] \;\;=\; \;4\bigg[1 - 3\sin^2\!x\cos^2\!x\bigg]

    . . =\;\;4 - 12\sin^2\!x\cos^2\!x \;\;=\;\;4-12\left(\frac{1-\cos2x}{2}\right)\left(\frac{1+\cos2x}{2}\right)

    . . = \;\;4 - 12\left(\frac{1-\cos^2\!2x}{4}\right) \;\;=\;\;4 - 3(1-\cos^2\!2x)

    . . =\;\;4-3+3\cos^2\!2x \;\;=\;\;1 + 3\cos^2\!2x \quad\hdots There!

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  5. #5
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    Thank You!
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