# The values of trigonometric ratios

Today I explained to my students how the value of $\sin(90°)$ is 1 by first drawing a right angled triangle and taking $90°$ as a reference angle, and told them that now $\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$. Then I used pythagorean relation to show that $\cos(90°)$ is 0, i.e $\sin^2(\theta)=1-\cos^2(\theta)$. And finally I used quotient relation to show how $\tan(90°)$ is infinity. The students didn't seem much convinced by my methods as they were asking why we couldn't find the value of $\cos(90°)$ and $\tan(90°)$ the same way we did to find the value of $\sin(90°)$. Is there any other ideas to show the value of $\sin(90°)$, $\cos(90°)$ and $\tan(90°)$ understandble to them?

• "to show how $\tan(90^\circ)$ is infinity" Technically, this is incorrect. Although the limit of $\tan\theta$ as $\theta$ approaches $90^\circ$ from below is infinity, what is the limit as $\theta$ approaches $90^\circ$ from above?
– JRN
Commented Jun 1, 2014 at 23:46

I don't know if, in good faith, you can really assert than $\sin(90^\circ)=1$ without your students having a very firm grasp of the unit circle or at least some kind of limiting process. But, then, if the students have a firm grasp of the unit circle, your argument for why $\cos(90^\circ)=0$ should be seen as unnecessarily complicated. And then, as mentioned above, you shouldn't say that $\tan(90^\circ)$ is infinite but you should really say that it's undefined (careful, though, if a curious student enters $\tan(\pi/2)$ into a certain free online calculator, they'll get something like $1.63317787 \times 10^{16}$).
• Using the double angle prove that $\sin(90^\circ)=1$
• Using the Pythagorean identity, prove $\cos(90^\circ)=0$.
• By evaluating $\tan(x)$ using the definition at $x=89^\circ,\, 89.9^\circ,\, 89.99^\circ,\dots$ and at $x=91^\circ,\, 90.1^\circ,\,90.01^\circ,\dots$ try to explain a little more about they sense in which $\tan(90^\circ)$ is undefined. Now try to explain it using the unit circle.