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Math Help - Trigonometry help

  1. #1
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    Trigonometry help

    I have done some of these and attempted others that I am stuck on....I have to simplify so that both sides are equal (RS and LS)...is there anyone who can take a look at my work and give me a little guidance?
    Last edited by aikenfan; November 28th 2007 at 01:24 PM.
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  2. #2
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    Hello, aikenfan!

    #3 and #6 are fine . . . good work!


    9)\;\;\sin^2\!a-\sin^4\!a \:=\:\cos^2\!a - \cos^4\!a
    Factor the left side: . \sin^2\!a\underbrace{(1-\sin^2\!a)}

    . . . . . . . . . . . . = \quad\;\;\sin^2\!a\cdot\cos^2\!a

    . . . . . . . . . . . . =\;\overbrace{(1-\cos^2\!a)}\cos^2\!a

    . . . . . . . . . . . . = \;\cos^2\!a - \cos^4\!a



    What is the right side of #12?


    15)\;\;\sin^{\frac{1}{2}}\!x\cos x = \sin^{\frac{5}{2}}\!x\cos x \:=\:\cos^3\!x\sqrt{\sin x}
    Factor the left side: . \sin^{\frac{1}{2}}\!x\cos x\left(1-\sin^2\!x\right) \;=\; \sin^{\frac{1}{2}}\!x\cos x\cdot\cos^2x

    . . =\;\cos^3\!x\sin^{\frac{1}{2}}\!x \;=\;\cos^3\!x\sqrt{\sin x}



    18)\;\;\frac{\sec x - 1}{1-\cos x} \:=\:\sec x

    The left side is: . \frac{\dfrac{1}{\cos x} - 1}{1-\cos x}

    Multiply by \frac{\cos x}{\cos x}\!:\;\;\frac{\cos x}{\cos x}\cdot\frac{\frac{1}{\cos x} - 1}{1 - \cos x} \;=\;\frac{1-\cos x}{\cos x(1-\cos x)} \;=\;\frac{1}{\cos x} \;=\;\sec x



    21)\;\;\csc x\cos x \:=\:\sin x\tan x
    This one isn't true . . . Is there a typo?

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  3. #3
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    Wow! Thank you very much for looking at my work...on number 21, it is minus on the left side not times.
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