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Math Help - Trigonometric proof

  1. #1
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    Trigonometric proof

    I would like to prove that

    sinX^4 - cosX^4 + 2cosX^2

    equals to 1 using sin^2 theta+ cos^2 theta=1 and a^2-b^2=(a-b)(a+b).

    I am completely stuck though, and do not know where to begin. Any advice would be appreciated.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello

    Note that \sin^4x-\cos^4x=(\sin^2 x)^2-(\cos ^2 x)^2. Can you factor this using a^2-b^2=(a-b)(a+b) ?

    Hint (highlight to read) : * let a=(sin x)^2 and b=(cos x)^2 *
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Pandora View Post
    I would like to prove that

    sinX^4 - cosX^4 + 2cosX^2

    equals to 1 using sin^2 theta+ cos^2 theta=1 and a^2-b^2=(a-b)(a+b).

    I am completely stuck though, and do not know where to begin. Any advice would be appreciated.
    \sin^4(x)-\cos^4(x)+2\cos^2(x)=1

    Let's focus on this term first:

    \sin^4(x)-\cos^4(x)

    using the formula a^2-b^2=(a+b)(a-b), we see that

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

    Now, we see that the equation now becomes

    \sin^2(x)-\cos^2(x)+2\cos^2(x)=1

    Simplifying, we get

    \sin^2(x)+\cos^2(x)=1\implies 1=1

    Does this make sense?

    --Chris
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  4. #4
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    Thank you

    Thank you both. It seems pretty clear now =) This forum is so efficient, and I will be definetly using it in the future.
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