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Math Help - Simplifying trigonometric functions

  1. #1
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    Simplifying trigonometric functions

    Where do I go on a problem like this? I am supposed to simplify them...I have done about 10 of these earlier today, but I am stuck on these last two...I know that there is probably more than one way to go on these, but I am having a hard time getting started
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  2. #2
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    for 66, I have done:
    Common denominator

    (Sec x - 1) - (sec x + 1)/ (Sec x - 1)(sec x + 1)
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  3. #3
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    Quote Originally Posted by aikenfan View Post
    for 66, I have done:
    Common denominator

    (Sec x - 1) + (sec x + 1)/ (Sec x - 1)(sec x + 1)
    that is wrong. you were on the right track, but you messed up the signs

    \frac 1{\sec x + 1} - \frac 1{\sec x - 1} = \frac {\sec x - 1 - \sec x - 1}{(\sec x + 1)(\sec x - 1)}

    note that the denominator is the difference of two squares, thus we have:

    \frac {-2}{\sec^2 - 1}

    now what?


    continue the same way for the next problem. you don't have many options besides combining the fractions, follow your nose
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  4. #4
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    this would make it -2/tan^2x
    because of Pythagorean Identities: tan^2x = sec^2x - 1
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    Quote Originally Posted by aikenfan View Post
    this would make it -2/tan^2x
    because of Pythagorean Identities: tan^2x = sec^2x - 1
    yes. but what is 1/tan^2(x) ?
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    cot^2x
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    Quote Originally Posted by aikenfan View Post
    cot^2x
    yes, so we have -2cot^2(x)

    now, move on to the second
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    So far, I have got:
    (cosx(1 + sinx) + 1 + sinx(1 + sinx))/(1 + sinx)(cosx)
    I am guessing that I have to distribute these
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    Quote Originally Posted by aikenfan View Post
    So far, I have got:
    (cosx(1 + sinx) + 1 + sinx(1 + sinx))/(1 + sinx)(cosx)
    I am guessing that I have to distribute these
    yup, that would be a good idea. again, you don't have many other options, so just follow your nose
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  10. #10
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    I get:
    (cosx + cosx(sinx)) + (1 + sinx + sinx + sinx^2)/ (1 + sinx)(cosx)

    = (cosx + cosx(sinx)) + (1 + 2sinx + sinx^2)/ (1 + sinx)(cosx)
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  11. #11
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    Quote Originally Posted by aikenfan View Post
    I get:
    (cosx + cosx(sinx)) + (1 + sinx + sinx + sinx^2)/ (1 + sinx)(cosx)

    = (cosx + cosx(sinx)) + (1 + 2sinx + sinx^2)/ (1 + sinx)(cosx)
    where did cosx(sinx) come from, you shouldn't have that
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  12. #12
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    Hello, aikenfan!

    There is a trick we can use on the second problem . . .


    68)\;\;\frac{\cos x}{1 + \sin x} + \frac{1 + \sin x}{\cos x}
    Multiply the first fraction by \frac{1-\sin x}{1-\sin x}

    . . We have: . \frac{\cos x}{1 + \sin x}\cdot{\color{blue}\frac{1-\sin x}{1-\sin x}} \;\;=\;\;\frac{\cos x(1 -\sin x)}{1-\sin^2\!x} \;\;=\;\;\frac{\cos x(1-\sin x)}{\cos^2\!x} \;\;=\;\;\frac{1-\sin x}{\cos x}


    The problem becomes: . \frac{1-\sin x}{\cos x} + \frac{1+\sin x}{\cos x} \;\;=\;\;\frac{2}{\cos x} \;\;=\;\;2\sec x

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  13. #13
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    Thank you both very much for all of your time and help!
    Last edited by aikenfan; November 19th 2007 at 12:40 PM.
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