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Math Help - Substitute x= 3 sec(theta) and simplify

  1. #1
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    Substitute x= 3 sec(theta) and simplify

     \frac{\sqrt{x^{2}+9}}{x^{2}} . Help write in terms of sines and cosines can anyone give me steps on how to do this problem
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    Re: Substitute x= 3 sec(theta) and simplify

    Quote Originally Posted by mathisfun26 View Post
     \frac{\sqrt{x^{2}+9}}{x^{2}} . Help write in terms of sines and cosines can anyone give me steps on how to do this problem
    Well if
    x = 3~sec( \theta) then x^2 = 9~sec^2( \theta )

    So
    \frac{\sqrt{x^2 + 9}}{x^2} = \frac{\sqrt{9~sec^2( \theta ) + 9}}{9~sec^2( \theta)}

    How do you simplify this?

    -Dan
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    Re: Substitute x= 3 sec(theta) and simplify

    Quote Originally Posted by mathisfun26 View Post
     \frac{\sqrt{x^{2}+9}}{x^{2}} . Help write in terms of sines and cosines can anyone give me steps on how to do this problem
    seems easier to use x = 3\tan{\theta}
    Thanks from topsquark
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    Re: Substitute x= 3 sec(theta) and simplify

    I'm not sure how to simplify it
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    Re: Substitute x= 3 sec(theta) and simplify

    Are you specifically instructed to use the given substitution? The substitution suggested by skeeter makes much more sense to use.
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    Re: Substitute x= 3 sec(theta) and simplify

    Yes I was given that as the instruction
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    Re: Substitute x= 3 sec(theta) and simplify

    Okay, look at what topsquark posted. Try factoring the radicand, and dividing out factors common to the numerator and denominator. What do you get?
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    Re: Substitute x= 3 sec(theta) and simplify

     \frac{3 csc\Theta }{{}9 sec\Theta ^{^{2}}}
    im not sure this is right
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    MHF Contributor MarkFL's Avatar
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    Re: Substitute x= 3 sec(theta) and simplify

    How did you arrive at that result?
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    Re: Substitute x= 3 sec(theta) and simplify

    i factored out a 9 then it became 3.. and secant +1 became csc help me please
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    MHF Contributor MarkFL's Avatar
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    Re: Substitute x= 3 sec(theta) and simplify

    Factoring out the 9, which the square root makes it a 3 is good:

    \frac{\sqrt{9\sec^2(\theta)+9}}{9\sec^2(\theta)}= \frac{\sqrt{9(\sec^2(\theta)+1)}}{9\sec^2(\theta)}  =\frac{3\sqrt{\sec^2(\theta)+1}}{9\sec^2(\theta)}=

    \frac{\sqrt{\sec^2(\theta)+1}}{3\sec^2(\theta)}

    However, \sec^2(\theta)+1\ne\csc^2(\theta). You are further instructed to write in terms of sines and cosines. How can you do this with the secant function?
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  12. #12
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    Re: Substitute x= 3 sec(theta) and simplify

    secant= 1/ cosine

     \frac{\sqrt{{\frac{1}{cos\Theta ^{2}} +1}}}{3\frac{1}{cos\Theta ^{2}}+1}
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    Re: Substitute x= 3 sec(theta) and simplify

    Where did the + 1 in the denominator come from?
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    Re: Substitute x= 3 sec(theta) and simplify

    my bad my mistake..
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    MHF Contributor MarkFL's Avatar
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    Re: Substitute x= 3 sec(theta) and simplify

    Also, I would use standard notation for the square of a trig function, i.e.:

    \cos^2(\theta)

    The way you have it written implies the argument (angle) is being squared.

    So, we now have:

    \frac{\sqrt{\frac{1}{\cos^2(\theta)}+1}}{\frac{3}{  \cos^2(\theta)}}

    How can we rewrite the division as a multiplication?
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