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Thread: tan 89 or tan 89.9

  1. #1
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    tan 89 or tan 89.9

    Can anyone see an easy way to explain why tan 89.9 is approximately the same as 10x tan 89?
    Just messing around on the calculator and they are remarkably close..
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  2. #2
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    Re: tan 89 or tan 89.9

    Quote Originally Posted by rodders View Post
    Can anyone see an easy way to explain why tan 89.9 is approximately the same as 10x tan 89?
    Just messing around on the calculator and they are remarkably close..
    with
    Suggestion: "Play around" at this website.
    Compare with this change: see here.

    Recall that $\large\tan(\theta)=\dfrac{\sin(\theta)}{\cos( \theta)}$
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  3. #3
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    Re: tan 89 or tan 89.9

    Consider the complimentary angle to theta, which we'll call alpha and is equal to theta - pi/2. We can use the fact that cos(theta) = sin (alpha), and also the fact that as alpha approaches 0 the value of sin(alpha) is very close to alpha itself. Hence for small values of alpha the value of tan(theta) is very close to 1/alpha. So, since the angle alpha is ten times larger for theta= 89 degrees than for theta = 89.9 degrees, the tangent will be approximately 1/10 as large for theta = 89 degrees compared to theta = 89.9 degrees.
    Last edited by ChipB; Jan 6th 2018 at 08:45 AM.
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    Re: tan 89 or tan 89.9

    Hence for small values of alpha the value of tan(theta) is very close to 1/alpha.

    Can you clarify.?

    if tan (theta) = sin (theta)/cos(theta) = sin(theta)/ sin(pi/2-theta) = theta/pi/2-theta for small values of theta???? except the denominator is no longer small?
    Sorry i am getting confused?
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  5. #5
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    Re: tan 89 or tan 89.9

    Quote Originally Posted by Plato View Post
    with
    Suggestion: "Play around" at this website.
    Compare with this change: see here.

    Recall that $\large\tan(\theta)=\dfrac{\sin(\theta)}{\cos( \theta)}$
    I can see that tan89 =cot (1) here but not fully sure why?
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  6. #6
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    Re: tan 89 or tan 89.9

    Quote Originally Posted by rodders View Post
    Hence for small values of alpha the value of tan(theta) is very close to 1/alpha.

    Can you clarify.?

    if tan (theta) = sin (theta)/cos(theta) = sin(theta)/ sin(pi/2-theta) = theta/pi/2-theta for small values of theta???? except the denominator is no longer small?
    Sorry i am getting confused?
    Not quite. The angle theta is close to pi/2, so itís not small. But sin(pi/2 - theta) is small, so appromimately equal to pi/2-theta. So for theta close to pi/2 we have tan(theta) is approximately equal to sin(pi/2) / sin(pi/2 - theta), which is approximately equal to 1/(pi/2-theta). For theta equal to 89 or 89.9 degrees the denominator is pi/180 or pi/1800, respectively (in radians). Hence the tangent is approximately 180/pi or 1800/pi, respectively.
    Last edited by ChipB; Jan 7th 2018 at 07:12 AM.
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    Re: tan 89 or tan 89.9

    Quote Originally Posted by ChipB View Post
    Not quite. The angle theta is close to pi/2, so it’s not small. But sin(pi/2 - theta) is small, so appromimately equal to pi/2-theta. So for theta close to pi/2 we have tan(theta) is approximately equal to sin(pi/2) / sin(pi/2 - theta), which is approximately equal to 1/(pi/2-theta). For theta equal to 89 or 89.9 degrees the denominator is pi/180 or pi/1800, respectively (in radians). Hence the tangent is approximately 180/pi or 1800/pi, respectively.
    Excellent! THANKS
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  8. #8
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    Re: tan 89 or tan 89.9

    I think i have worked out a simpler way..

    tan 1 is approx equal to 10tan(0.1)

    so given tan 89 = 1 /tan1 and tan 89.9 = 1/(tan 0.1) the result follows?

    Does that feel ok?

    It does rely on tan x = 1/tan(90-x) , which i am not sure how to prove yet..
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