Are there 3 trig functions or are there 6 trig functions?

Solution 1:

It depends on how you look at it I guess, but:

$$\cot(x) = \frac{1}{\tan(x)}$$

$$\csc(x) = \frac{1}{\sin(x)}$$

$$\sec(x) = \frac{1}{\cos(x)}$$

So the three "extra" functions your friend told you about are just derived from the three you know. But if that's the rule, then two of the ones you know,

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

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

are also just derived functions. Hence we would say there is only one trigonometric function, for example $\sin{x}$.

(As others have mentioned, this statement works even counting hyperbolic functions, because of properties like $\cosh(x) = \cos(ix)$ and so on, or using $e^{i\theta} = \cos{\theta} + i\sin{\theta}$. But since you don't appear to be at this level of math yet, I won't go into detail about that.)

Bottom line: We only need one trigonometric function, but for practical reasons, there are more.

Solution 2:

I strongly recommend this blog post by John Cook: as he says, the calculator answer is 3, the college textbook answer is 6, and the historical answer is at least 12

Solution 3:

The three sides $a,b,c$ of a right-angled triangle can make nine ratios $$\frac aa, \frac ab, \frac ac, \frac ba, \frac bb, \frac bc, \frac ca, \frac cb, \frac cc$$

Three of these ratios are trivially equal to $1$, and the other six all have names (depending where the right-angle is).

If $r$ is a ratio, then so is $\frac 1r$, so the ratios come in pairs. So sometimes people think of three basic ratios sine, cosine and tangent, because the others are their reciprocals, and these three are sufficient because you can divide by them if necessary.

As it happens when it comes to calculus (particularly integration) it turns out that the other ratios come in handy, so even though the names are not much used in early work, they are useful to know later.

Since tan=sin/cos and in a right-angled triangle $\sin^2+\cos^2=1$ we can write everything in terms of the sine function if we like (and don't mind taking square roots).

As it happens one of the most useful ways of writing these functions in terms of a single function is to use $t=\tan \frac {\theta}2$ when we get the nice expressions $$\sin\theta=\frac {2t}{1+t^2}, \cos \theta=\frac {1-t^2}{1+t^2}$$

Solution 4:

To be a little silly, there is only one "trig" function. By Euler's formula, $e^{i \theta} = \cos \theta + i \sin \theta$, for real valued angles $\theta$. This means that $$e^{-i\theta} + e^{i \theta} = (\cos \theta + i \sin \theta) + (\cos (-\theta) + i \sin (-\theta)) = 2 \cos \theta$$ since $\cos \theta$ is even and $\sin \theta$ is odd.

Using something similar for $\sin \theta$, we can write $$\cos \theta = \frac{e^{i \theta} + e^{-i \theta}}{2},$$ $$\sin \theta = \frac{e^{i \theta} - e^{-i\theta}}{2i}.$$

From the identity $\tan \theta = \frac{\sin \theta} {\cos \theta}$, we can write

$$\tan \theta = \frac{e^{i \theta} - e^{-i\theta}}{i(e^{i \theta} + e^{-i \theta})}.$$

And the additional trig functions follow easily from the above and we write: $$\sec \theta = \frac{2}{e^{i \theta} + e^{-i \theta}},$$ $$\csc\theta = \frac{2i}{e^{i \theta} - e^{-i\theta}},$$ $$\cot \theta = \frac{i(e^{i \theta} + e^{-i\theta})}{e^{i \theta} - e^{-i \theta}}.$$