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In high school, it said that trigonometric functions are defined on right triangles, and sine is given by the formula $\sin(x) = \frac{\text{opposite side} }{\text{hypotenuse} }$. If we evaluate this formula at the ninety degrees vertex, we get a $1$ (opposite side is hypotenuse). But, how would we evaluate cosine at the ninety degree vertex to show that it is $0$?

N. F. Taussig
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tryst with freedom
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  • Question came up when explaining tenth grade trignometry to my sister. I am trying to figure out a simple answer for this so I can explain to her.. – tryst with freedom Apr 19 '23 at 19:57
  • See, for instance, this old answer of mine. You can also try a site search; this question arises pretty often. – Blue Apr 19 '23 at 19:58
  • Your argument for $\sin (90)$ isn't good. You can't use the fixed right angle to compute $\sin (\theta)$, you need a right triangle containing $\theta$ in addition to the right angle. – lulu Apr 19 '23 at 20:01
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    The hand waving argument is: if $\theta=89.999$ then your "triangle" essentially consists of two very nearly parallel lines which meet far from the fixed leg. Or, if you fix the hypotenuse, then the hypotenuse lines up along one leg, save for an infinitesimal gap at the base. – lulu Apr 19 '23 at 20:04
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    opposite/adjacent = sin? that’s tangent. anyway, I would just handwave and invoke continuity without saying so, to convey the idea that $\sin(\pi/2)=1$ and $\cos(\pi/2)=0$. If a little more advanced, I would introduce the idea of the unit circle and define the trig functions that way, rather than the right-angled triangle, which really is only good for acute angles. – peek-a-boo Apr 19 '23 at 20:04
  • The unit circle definition is much much better. It’s still extremely related to SohCahToa but generalises to arbitrary angles. – FShrike Apr 19 '23 at 20:10
  • @peek-a-boo mistake. – tryst with freedom Apr 19 '23 at 21:55
  • High school is a trap. – Gerry Myerson Apr 20 '23 at 00:22

3 Answers3

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Figure: Sharp right triangle.

$\angle B=\angle C=90^\circ$, $CB=a$, $AC=\infty$: $$\cos\angle C=\frac{CB}{AC}=\frac a\infty=0.$$

Bob Dobbs
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Use the Law of Cosines in conjunction with the Pythagorean Theorem.

Letting $~C~$ represent the hypotenuse, the angle opposite side $~C~$ is $~90^\circ.$

Then, you have the following two results:

  • $C^2 = A^2 + B^2.$
    This is based on the Pythagorean Theorem.

  • $C^2 = A^2 + B^2 - 2AB\cos(90^\circ).$
    This is based on the Law of Cosines.

As a result of the two bullet points above, you must have that (in general, for any right triangle), $~2AB\cos(90^\circ) = 0.$

The only way that this is possible is if $~\cos(90^\circ) = 0.$

user2661923
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The angles the trig functions deal with are never the right angle, itself, but the other two. If one of those angles was very close to 90 degrees then the opposite side and the hypotenuse would be nearly equal and very long. The sine of it would be close to 1. The angle can't be 90 degrees: a triangle can't have two right angles.

tryst with freedom
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stretch
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