# Is there a relationship between trigonometric functions and their “co” functions?

We all know that sine is one over cosecant, cosine is one over secant, etc.

But is there any relationship between sine and cosine, secant and cosecant, and tangent and cotangent?

What I am asking here is the meaning of the "co-" prefix.

Thanks.

There is; they are the function applied to complement of the given angle:

$$\cos(x) = \sin(\frac{\pi}{2} - x)$$

$$\csc(x) = \sec(\frac{\pi}{2} - x)$$

$$\cot(x) = \tan(\frac{\pi}{2} - x)$$

These hold for all $x$, not just for $x < \frac{\pi}{2}$. Also note that these relationships hold in the reverse direction, since taking the complement of a complement results in the original angle.

• cosinus meant indeed complementaris sinus (sine of the complementary) and so on. – egreg Apr 9 '15 at 22:55
• gasp! How can it be that in decades of using (and indeed on occasion teaching) these functions, I've never realised this? And it's obvious now you point it out. – Norman Gray Apr 9 '15 at 23:24
• @JohnColanduoni True, true. But what makes it worse is that in the context of spherical trig, the colatitude is often used (90 degrees minus the latitude), or zenith distance (which one could call co-altitude), so the clue about the ‘co-’ prefix has always been right there in front of me. Hah! – Norman Gray Apr 9 '15 at 23:32
• And in case it isn't obvious, the two non-right angles in a right-angled triangle are each other's complements. So you can also construe the "co-" versions as exchanging opposite for adjacent edge, at least in the first quadrant. – Steve Jessop Apr 9 '15 at 23:52
• To elaborate on @SteveJessop's comment (see Wikipedia Right-angled triangle definitions): We can first define sine as $\frac{\text{opposite}}{\text{hypotenuse}}$, define tangent as $\frac{\text{opposite}}{\text{adjacent}}$, and secant as $\frac{\text{hypotenuse}}{\text{adjacent}}$. After that we can define the three "co-" functions by exchanging the words "opposite" and "adjacent". As Steve says, going "co-" corresponds to shifting to the other acute angle in the right triangle. – Jeppe Stig Nielsen Apr 10 '15 at 12:55

For acute angles we can picture this relationship inside a right-angled triangle.

The two angles $\theta$ and $\phi$ are complementary, i.e. sum to a right angle, since the angles in a triangle must sum to two right angles. But depending on which angle we choose as our angle of interest, what side counts as "adjacent" and which as "opposite" changes. So we have:

$$\sin \theta \equiv \frac{\text{opp}(\theta)}{\text{hyp}} \equiv \frac{\text{adj}(\phi)}{\text{hyp}} \equiv \cos \phi \equiv \cos \left(\frac {\pi}{2} - \theta \right)$$

$$\tan \theta \equiv \frac{\text{opp}(\theta)}{\text{adj}(\theta)} \equiv \frac{\text{adj}(\phi)}{\text{opp}(\phi)} \equiv \cot \phi \equiv \cot \left(\frac {\pi}{2} - \theta \right)$$

$$\sec \theta \equiv \frac{\text{hyp}}{\text{adj}(\theta)} \equiv \frac{\text{hyp}}{\text{opp}(\phi)} \equiv \csc \phi \equiv \csc \left(\frac {\pi}{2} - \theta \right)$$

For readers unfamiliar with radian measure, replace the mysterious $\frac{\pi}{2}$ by $90°$. This is why, for instance, $\sin 30° = \cos 60°$. For negative angles or those beyond a right angle, it is necessary to consider the unit circle rather than SOHCAHTOA as a definition for the trigonometric functions (image source):

The radius making an angle $\theta$ with the positive $x$-axis intersects the circumference at our point of interest (marked red in the diagram). The $y$-coordinate of this point is defined as $\sin \theta$. For acute angles this is the length of the vertical line segment from the point to the $x$-axis, but for angles in the third and fourth quadrants (when the point is below the $x$-axis, so the $y$-coordinate is negative) $\sin \theta$ is the negative of this length.

Extending the radius to meet the (vertical) tangent to the circle at $(1,0)$ gives the secant line. The $y$-coordinate of the point of intersection between tangent and secant is $\tan \theta$ — can you see the problem with $\tan 90°$ and $\tan 270°$? For acute angles this is simply the length of that segment of the tangent, but for angles in the second and fourth quadrants we find that $\tan \theta$ is negative. Similarly, for acute angles, $\sec \theta$ is the length of the segment of the secant from the origin to the tangent, but for angles in the second and third quadrants it is the negative of that length (these are the cases in which the secant has to "come out of the back of the circle": the segment we measure intersects the circumference not at our original point of interest but at its antipodal point).

