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We know $$\sin\theta = \frac{\text{perpendicular}}{\text{hypotenuse}}$$ then $\sin 90^\circ = 1$ refers in this case perpendicular = hypotenuse. But, then, the base becomes $0$ according to the Pythagorean triplet law. How is this possible?

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  • $\begingroup$ Perhaps this answer of mine will be helpful to you. $\endgroup$ – Blue Feb 3 '18 at 2:42
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    $\begingroup$ @Blue first time I have seen someone answer two different questions with only $1$ answer :) $\endgroup$ – Mr Pie Feb 3 '18 at 2:43
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    $\begingroup$ SOH-CAH-TOA has its shortcomings as a definition, and this is one of the reasons why. But you are absolutely correct, as $\theta$ approaches $0,$ your triangle become degenerate, and becomes a vertical line, with the opposite edge and hypotenuse as the same line. $\endgroup$ – Doug M Feb 3 '18 at 2:50
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    $\begingroup$ Generally the Sinus of an angle is the ratio of the side opposite to it on hypotenuse. The side opposite to angle 90 degree is hypotenuse itself, that is why the ratio is one. $\endgroup$ – sirous Feb 3 '18 at 3:02
  • $\begingroup$ Would you explain this phenomenon with a figure @sirous $\endgroup$ – Time rub Feb 3 '18 at 3:11
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The Pythagorean Theorem asserts that $a^2 + b^2 = c^2$ such that $c$ represents the hypotenuse, namely the side of a triangle that is directly opposite the $90^\circ$ angle. There has to be a $90^\circ$ angle because this theorem only applies to right triangles. If $\sin 90^\circ = 1$, since $$\sin(\cdot) = \frac{\text{opposite}}{\text{hypotenuse}}$$ then we have $\text{opp} = \text{hyp}$. Let $a = \text{opp}$ then we have $$a^2 + b^2 = a^2$$ and therefore $b^2 = 0\Leftrightarrow b = 0$. This means that we don’t have a right triangle because the side $b$ has no length (it has length $0$) and so it does not exist. We simply have a line $a = c$. And since $a/c = 1$ then $\sin 90^\circ = 1$.

Consider for example, $$x^2 + y^2 = z^2.$$ Let $z = y + 1$, then we have $y = \dfrac{x^2 - 1}{2}$. If $x = z$, then $$y = \frac{(y+1)^2 - 1}{2}\Leftrightarrow 1 = \frac{y + 2}{2}$$ and therefore $y = 0$. We can’t escape the fact that $y = 0$ or $b = 0$ and so this is simply just a special case of $\sin\theta$ such that $\theta=90^\circ$.

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  • $\begingroup$ Thanks for the answer $\endgroup$ – Time rub Feb 3 '18 at 3:16
  • $\begingroup$ @Timerub No problem :) $\endgroup$ – Mr Pie Feb 3 '18 at 3:16
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It is a degenerate case or limit case with $$h\geq b\gg a$$

enter image description here

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Usually, the trigonometric relations, when defined using triangles, are defined only for the angles in the range $0<\theta<π/2$. However, it's still possible to see why $\sin π/2$ is taken to be $1$ from a right triangle. Recall that $$\sin(\cdot)=\frac{\text{opposite}} {\text{hypotenuse}}. $$ But when the angle is $π/2$ then $\text{opposite}=\text{hypotenuse} $, so that the result follows.

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