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Math Help - trigonometry, inequation

  1. #1
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    trigonometry, inequation

    This is the question from memory, i may have potentially made a mistake, but i do believe it is correct. This should all be done without a calculator.

    For the equation y = \arctan(x-1) - a\tan(\frac{\pi}{8}), find the minimum value for 'a' where y > 0 for all of x. Leave in exact answer form.

    Note, on a previous question i was asked to show (without a calculator) that \tan(\frac{\pi}{8}) = \sqrt{2} - 1, so that may be relevant.

    In fact, i would appreciate if you could explain the second question too.

    Thanks
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  2. #2
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    Quote Originally Posted by scorpion007 View Post
    This is the question from memory, i may have potentially made a mistake, but i do believe it is correct. This should all be done without a calculator.

    For the equation y = \arctan(x-1) - a\tan(\frac{\pi}{8}), find the minimum value for 'a' where y > 0 for all of x. Leave in exact answer form.

    Note, on a previous question i was asked to show (without a calculator) that \tan(\frac{\pi}{8}) = \sqrt{2} - 1, so that may be relevant.

    In fact, i would appreciate if you could explain the second question too.

    Thanks
    Let b=a\tan (\pi/8)
    You want the range of the strictly monotomic function (an increasing function) of \tan^{-1}(x-1)+b>0

    The regular arc-tangent function -\pi/2<y<\pi/2 has this range. If you add \pi/2 you have, 0<y+\pi/2<\pi
    Thus,
    \tan^{-1}(x-1)+\pi/2
    That means,
    b=\pi/2=a(\sqrt{2}-1)
    Thus,
    a=\frac{\pi}{2(\sqrt{2}-1)}
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  3. #3
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    could you help me prove that
    \tan(\frac{\pi}{8}) = \sqrt{2} - 1 ?
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  4. #4
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    Quote Originally Posted by scorpion007 View Post
    could you help me prove that
    \tan(\frac{\pi}{8}) = \sqrt{2} - 1 ?
    You need to know that \tan(\pi/4)=1,\ \sin(\pi/4)=\cos(\pi/4)=1/\sqrt{2}, then use the 1/2-angle formula for 1st quadrant angles:

    <br />
\tan(A/2) = \frac{\sqrt{1-\cos(A)}}{\sqrt{1+\cos(A)}}=\frac{\sin(A)}{1+\cos(  A)}<br />

    Put A=\pi/4 and substitute in the know values of the trig functions for this angle, and then rationalise.

    RonL
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  5. #5
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    hmm, that method works wonderfully.
    Unfortunately, i never learned that formula (its not in my textbook). Is it anything like this one?

    \tan(2x) = \frac{2\tan(x)}{1-\tan^2(x)}

    Is it possible to use that one to solve it?
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  6. #6
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    Quote Originally Posted by scorpion007 View Post
    ...Is it anything like this one?

    \tan(2x) = \frac{2\tan(x)}{1-\tan^2(x)}
    Is it possible to use that one to solve it?
    No,

    that's the double-angle formula. CaptBlack used the half-angle formula:
    <br />
\tan(A/2) = \frac{\sqrt{1-\cos(A)}}{\sqrt{1+\cos(A)}}=\frac{\sin(A)}{1+\cos(  A)}<br />
= \frac{1-\cos(A)}{\sin(A)}

    Your formula can be completed to:

    \tan(2x) = \frac{2\tan(x)}{1-\tan^2(x)}= \frac{2}{\cot(x)-\tan(x)}

    (If you solve your equation for tan(x) you will probably come out with CaptBlack's formula)

    EB
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