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Math Help - Trig Review Help

  1. #1
    Junior Member
    Joined
    Aug 2009
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    Trig Review Help

    1.
    Simplify the following expression in terms of x
    sq(4tan^2x+4)

    I did:
    sq(4(tan^2x+1))
    sq(4sec^2x)
    =+/- 2secx

    The correct answer is 2abs(secx) and I don't understand why, so I'd appreciate it if someone would explain it to me.


    2.
    Write tan(arc cos(x/3)) in algebraic form
    I don't remember how to do this at all, so detailed explanation on what to do would be extremely helpful.

    Any help is greatly appreciated.
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  2. #2
    MHF Contributor
    Grandad's Avatar
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    Hello Naples
    Quote Originally Posted by Naples View Post
    1.
    Simplify the following expression in terms of x
    sq(4tan^2x+4)

    I did:
    sq(4(tan^2x+1))
    sq(4sec^2x)
    =+/- 2secx

    The correct answer is 2abs(secx) and I don't understand why, so I'd appreciate it if someone would explain it to me.
    You've done all the hard work! It's simply that the original expression, \sqrt{4\tan^2x+4}, is written as a positive square root. Hence it becomes the positive square root of 4\sec^2x, which is written as an absolute value: 2|\sec x|, so that when x would make \sec x negative, it still gives the corresponding positive value.

    2.
    Write tan(arc cos(x/3)) in algebraic form
    I don't remember how to do this at all, so detailed explanation on what to do would be extremely helpful.

    Any help is greatly appreciated.
    Let y = \arccos(x/3). Then we shall need to find \tan y.

    \cos y = \frac x3

    \Rightarrow \sec y = \frac3x

    \Rightarrow \sec^2 y = 1+\tan^2 y= \frac{9}{x^2}

    \Rightarrow \tan^2y= \frac{9}{x^2}-1=\frac{9-x^2}{x^2}

    \Rightarrow \tan y = \tan(\arccos (x/3) = \sqrt{\frac{9-x^2}{x^2}}

    Grandad
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