So, a 45 degree angle in the unit circle has a tan value of 1. Does that mean the slope of a tangent line from that point is also 1? Or is something different entirely?
Is the tangent function (like in trig) and tangent lines the same?
3 Answers
Have a look at this drawing from Wikipedia: Unit Circle Definitions of Trigonometric Functions.
When viewed this way, the tangent function actually represents the slope of a line perpendicular to the tangent line of that point (i.e. the slope of the radius that touches the angle point).
However, you can actually see that the "tangent line", consisting the values of the tangents, is the actual tangent line of the circle at the point from which the angles are measured, and I would guess that this is the source of the name.
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1The latin *tangere* means "to touch". For the calculus definition at least, this probably relevant. – 2012-12-27
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0Interesting! Never thought of that... like "tangible".[Here](http://www.algebralab.org/lessons/lesson.aspx?file=Trigonometry_TrigNameOrigins.xml) is an (unsourced) description of origins of the names of the trigonometric functions. – 2012-12-27
Yes and no, resp.: yes, any line in the plane that forms an angle of $\,45^\circ\,$ with the positive direction of the $\,x-$axis has a slope of $\,\tan 45^\circ=1\,$, and no: it isn't something different.
It is not completely clear though what you mean by "tangent line"...perhaps you meant "tangent line at some point on the graph of a (derivable) function"?
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0I don't know to reply directly to you, Don. Anyway, what I mean is this. tan(60 degrees) is sqrt(3). What's that mean though? Does it mean the slope of a tangent line at the point, let's use the unit circle, (1/2, sqrt(3)/2) is sqrt(3)? My confusion is what that sqrt(3) is meant to represent. – 2012-12-27
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0Yes, of course...in fact, that is the direct definion from trigonometry and a straight angle triangle: $$\tan 60^\circ=\frac{\text{opposite leg}}{\text{adjacent leg}}=\frac{\frac{\sqrt 3}{2}}{\frac{1}{2}}=\sqrt 3$$ The $\,\sqrt 3\,$ represents the ration of the lengths of the two legs in that triangle. – 2012-12-27
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0That's what I thought. Thank you, but I have a follow up question. If tan if y/x (unit circle), then tan at 0 degrees is 0/1, which is zero. But wouldn't a tangent line from that point just be a vertical line, which has an undefined slope? Similarly, at a 90 degree angle tan = 1/0, which is undefined, but a tangent line from that point would be horizontal, no? Here's an image that I hope will show the contradiction that I'm trying to get cleared up. http://i.imgur.com/eNm0j.jpg – 2012-12-27
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0Remember that a horizontal line (=parallel to the $\,-$axis) is the one that has zero slope = $\,\tan 0\,$ , as the difference between the $\,y-$coordinates is zero. A vertical line has no defined slope, though later (much later) one could assign it, under certain conditions, an infinite slope. BTW, your drawing shows you're confusing the definition of tangent: it is "difference of y-coordinates divided by difference of x-coordinates", as long as the later is non-zero – 2012-12-27
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0How would you draw that on the unit circle? – 2012-12-27
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0Just as you did but with the $\,\frac{1}{0}\,\,,\,\,\frac{0}{1}\,$ interchanged... – 2012-12-27
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0Why are they being changed? This is why I am confused. Please, let me try to clarify. tan = sin/cos. tan(0 degrees) = 0/1 because sin(0 degrees) = 0; cos(0 degrees) = 1. Therefore, tan is zero. But visually (my picture I posted), the tangent line is vertical. Wouldn't the slope then be undefined (or infinity, but I'm not sure we're dealing with the certain conditions you were referring to; either way, it wouldn't be zero). Same with tan of 90 degrees: sin = 1; cos = 0; so tan is undefined, but visually it's a horizontal line, which has a slope of zero, not undefined. – 2012-12-27
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0No. "Visually", the line is a *horizontal* one, not a vertical one! You have the two things confused: $\,\tan 0 = 0\,$ is the slope of a HORIZONTAL LINE, this is what you *must* denote by $\,\frac{0}{1}\,$...just check where you have the difference of y-coordinates zero! – 2012-12-27
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0I think you two are talking about two different lines. Don is talking about the radius from the center to the point on the circle. @user is talking about the tangent to the circle at that point. The slope of the radius is $\tan\theta$, but the slope of the tangent is $1/\tan\theta$. – 2012-12-27
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0Nop, I've been talking all the time about the two colored tangent lines (blue and red) to that circle that the OP drew in that link he posted. – 2012-12-27
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0OK, in that case, perhaps you disagree about which angle you are taking the tangent of. From @user54350's first two comments, I gather they are talking about the tangent at the point $(\cos\theta,\sin\theta)$ on the unit circle. You seem to be talking about the angle made by that line with the horizontal axis, which is not the same. – 2012-12-27
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0Forget the comments: take a peek at the diagram drawn in the link the OP sent. – 2012-12-27