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Math Help - Math Trig Identity

  1. #1
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    Math Trig Identity

    Please help me prove this trig identity (please include justification)

    (sinA^4 - cosA^4) secA^2 =secA^2-2

    appreciated
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    Re: Math Trig Identity

    Quote Originally Posted by danny88 View Post
    (sinA^4 - cosA^4) secA^2 =secA^2-2
    Are you sure that that's not supposed to be \left(\sin^4A-\cos^4A\right)\sec^2A = \sec^2A-2\mathrm?

    If that is what you are trying to prove, begin by factoring the \sin^4A-\cos^4A as a difference of squares. Then use the Pythagorean identity and the rest should be obvious.
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    Re: Math Trig Identity

    I would really appreciate if you would show me step by step because I seem to be making an error in my steps.
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    Re: Math Trig Identity

    Quote Originally Posted by danny88 View Post
    I would really appreciate if you would show me step by step because I seem to be making an error in my steps.
    Well, how about you post the work that you have done so that I can see where you went wrong?
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    Re: Math Trig Identity

    (sin^2 A+cos^2 A)(sin^2A-cos^2A) * sec^2A
    (1)*(sin^2A-cos^2A) * sec^2A
    1*(sinA-cosA)^2 * sec^2A
    1*(1-2sinAcosA)* sec^2A
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    Re: Math Trig Identity

    Thank you.
    Quote Originally Posted by danny88 View Post
    1*(sinA-cosA)^2 * sec^2A
    This is your error. We cannot distribute powers over addition or subtraction and, likewise, we cannot "factor out" a power like that. a^2+b^2 is not necessarily equal to (a+b)^2. Instead, convert the secants to cosines:

    \left(\sin^4A - \cos^4A\right)\sec^2A

    =\left(\sin^2A + \cos^2A\right)\left(\sin^2A - \cos^2A\right)\sec^2A

    =\left(\sin^2A - \cos^2A\right)\sec^2A

    =\frac{\sin^2A - \cos^2A}{\cos^2A}

    =\frac{\sin^2A}{\cos^2A} - \frac{\cos^2A}{\cos^2A}

    Can you continue from here?
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    Re: Math Trig Identity

    sin^2 A/ cos^2A =tan^2A?
    then -cos^2A/cos^2A= -1 ?
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    Re: Math Trig Identity

    Quote Originally Posted by danny88 View Post
    sin^2 A/ cos^2A =tan^2A?
    then -cos^2A/cos^2A= -1 ?
    Correct. So we have \tan^2A-1.

    Is there an identity we can use that relates \tan^2A with \sec^2A\,?
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    Re: Math Trig Identity

    1= sec^2A-tan^2A
    but how does that work out?
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    Re: Math Trig Identity

    Quote Originally Posted by danny88 View Post
    1= sec^2A-tan^2A
    but how does that work out?
    Solve for \tan^2A:

    \tan^2A+1=\sec^2A\Rightarrow\tan^2A=\sec^2A-1


    Now substitute that into our expression:

    \tan^2A-1

    =\left(\sec^2A-1\right)-1

    =\sec^2A-2.
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    Re: Math Trig Identity

    thanks a lot of the help, I really appreciate it.
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