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?
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3 Answers
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The $\tan$ function can be described four different ways that I can describe and each adds to a fuller understanding of the tan function.
First, the basics: the value of $\tan$ is equal to the value of $\sin$ over $\cos$.
$$\\tan(45^\circ)=\frac{\sin(45^\circ)}{\cos(45^\circ)}=\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}=1$$So, the $\tan$ function for a given angle does give the slope of the radius, but only on a unit circle or only when the radius is one. For instance, when the radius is 2, then $2\tan(45^\circ)=2$, but the slope of the 45 degree angle is still 1.
The value of the $\tan$ for a given angle is the length of the line, tangent to the circle at the point on the circle intersected by the angle, from the point of intersection (A) to the $x$-axis (E).
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- The value of the tangent line can also be described as the length of the line $x=r$ (which is a vertical line intersecting the $x$-axis where $x$ equals the radius of the circle) from $y=0$ to where the vertical line intersects the angle.
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The explanations in examples 3 and 4 might seem counter intuitive at first, but if you think about it, you can see that they are really just reflections across a line of half the specified angle. Image to follow.
The images included are both from Wikipedia.
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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.
answered Dec 27, 2012 at 2:43
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2$\begingroup$ The latin tangere means "to touch". For the calculus definition at least, this probably relevant. $\endgroup$ Commented Dec 27, 2012 at 3:51
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$\begingroup$ Interesting! Never thought of that... like "tangible".Here is an (unsourced) description of origins of the names of the trigonometric functions. $\endgroup$ Commented Dec 27, 2012 at 4:52
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$\begingroup$ @AlfonsoFernandez: Slight correction to the linked page: Arab jayb, straight bow string, was reused by Arab mathematicians to denote the projected sine. In Latin, sine meant something curved (sail/toga fabric undulations). Europeans looking at Arabs sine curve thought they used jayb for the undulation idea, and associated it with their sine. However the bow idea was also kept: The circle arc was associated with Latin arcus (bow, arc in Fr.), and the bow string idea gave chorda (bowel), the string used for bows and musical instruments (corde in Fr.) $\endgroup$– minsCommented Oct 30, 2022 at 8:08
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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"?
answered Dec 27, 2012 at 2:34
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$\begingroup$ I 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. $\endgroup$ Commented Dec 27, 2012 at 2:44
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$\begingroup$ Yes, 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. $\endgroup$ Commented Dec 27, 2012 at 2:46
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$\begingroup$ That'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. i.imgur.com/eNm0j.jpg $\endgroup$ Commented Dec 27, 2012 at 2:57
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$\begingroup$ Remember 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 $\endgroup$ Commented Dec 27, 2012 at 3:01
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$\begingroup$ How would you draw that on the unit circle? $\endgroup$ Commented Dec 27, 2012 at 3:08
