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Math Help - Trigonometry QUESTION?

  1. #1
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    Trigonometry QUESTION?

    PROVE that sin^4 x + cos^4 x= 1/4(3+cos4x). hence find the largest possible value and smallest possible value of sin^4 x+cos^4 x



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    Last edited by JohnRamey; November 26th 2010 at 09:26 PM.
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  2. #2
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    Hello, John!

    I'm delighted that you solved it.
    Here's my approach . . .


    \text{Prove that: }\:\sin^4\!x + \cos^4\!x \:=\: \frac{1}{4}(3+\cos4x)

    We have: . \left(\sin^2\!x\right)^2 + \left(\cos^2\!x\right)^2 \;=\;\left(\dfrac{1-\cos2x}{2}\right)^2 + \left(\dfrac{1+\cos2x}{2}\right)^2

    . . . . . . \displaystyle =\;\frac{1 - 2\cos2x + \cos^2\!2x}{4} + \frac{1 + 2\cos2x + \cos^2\!2x}{4}

    . . . . . . =\;\displaystyle \frac{2 + 2\cos^2\!2x}{4} \;=\;\frac{1+\cos^2\!2x}{2} \;=\;\frac{1}{2}(1 + \cos^2\!2x)

    . . . . . . \displaystyle=\;\frac{1}{2}\left(1 + \frac{1+\cos4x}{2}\right) \;=\;\frac{1}{2}\left(\frac{3 + \cos4x}{2}\right) \;=\;\dfrac{1}{4}(3 + \cos 4x)





    \text{Hence, find the largest and smallest possible value of }\sin^4\!x+\cos^4\!x

    The maximum value of \cos4x is 1.
    The minimum value of \cos4x is -1.


    Therefore: . \frac{1}{4}(3 - 1)\;\leq \;\sin^4\!x + \cos^4\!x \;\leq \;\frac{1}{4}(3+1)

    . . . . . . . . . . . . . \frac{1}{2}\;\le \;\sin^4\!x+\cos^4\!x \;\le \; 1

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