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Math Help - trig identity

  1. #1
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    Exclamation trig identity

    here are the problems I am having problem with... no pun intended...

    (1 + tan^2X)(1 + cos2x) = 2 (prove identity)

    (1 + sinA - cos2A) / (cosA + sin2A) = tanA

    and Solve the equation for 0 < or = x < or = 2pi

    cosx + cos2x + 2sin^2x = 0

    your help is much appreciated!
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  2. #2
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    Quote Originally Posted by math:/ View Post
    here are the problems I am having problem with... no pun intended...

    (1 + tan^2X)(1 + cos2x) = 2 (prove identity)

    (1 + sinA - cos2A) / (cosA + sin2A) = tanA

    and Solve the equation for 0 < or = x < or = 2pi

    cosx + cos2x + 2sin^2x = 0

    your help is much appreciated!
    Hint, whenever you see \sin{2x} or \cos{2x} you usually have to use double-angle identities.

    2. Use the identities \cos{2A} = \cos^2{A} - \sin^2{A}, \sin{2A} = 2\sin{A}\cos{A} and \sin^2{A} + \cos^2{A} = 1.


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

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

     = \frac{\sin^2{A} + \sin^2{A} + \sin{A}}{\cos{A}(2\sin{A} + 1)}

     = \frac{2\sin^2{A} + \sin{A}}{\cos{A}(2\sin{A} + 1)}

     = \frac{\sin{A}(2\sin{A} + 1)}{\cos{A}(2\sin{A} + 1)}

     = \frac{\sin{A}}{\cos{A}}

     = \tan{A}.
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  3. #3
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    Quote Originally Posted by math:/ View Post
    here are the problems I am having problem with... no pun intended...

    (1 + tan^2X)(1 + cos2x) = 2 (prove identity)

    (1 + sinA - cos2A) / (cosA + sin2A) = tanA

    and Solve the equation for 0 < or = x < or = 2pi

    cosx + cos2x + 2sin^2x = 0

    your help is much appreciated!
    3. Use the identities \cos{2x} = \cos^2{x} - \sin^2{x} and \sin^2{x} + \cos^2{x} = 1.


    \cos{x} + \cos{2x} + 2\sin^2{x} = 0

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

    \cos{x} + \cos^2{x} + \sin^2{x} = 0

    \cos{x} + 1 = 0

    \cos{x} = -1

     x = \pi + 2n\pi, n \in \mathbf{Z}.


    In the domain  0 \leq x \leq 2\pi, the only solution is

     x = \pi.
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  4. #4
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    1. use the identities:
    \frac{1+\cos{2x}}{2} = \cos^2{x}

    1+\tan^2{x} = \sec^2{x}

    ==========

    Use 2 to multiply by 1 to get:

    2(1+tan^2{x})\left(\frac{1+\cos{2x}}{2}\right)

    This simplifies to 2(\sec^2{x})(\cos^2{x}) and the result follows.
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  5. #5
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    Quote Originally Posted by math:/ View Post
    here are the problems I am having problem with... no pun intended...

    (1 + tan^2X)(1 + cos2x) = 2 (prove identity)

    (1 + sinA - cos2A) / (cosA + sin2A) = tanA

    and Solve the equation for 0 < or = x < or = 2pi

    cosx + cos2x + 2sin^2x = 0

    your help is much appreciated!
    1. Use the identities \frac{\sin{x}}{\cos{x}} = \tan{x} and \cos{2x} = \cos^2{x} - \sin^2{x}.


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

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

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

     = \frac{1}{\cos^2{x}}(2\cos^2{x})

     = 2.
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  6. #6
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    Quote Originally Posted by Chop Suey View Post
    1. use the identities:
    \frac{1+\cos{2x}}{2} = \cos^2{x}

    1+\tan^2{x} = \sec^2{x}

    ==========

    Use 2 to multiply by 1 to get:

    2(1+tan^2{x})\left(\frac{1+\cos{2x}}{2}\right)

    This simplifies to 2(\sec^2{x})(\cos^2{x}) and the result follows.
    Very nice
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