If we define the sine and cosine functions with infinite series, how can one prove that for instance the sine of an angle between $0$ and $90$ is equal to the ratio of the opposite and the hypotenuse of a right-angled triangle?
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How do you prove it if it's not a series definition? – naslundx Dec 20 '17 at 19:25
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@naslundx What? – ClassicEndingMusic Dec 20 '17 at 19:30
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Related: https://math.stackexchange.com/questions/2230596/relationship-between-sine-as-a-series-and-sine-in-triangles. – arseniiv Dec 20 '17 at 19:47
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Related, too: https://math.stackexchange.com/a/2549714/44121 – Jack D'Aurizio Dec 20 '17 at 21:46
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Also related: "Different definitions of trigonometric functions". – Blue Dec 21 '17 at 12:40
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@ClassicEndingMusic If you are ok, you can accept the answer and set as solved. Thanks! https://math.meta.stackexchange.com/questions/3286/how-do-i-accept-an-answer – user Dec 22 '17 at 13:04
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A link between series and geometric meanings can be established by Euler’s identity
$$e^{ix}=\cos x+i\sin x$$
$$\begin{align} e^{i\theta} &{}= 1 + i\theta + \frac{(i\theta)^2}{2!} + \frac{(i\theta)^3}{3!} + \frac{(i\theta)^4}{4!} + \frac{(i\theta)^5}{5!} + \frac{(i\theta)^6}{6!} + \frac{(i\theta)^7}{7!} + \frac{(i\theta)^8}{8!} + \cdots \\[8pt] &{}= 1 + i\theta - \frac{\theta^2}{2!} - \frac{i\theta^3}{3!} + \frac{\theta^4}{4!} + \frac{i\theta^5}{5!} - \frac{\theta^6}{6!} - \frac{i\theta^7}{7!} + \frac{\theta^8}{8!} + \cdots \\[8pt] &{}= \left( 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + \frac{\theta^8}{8!} - \cdots \right) + i\left( \theta- \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \frac{\theta^7}{7!} + \cdots \right) \\[8pt] &{}= \cos \theta + i\sin \theta . \end{align} $$
Another way is to show that $\sin^2 x+ \cos^2 x=1$ holds by series:
user
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The simplest series definitions for sine and cosine deal with angles measured in radians (that is why radians are used).
It is possible to prove quite easily that $\cos 1 \lt 0$ and that the cosine function is strictly decreasing in $[0,1]$ so there is a unique value $0\lt r \lt 1$ with $\cos r=0$ ($r$ for right-angle). If $0\le t \le 1$ there is also a unique $x\in [0,r]$ with $\cos x =t$
It is not difficult to prove from the series that $\cos^2 x+\sin^2 x =1$
Now consider a right-angled triangle with $a^2+b^2=c^2$ or $\left(\frac ac\right)^2+\left(\frac bc\right)^2=1$
There is a unique $x\in [0,r]$ with $\cos x = \frac ac$ and since $\sin x \ge 0$ for these values (another fact to be checked), we have $\sin x = \frac bc$
When we choose the units for $r$ we locate also the units for $x$.
Mark Bennet
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