The cosine, cosecant and cotangent are defined in an analogous way, but switching "horizontal" and "vertical", $x$- and $y$-coordinates and axes, and the tangent at $(1,0)$, where $\theta$ is zero, with the tangent at $(0,1)$, where $\theta$ is a right-angle. This is essentially a reflection across the 45° line. This shows how the three trigonometric identities above hold for angles which are not acute also. You may want to draw a sketch graph of each trigonometric function and then sketch its reflection across $\theta = 45°$ (which will be a vertical line in your graph). For instance, here the reflection of the sine graph (blue) gives us the cosine graph (red). Note that in general the transformation from $y=f(x)$ to $y=f(2k-x)$ is a reflection across $x=k$, where in our case $k=45°$.

I want to give a couple of practical examples of why we might care about the complementary trigonometric identities, but first an aside. You may have noticed that in my definition of the trig functions from the unit circles I didn't make use of the reciprocal trig identities (that $\sec$ is the reciprocal of $\cos$ etc). In fact, inspection of the two diagrams reveals three similar right-angled triangles yielding:

$$\cos \theta : \sin \theta : 1 \equiv \cot \theta : 1 : \csc \theta \equiv 1 : \tan \theta : \sec \theta$$

From this ratio, we can deduce the reciprocal identities, and facts such as $\frac{\sin \theta}{\cos \theta} \equiv \frac{1}{\cot \theta} \equiv \tan \theta$ (divide the second term by the first). If I don't have time to teach the whole unit circle, my cut-down explanation is that the $x$-coordinate of the point of interest is $\cos \theta$, the $y$-coordinate is $\sin \theta$ and the gradient of the radius is $\tan \theta$. This obscures the geometric significance of the reciprocal trig functions, and why $\tan$ is short for "tangent", but does emphasise the $\frac{\sin \theta}{\cos \theta}$ identity and gives a neat way to see the problem with $\tan 90°$. Sadly the gradient interpretation for $\tan$ doesn't extend nicely to $\cot$, since what was $\frac{\Delta y}{\Delta x}$ becomes $\frac{\Delta x}{\Delta y}$ after reflection across the 45°-line (because the $x$ and $y$ swap), which is the reciprocal of a gradient rather than a gradient, so is harder to interpret. However, that does suffice to show that $\cot \theta$ (i.e. the $\tan$ of the complementary angle) is the reciprocal of $\tan \theta$, without drawing tangent and secant lines onto the unit circle or considering the similar triangles.

Get two trig identities for the price of one

Suppose we have deduced an identity such as $\tan^2 \theta + 1 \equiv \sec^2 \theta$ (for instance, by dividing $\sin^2 \theta + \cos^2 \theta \equiv 1$ by $\cos^2 \theta$, or by performing Pythagoras on one of those similar right-angled triangles I mentioned earlier). Since this is true for any value of $\theta$ we can replace $\theta$ by $90° - \theta$. We then obtain $\tan^2 (90° - \theta) + 1 \equiv \sec^2 (90° - \theta)$ and hence $\cot^2 \theta + 1 \equiv \csc^2 \theta$.

Similarly from $\tan \theta \equiv \frac{\sin \theta}{\cos \theta}$ we deduce $\tan (90° - \theta) \equiv \frac{\sin (90° - \theta)}{\cos (90° - \theta)}$ and hence $\cot \theta \equiv \frac{\cos \theta}{\sin \theta}$. Another way to show that $\cot$ is the reciprocal of $\tan$!

The results are not always so illuminating. Try it on $\sin^2 \theta + \cos^2 \theta \equiv 1$ to see why.

Get two trig derivatives for the price of one

If you have battled your way through the derivation of the derivative of sine, you may not be looking forward to demonstrating the derivative of cosine. But we can use the complementary angle identity and then the chain rule:

$$\frac{\text{d}}{\text{d}x} \cos x = \frac{\text{d}}{\text{d}x} \sin \left(\frac{\pi}{2} - x \right) = - \cos\left(\frac{\pi}{2} - x \right) = - \sin x$$

The same trick shows why pairs of complementary trig functions have derivatives of a similar form but reversed sign and $\sin \leftrightarrow \cos$, $\sec \leftrightarrow \csc$ and $\tan \leftrightarrow \cot$. For example:

$$\frac{\text{d}}{\text{d}x} \tan x = \sec^2 x \implies \frac{\text{d}}{\text{d}x} \cot x = -\csc^2 x$$

$$\frac{\text{d}}{\text{d}x} \sec x = \sec x \tan x \implies \frac{\text{d}}{\text{d}x} \csc x = -\csc x \cot x$$

For example we can read off from the left-most column that $\sin (12° 42') = 0.21985$, and from the right-most column of the same row we see that $\sin (77° 18') = 0.97533$. We have immediately that $\cos (12° 42') = 0.97533$, and that $\cos (77° 18') = 0.21985$. Incidentally, both of these figures matched the answer on my electronic calculator to the specified level of accuracy.