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I know that sine and cosine can be defined as certain ratios between the sides of a right-angled triangle. I use the laws very often, but this morning, I came up with the ascertainment that I don't know what justifies the definitions.

For instance, for sine to make sense as the ratio "opposite-over-hypotenuse", I need to be sure that, if I fix an angle and double the hypotenuse, the leg opposite the angle also doubles. What's the proof for this fact?

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    $\begingroup$ Which laws/facts in particular do you mean? $\endgroup$ – Matti P. Apr 6 '18 at 6:44
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    $\begingroup$ If you mean, for example, proof that "sine of an angle is the ratio of the length of the side that is opposite that angle to the length of the longest side of the triangle (the hypotenuse)" ... I would take this as a definition which does not need a proof. $\endgroup$ – Matti P. Apr 6 '18 at 6:45
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    $\begingroup$ One cannot prove a definition. $\endgroup$ – user Apr 6 '18 at 6:51
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    $\begingroup$ Just similar triangles $\endgroup$ – CY Aries Apr 6 '18 at 6:58
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    $\begingroup$ @Qwerto: The definitions of sine and cosine do indeed rely on the fact that triangles have equal corresponding angles if and only if they have proportional corresponding sides. Proof of the fact goes all the way back to Euclid's Elements: Book VI, Proposition 4 proves that equal angles implies proportional sides; Proposition 5 proves the converse. (FYI, in non-Euclidean geometry, the fact doesn't hold, so sine & cosine aren't defined as ratios of lengths.) $\endgroup$ – Blue Apr 6 '18 at 8:00
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Suppose that we have right-angled triangles $ABC$ and $A_1B_1C_1$ with $\angle C=\angle C_1=90^\circ$ and $\angle A=\angle A_1$. The two triangles are similar.

So. we have $\displaystyle \frac{BC}{AB}=\frac{B_1C_1}{A_1B_1}$, $\displaystyle \frac{AC}{AB}=\frac{A_1C_1}{A_1B_1}$ and $\displaystyle \frac{BC}{AC}=\frac{B_1C_1}{A_1C_1}$

Say if I draw a $20^\circ$-$70^\circ$-$90^\circ$, measure and calculate the ratio of the side opposite to the $20^\circ$ to the hypotenuse. This ratio is not same no matter how big or small is my triangle. We can define it as $\sin20^\circ$.

If $\angle A=\theta$ and $\angle C=90^\circ$, we can define $\displaystyle \sin\theta=\frac{BC}{AB}$, $\displaystyle \cos\theta=\frac{AC}{AB}$ and $\displaystyle \tan\theta=\frac{BC}{AC}$. The size of the triangle does not matter.

We can actually construct a table for the trigonometric ratios by drawing and measuring right-angled triangles of different angles.

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