# Trigonometric ratios for angles greater than $90^\circ$?

The trigonometric ratios of an angle greater than $90^\circ$ are equal to the supplementary angle's ratios.

I'm just clarifying this, but the ratios don't actually exist for angles greater than $90^\circ$ right? (since by definition these ratios are of a right angle triangle). Is it just mathematical custom to assume the ratios of an angle greater than $90^\circ$ to be equivalent to its supplementary angle's ratios? For convenience?

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It helps to visualize a circle of radius 1 in the $xy$ plane. The sine of the angle, which is measured counter-clock-wise, with the $x$ axis being zero, is $y$ and the cosine is $x$. As we pass 90 degrees, we cross the $y$ axis ans $x$ (i.e. sine) swings negative. – Kaz Apr 25 '12 at 1:07

Instead of ratios in right triangles (which as you notice make sense only for acute angles), one can consider the cosine and sine defined as the $x$ and $y$ coordinate of a point that moves around a unit circle. This works for all angles -- and for acute angles you can inscribe a right triangle in the first quadrant of the unit circle and see that the unit-circle definition matches the right-triangle one.

After 90°, the cosine becomes negative, because the point is now to the left of the $y$ axis (so the $x$ coordinate is negative).

After 180°, the sine becomes negative too -- both coordinates of the moving point are now negative.

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Wouldn't the tangent become negative too? Considering it too takes the y coordinate into account? – user26649 Apr 25 '12 at 1:14
Yes, the tangent becomes (suddenly) negative after 90° too, and then smoothly rises to become zero at 180° and positive between 180° and 270°. – Henning Makholm Apr 25 '12 at 1:16
Awesome, thank you. – user26649 Apr 25 '12 at 1:54

There are two possible definitions of the trigonometric ratios:

• The trigonometric ratios can be defined for angles greater than $0^\circ$ and less than $90^\circ$ using right triangles. In particular, $\sin(\theta)$ is defined as the ratio of the lengths of the opposite leg and the hypotenuse, and $\cos(\theta)$ is defined as the ratio of the lengths of the adjacent leg and the hypotenuse.

• The trigonometric ratios can be defined for any angle using the unit circle. In this definition, $\sin(\theta)$ is the $y$-coordinate of a point on the unit circle with angle $\theta$, and $\cos(\theta)$ is the $x$-coordinate of a point on the unit circle with angle $\theta$.

The unit circle definition is the same as the triangle definition for angles between $0$ and $90^\circ$, but is more general since it works for any angle. The following picture from Wikipedia illustrates this definition:

For each point, the $x$-coordinate is the cosine, and the $y$-coordinate is the sine.

This picture only shows angles between $0^\circ$ and $360^\circ$, but you can extend to less than $0^\circ$ by continuing clockwise around the circle, or to greater than $360^\circ$ by continuing counterclockwise.

The following pictures show graphs of $\sin(x)$ and $\cos(x)$ for $-2\pi\leq x\leq2\pi$. (The $x$-axis is the angle measured in radians.)

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Yes, except the cosine gets a negative sign. You're right - you can't make a right triangle containing one of these angles and so the classical definition of sine and cosine as ratios in right triangles does not make sense for them. You can take it as a "convention" that the angles get the definition they do, or you can think of it a different way.

An alternative interpretation of trig functions that makes sense for any angle, even negative angles or angles above 360, is that the cosine of $\theta$ is the $x$-coordinate of the point on the unit circle (the circle of radius 1 with center at the origin) whose counterclockwise angle from the positive $x$ axis is $\theta$, and the sine of $\theta$ is the $y$-coordinate of that point. If you draw the picture you'll see why the two definitions agree for $\theta$ between 0 and 90.

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Thanks! Just one thing - I assume the cosine becomes negative because the x coordinate of the supplementary angle becomes negative - but then why isn't the tangent negative too? Since its (y/x) – user26649 Apr 25 '12 at 1:06

You have the following relationships, where two have a sign change:

• $\sin(180^\circ - x^\circ) = \sin(x^\circ)$
• $\cos(180^\circ - x^\circ) = -\cos(x^\circ)$
• $\tan(180^\circ - x^\circ) = -\tan(x^\circ)$

One way to think about angles greater than $90^\circ$ is with Cartesian co-ordinates with $x$ and $y$ axes (so some of $x$ and $y$ values can be negative, but the radius or hypotenuse to the origin is non-negative) and the angle is measured anti-clockwise from the $x$ axis.

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