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Math Help - Trigonometric Identities

  1. #1
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    Trigonometric Identities

    Establish each identity below.

    (1) 9 sec^2 (x) - 5 tan^2 (x) = 5 + 4 sec^2 (x)

    (2) cos(x)/[1 + sin(x)] + [1 + sin(x)]/cos(x) = 2 sec (x)
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  2. #2
    o_O
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    #1:
    Note that \tan^{2} x = \sec^{2} x - 1. So substitute it in and simplify.

    #2:
    Combine into a single fraction:
    \frac{\cos x}{1+\sin x} {\color{blue}\cdot \frac{\cos x}{\cos x} } \: + \: \frac{1+ \sin x}{\cos x} {\color{blue} \cdot \frac{1+ \sin x}{1 + \sin x} }

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

    Expand the numerator and simplify.

    Post back if you have any troubles!
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  3. #3
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    Where???

    Quote Originally Posted by o_O View Post
    #1:
    Note that \tan^{2} x = \sec^{2} x - 1. So substitute it in and simplify.

    #2:
    Combine into a single fraction:
    \frac{\cos x}{1+\sin x} {\color{blue}\cdot \frac{\cos x}{\cos x} } \: + \: \frac{1+ \sin x}{\cos x} {\color{blue} \cdot \frac{1+ \sin x}{1 + \sin x} }

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

    Expand the numerator and simplify.

    Post back if you have any troubles!
    What in the question indicated that you needed to make cosx/cosx
    on the left side and 1 + sinx/1 + sinx on the right side?
    Last edited by magentarita; August 23rd 2008 at 04:28 PM. Reason: Rewrite
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  4. #4
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    What in the question indicated that you needed to make cosx/cosx
    on the left side and 1 + sinx/1 + sinx on the right side?
    basic addition of fractions ... you need the common denominator, \cos{x}(1 + \sin{x}), to add the two fractions.
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  5. #5
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    I see..

    Quote Originally Posted by skeeter View Post
    basic addition of fractions ... you need the common denominator, \cos{x}(1 + \sin{x}), to add the two fractions.
    It's searching for the LCD, right?
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  6. #6
    o_O
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    An analogy:

    \frac{1}{2} + \frac{2}{3}

    = \frac{1}{2} {\color{blue}\cdot \frac{3}{3} } + \frac{2}{3} {\color{blue}\cdot \frac{2}{2}}

    = \frac{3}{6} + \frac{4}{6}

    =\frac{7}{6}
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  7. #7
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    Great...

    Great analogy!

    Thanks
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