How can two angles of a triangle be equal to $90°$? If two angles were $90°$, this would mean that the two sides would be parallel and the angle of the third side would be equal to 0. Thus, there would be only two vertices and this wouldn't be a triangle at all, ultimately making $\sin 90° = 1$ impossible.
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19There's more to the sine function than triangles. – Angina Seng Aug 03 '19 at 06:15
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See here the definition of the $\sin$: https://en.wikipedia.org/wiki/Unit_circle – Michael Rozenberg Aug 03 '19 at 06:16
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1Trigonometric functions can be defined beyond triangles by invoking rotation along the unit circle, and even more, independently of geometry. For instance, one can define sine function by $$\sin(x^{\circ}) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \left(\frac{\pi}{180}x\right)^{2n+1} $$ for any complex number $x$. – Sangchul Lee Aug 03 '19 at 06:21
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8I applaud this very good first question from a new user – gen-ℤ ready to perish Aug 03 '19 at 06:35
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1If you want to stick to triangles, you may think of $\sin(90)$ as the degenerate case for a triangle. Like, $\sin(85) = 0.996$, $\sin(86) = 0.997, \sin(89) = 0.9998\ldots$. So it seems $\sin(90) = 1$ – AgentS Aug 03 '19 at 06:40
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3@rsadhvika Well, I think this fact is known by the OP. He's just trying to figure out shouldn't $\sin (90^o)$ be undefined if there is no triangle at all? – Tony Aug 03 '19 at 06:54
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4Isn't it more natural to ask "how is $\sin 0° = 0$ possible?" Good question, by the way. – user1551 Aug 03 '19 at 06:55
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3I like to think that my answer to the question "When the trig functions moved from the right triangle to the unit circle?" could be helpful to you. – Blue Aug 03 '19 at 06:59
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With regards to the first comment, we need the unit circle definition to compute sines of angles greater than $90º$ and negative angles. Otherwise, there would be no way to prove basic theorems such as $\sin(-\theta) = - \sin(\theta)$. – Toby Mak Aug 03 '19 at 12:47
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Consider polar coordinates, $(r \cos\theta, r \sin \theta)$ in a unit circle such that $r=1$. Then, we can see that, the mapping in the first quadrant inside the unit circle is just $(\cos \theta, \sin \theta)$. Now, consider a moving point $A$ starting from $\theta=0°$ to $\theta=90°$ on the circumference of the circle. Now, $OA$ is the hypotenuse of the right angle triangle inside the circle. Now, it is clear that at $\theta=90°$, the hypotenuse and the perpendicular are the same line in the $x$-axis (i.e., they coincide). So, $\sin(90°)=\frac{p}{h}=1$. As per your argument, the case is two sides coinciding rather than being parallel, because by Pythagoras theorem, we have $h^2=p^2+b^2$, so if $p$ increases then $b$ must decrease, and $h=p$ iff $b=0$.
Where, h is the radius of the unit circle, p is the perpendicular drawn from (aka height) the point in the circumference to the x-axis, and b is the distance of intersection of p and x-axis from the origin.
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1I made some edits, but we still need to know what $h,p,b$ are – gen-ℤ ready to perish Aug 03 '19 at 06:41
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Thank you. $h$ is the radius of the unit circle, $p$ is the perpendicular drawn from (aka height) the point in the circumference to the x-axis, and $b$ is the distance of intersection of $p$ and x-axis from the origin. – problem_cracker Aug 03 '19 at 06:43
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3@problem_cracker It would be good if mentioned what $h,p,b$ are in the answer itself. – Tony Aug 03 '19 at 06:52
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You can very well admit the concept of flat triangles (coming in two flavors, with a $0°$ and a $180°$ angle).
Anyway, the usual definition of the sine does not involve a triangle. Rather, the trigonometrical circle and the projection of a point at a given angle to the vertical axis. Indisputably, $\sin90°=1$. Simlarly, $\cos90°=0$.
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A trivial Pythagorean triple, generated by Euclid's formula where $m=n$ such as $(A,B,C)=(0,2,2)$, is a vertical line $(A=0)$ on the $y$ axis. Since $\frac{C}{B}=1, \sin90^\circ=1$.
poetasis
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As noted in a comment, this answer to a previous similar question may well help.
To answer your question directly: I offer that mathematicians made $\sin(x)$ “well defined” for values of $x$ that are smaller than $0^{\circ}$ or larger than $90^{\circ}$ (that is, outside the range of values that are valid for an interior angle of a triangle) when it became useful to do so. “Well defined” is a technical term that means, roughly, that everyone agrees on the answer.
As a simple example, imagine an object that is moving in the plane at constant velocity. The velocity can be described as a speed $v$ and an angle $\theta$ measured clockwise from the $x$ axis. Now suppose that we ask the question, “How fast is the object moving in the $y$ direction?”
If $\theta$ is less than $90^{\circ}$ then trigonometry gives the answer $v \sin(\theta)$. For values of $\theta$ equal or larger than $90^{\circ}$, it’s possible to calculate an answer by deducting an appropriate multiple of $90^{\circ}$ to bring $\theta$ back into the range of valid triangle interior angles. But that’s tedious. The answer “should” be $v \sin(\theta)$ for any direction $\theta$. To make this answer work, the definition of $\sin$ has to be extended from the range of valid triangle interior angles $(0^{\circ},90^{\circ})$ to the range of valid directions $[0^{\circ},360^{\circ})$. In this extension definition, $\sin(90^{\circ})=1$.
PtH
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