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Math Help - Trig identities

  1. #1
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    Trig identities

    1.cot^2x+sec^2x=tan^2x+csc^2x

    2.cos3x=4cos^3x-3cosx

    Can someone help guide me towards the answer but not actually give me the full solution to the answer. Thanks
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  2. #2
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    1. use the Pythagorean identities \cot^2{x} + 1 = \csc^2{x} and sec^2{x} = 1 + tan^2{x} in that order on the left side.
    They might need a slight rearrangement before you use them.

    2. use the sum identity \cos(a+b) = \cos{a}\cos{b} - \sin{a}\sin{b} plus the fact that \cos(3x) = \cos(2x+x).
    you'll also need the double angle identity for cosine \cos(2x) = 2\cos^2{x} - 1
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  3. #3
    Super Member 11rdc11's Avatar
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    K I'm not really sure how to show what I mean without solving it. This a long way but I thought I show you a different way to work the problem.

    Another way to see if the RHS is equal to the LHS is

    LHS-RHS = 0~\text{or}~RHS-LHS = 0

    \cot^2{x} + \sec^2{x} = \tan^2{x} + \csc^2{x}

    \cot^2{x} + \sec^2{x} - \tan^2{x} - \csc^2{x} = 0

    \frac{\cos^2{x}}{\sin^2{x}} + \frac{1}{\cos^2{x}} - \frac{\sin^2{x}}{\cos^2{x}} - \frac{1}{\sin^2{x}} = 0

    \frac{\cos^4{x} - \cos^2{x} + \sin^2{x} - \sin^4{x}}{\cos^2{x}\sin^2{x}} = 0

    \frac{\cos^2{x}(\cos^2{x} -1)}{\cos^2{x}\sin^2{x}} + \frac{\sin^2{x}(1-\sin^2{x})}{\cos^2{x}\sin^2{x}}= 0

    \frac{\cos^2{x} - 1}{\sin^2{x}} + \frac{\sin^2{x}\cos^2{x}}{\sin^2{x}\cos^2{x}} = 0

    \frac{\cos^2{x} - 1}{1-\cos^2{x}} +1 = 0

    -\bigg(\frac{-\cos^2{x} + 1}{1 - \cos^2{x}}\bigg) + 1 = 0

    -1 + 1 = 0

    0 = 0
    Last edited by 11rdc11; September 28th 2008 at 01:54 PM.
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  4. #4
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    Hello, johntuan!

    1)\;\;\cot^2\!x+\sec^2\!x\;=\;\tan^2\!x+\csc^2\!x
    \text{The left side is: }\:\overbrace{(\csc^2\!x - 1)}^{\cot^2\!x} + \overbrace{(\tan^2\!x + 1)}^{\sec^2\!x} \;=\; \tan^2\!x + \csc^2\!x

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  5. #5
    Super Member 11rdc11's Avatar
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    Another way to do it using skeeter method

    \sec^2{x} - \tan^2{x}~=~\csc^2{x} - \cot^2{x}

    which simplfies to

    1~=~1
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