Why do trigonometric ratios work for all values of $\theta$, while it is initially stated to be true only for $\theta<90°$

trigonometry

While trigonometric ratios such as $\sin{(90°+\theta)}$, $\cos{(180°-\theta)}$, and so on are shown to work for acute angle values of $\theta$ in textbooks, they hold for all values of $\theta$. Even certain proofs make use of the fact. For instance, the proof my textbook demonstrates for $\sin{(270°+\theta)}$ is like: $$\sin{(270°+\theta)}=\sin{\{180°+(90°+\theta)\}}=-\sin{(90°+\theta)}=-\cos{\theta}$$ Evidently $(90°+\theta)>90°$, and this proof is assuming that $\sin{(180°+\theta)}$ holds for all values of $\theta$. The same goes for $\sin{(A+B)}$. While it is initially stated to be true for $A+B<90°$, it works for all values of $A+B$. I guess that makes sense if the previous one does.

Now, isn't there like, a general proof or a logical argument showing why this is happening? Or why trigonometric ratios work for all values of $\theta$? Any help is appreciated.

Best Answer

We define two functions called sine and cosine for all real number angle measures using the unit circle. We look at where the line formed by some chosen angle intersects the unit circle; the $x$-coordinate is defined to be the cosine of that angle, and the $y$-coordinate to be the sine:

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Notice that the right triangle definitions for acute angles are just special cases of the broader unit circle definition (even though you might not have known that when you first learned about sine and cosine).

All the remaining trig functions other than sine and cosine can be defined in terms of those two functions.

EDIT: Some further explanation of what I've described above can be found at the following:

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