How can $\sin(90) = 1$?

Question

I know this question has been asked before, but the answers don't really make intuitive sense.

As angle $\theta$ approaches $90$ degrees, how can $\sin(90) = 1$ if there IS NO triangle?

This breaks all other ratios as well, because the other ratios can be written in the form of $\sin(\theta)$ as well.

A simple answer (if it exists) s'il vous plait.

Thanks in advance.

Answer 1

To spell out @Vercingetorix's point:

The radius-$1$ circle centred on the origin in $2$ dimensions has equation $x^2+y^2=1$ by the Pythagorean theorem. (If it's not obvious this holds when $x$ and/or $y$ is negative, reflect in an axis so they're not, and note $(-u)^2=u^2$.)

For acute $\theta$, a right-angled triangle exists containing $\theta$ with vertices $(0,\,0),(\cos\theta,\,0),\,(\cos\theta,\,\sin\theta)$. The first and last of these are the ends of the hypotenuse, which is a radius, and the side adjacent to $\theta$ is the radius ending in $(1,\,0)$. Rotating $\theta$ anticlockwise from that radius gives the hypotenuse.

Any radius can be used the same way, even if the resulting angle isn't acute, and it meets the circle at $(\cos\theta,\,\sin\theta)$. This defines the trigonometric functions for any $\theta\in\Bbb R$. The choice $\theta=90^\circ$ gets you to the radius ending at $(0,\,1)$, so $\cos90^\circ=0,\,\sin90^\circ=1$.

Answer 2

$\sin x$ is the $y$ coordinate on the unit circle and $\cos x$ is the $x$ coordinate. If you use your intuition a $90$ degrees $\cos x$ must be $0$ because its a straight line up and $\sin x$ must be $1$.

Answer 3

If you a want to know how is $\sin(90^{\circ})=1$ from a triangle's point of view, well you can take any right angled triangle and fix the base length. Now increase the perpendicular side. You will see that as you increase the side, length of hypotenuse also increases. If the angle opposite to the perpendicular side is $\theta$, so now as you increased that side, notice that $\theta \to 90^{\circ}$, and the length of hypotenuse tends to the length of the perpendicular side. So, $\sin \theta = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} \to 1 $ as $\theta \to 90^{\circ}$.