How to show that given two acute angles, the sine ratio of the greater angle is greater than the sine ratio of lesser angle?

Question

How do I show that for angles $\theta ,\psi \in [0^°,90^°$], if $\theta > \psi$, then $\sin \theta > \sin \psi$?

I thought of proving this by creating two right triangles with the same hypotenuse length $h$, one having an acute angle of $\theta$ and another right triangle having one of its acute angle as $\psi$. I denote the side opposite to $\theta$ and $\psi$ as $p$ and $q$, respectively.

$\theta$ and another triangle with its acute angle being $\psi$" />

Then by the greater side is opposite to greater angle in a triangle theorem: $$p>q$$

$$\frac{p}{h}> \frac{q}{h}$$

Therefore, $\sin \theta = \frac{p}{h}> \frac{q}{h} = \sin \psi$. However, I then realize that the 'greater side is opposite the greater angle' should be applied within a triangle. I also think that maybe I shouldn't assume that both right triangles have the same hypotenuse length. How would I modify this proof so that it is correct? Or is it that I'll have to find a completely different approach of proving this? I'm still new to trigonometry so I hope that the proofs given won't use any complicated theorems. Thank you.

Answer 1

Let us consider $\triangle{ABC}$ such that $\angle{ABC}=\psi$ and $\angle{ACB}=\theta$.

Let $D$ be a point on $BC$ such that $AD\perp BC$.

We use the following claim. A proof of the claim is written at the end of this answer.

Claim : In $\triangle{ABC}$, if $\angle{ACB}\gt \angle{ABC}$, then $AB\gt AC$.

Using the claim above, we have $$\begin{align}\theta\gt \psi&\implies AB\gt AC \\\\&\implies \frac{AD}{AC}\gt\frac{AD}{AB} \\\\&\implies \sin\theta\gt\sin\psi.\ \blacksquare\end{align}$$

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In the following, let us prove the following claim :

Claim : In $\triangle{ABC}$, if $\angle{ACB}\gt \angle{ABC}$, then $AB\gt AC$.

Proof :

Suppose that $AB=AC$. Then, $\angle{ABC}=\angle{ACB}$. This is a contradiction.

Suppose that $AB\lt AC$. Taking a point $E$ on $AC$ such that $AB=AE$, we have $\angle{ACB}\lt \angle{AEB}=\angle{ABE}\lt\angle{ABC}$. This is a contradiction.$\ \square$

Answer 2

Consider a circle of radius $1$ centered at the point $O$. Fix any point $A$ on the circle. Fix a point $B$ on the circle such that $\angle BOA=\frac{\pi}2-\psi$. Let $BB'$ be the altitude of the triangle $BOA$. Draw a radius $r$ from the point $O$ inside the angle $BOA$ such that the angle between $r$ and $OA$ is $\frac{\pi}2-\theta$. Then the ray $r$ ends and the point $C$ at the circle, which is contained in the arc $AB$, because $\theta>\psi$ . Let $CC'$ be the altitude of the triangle $COA$. Then $C'$ belongs to the segment $B'A$. Since $C\ne B$, we have $C'\ne B'$, so $$\sin\theta=|OC'|=|OB'|+|B'C'|=\sin\psi+|B'C'|>\sin\psi.$$

Answer 3

I used your image with some modifications: Let $AC=DF=1$ and $AB=H$. Since $\theta>\psi$ we have $p>q.$ But then $$p^2>q^2$$ $$p^2+p^2q^2>q^2+p^2q^2$$ $$p^2(1+q^2)>q^2(1+p^2)$$ and $$\frac p{\sqrt{1+p^2}}>\frac{q}{\sqrt{1+q^2}}$$ that is $$\sin\theta=\frac pH>\frac qh=\sin\psi.$$

Answer 4

Here I show the result is true even if neither the hypotenuses or the adjacent sides are not necessarily equal.

Let $a_1$ and $a_2$ be the sides adjacent to $\theta$ and $\psi$ respectively.
Let $h_1$ and $h_2$ be the hypotenuses for $\theta$ and $\psi$ respectively.

Let $p = q*x \ $ where $x>1$.
Let $a_1 = a_2*y \ $ where $0 < y < x$.

This range of $y$ is based on the fact that for a right angle triangle, if $\theta > \psi$ then $p/a_1 > q/a_2 \rightarrow a_1 < a_2 x. \ $ (If $a_1 < a_2 x$ and $a_1 = a_2 y \ $ then $y < x$). This is related to the fundamental idea that the largest angle is opposite the largest side.

Let's assume to the contrary that $\sin \theta \leq \sin \ \psi$ when p>q and $\theta > \psi$ and see if this leads to a contradiction.

If $\sin \theta \leq \sin \psi$ then by definition $\frac{p}{h_1} \leq \frac{q}{h_2}$

Using Pythagorus theorem,

$\frac p {\sqrt{a_1^2 +p^2}} \leq \frac q {\sqrt{a_2^2 +q^2}}$

Putting everything in terms of q and $a_2$ we get:
$\frac {q x} {\sqrt{a_2^2 y^2 +q^2 x^2}} \leq \frac q {\sqrt{a_2^2 +q^2}}$
$\frac {q^2 x^2} {(a_2^2 y^2 +q^2 x^2)} \leq \frac {q^2} {(a_2^2 +q^2)}$

$x^2 (a_2^2 +q^2) \leq (a_2^2 y^2 +q^2 x^2)$
$x^2 a_2^2 + x^2 q^2 \leq a_2^2 y^2 + q^2x^2$
$x^2 a_2^2 \leq a_2^2 y^2 $
$x^2 \leq y^2 $

which contradicts the initial assumption that y<x (if x and y are positive non zero numbers).