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I can’t find an easy explanation for negative values of either/or sine, cosine and tangent when applied to angles outside $0-90^0$.

I tried to reason using the cosine law where is obvious that instead of cosine of an obtuse angle the negative cosine of the supplementary (acute) angle is used, but that is more of a redo of the cosine law than a reason for the negative value of the cosine of an obtuse angle.

I’ve tried to rely on unit circle to imagine $sine=\frac{vertical}{radius}$ while tracking a star. This led to $cosine=\frac{horizontal}{radius}$. I’m not into astronomy though. Then I looked at vertical measurement as positive when measured from ground up and negative when measured from ground under (when star falls under horizon). As for horizontal measurement, I took it negative ( as in “opposite direction”) after the observer needed to turn around to keep tracking a descending star. I did this in order to fit the unit circle.

This could explain the negative values the trigonometric functions take for angles outside $0-90^0$

Is there an easier explanation to why the trigonometric functions sometimes return negative values?

WindSoul
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  • The unit circle is exactly the right way to think about it: https://en.wikipedia.org/wiki/File:Sinus_und_Kosinus_am_Einheitskreis_1.svg –  Jan 29 '20 at 04:42
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    Perhaps my answer here will be helpful. – Blue Jan 29 '20 at 04:45
  • I like your answer a lot. I have also used a similar way to determine the sine and cosine of sum and difference by using rotation, but I could not extrapolate from there to obtuse or negative angles. Your diagrams are so easy to understand and remember! This answers my question. – WindSoul Jan 29 '20 at 05:19
  • So you can plot on a graph. This little animation shows is helpful to visualise the negative values https://www.mathsisfun.com/algebra/icircle-triangle.html – Neil Jun 23 '20 at 11:49

2 Answers2

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In the case of the tangent function, $\tan(\theta)$ is simply the slope of the terminal side of $\theta$, when the angle is in standard position. Slopes can be negative, therefore tangent values can be negative.

As for cosines, if you believe the double-angle formula for cosine, apply it to a $60^\circ$ angle, and see what happens! With sines, use the identity for the sine of a difference to find $\sin(30^\circ - 60^\circ)$. These come out negative, so how are we supposed to save these formulas if the functions can only ever be positive?

A good definition that doesn't raise these questions is this: Place your angle $\theta$ in standard position, i.e., vertex at $(0,0)$, and initial side on the positive $x$-axis. If $(x,y)$ is a point on the terminal side of the angle, then define $r=\sqrt{x^2+y^2}$, which is always positive. Now we can define: $\sin(\theta)=\frac{y}{r}, \cos(\theta)=\frac{x}{r}, \tan(\theta)=\frac{y}{x}$.

G Tony Jacobs
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  • The issue is this: sine, cosine and tangent are defined using SOH-CAH-TOA. This requires the existence of the right triangle, but then the angle must be acute. How would one build opposite over hypothenuse for an obtuse angle? The school days consider the legs either positive or negative in length to solve the sign, but then why the hypothenuse is always considered positive? I can use the ratios of sums or differences of acute angles, but only when the result is also an acute angle. Beyound that, I have to consider right triangles with negative side lengths, but only for legs never for hyp – WindSoul Feb 02 '20 at 19:42
  • You either have to consider right triangles with "directed" sides, in the sense of allowing side lengths to be positive or negative according to which direction they point, or else you forget about right triangles, and use a more general definition that happens to agree with SOHCAHTOA when the angles are acute. If you think of directed sides, then the reason the hypotenuse is always positive is that it's a distance, while the other two are coordinates. – G Tony Jacobs Feb 08 '20 at 00:38
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    They don't have to be defined using SOHCAHTOA, and if they are, that definition has to be abandoned in favor of a more general one at some point. – G Tony Jacobs Feb 08 '20 at 00:40
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That's just how they are defined -- wlog as the rectangular Cartesian coordinates of points on the unit circle corresponding to the angle. And we get consistent results when the angles are acute.

Allawonder
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