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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.

HERO
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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:

Will
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  • By the “Right Triangle Definitions”, you mean those that are demonstrated in our textbooks? – HERO Nov 09 '19 at 03:45
  • I think so. I'm referring to the definitions based on side length ratios from a right triangle (sine as "opposite/hypotenuse," cosine as "adjacent/hypotenuse," etc.) – Will Nov 09 '19 at 03:47
  • Yes, then you mean those. I'll ask of one more favor though. This explanation that you provided—perhaps called the unit circle definition—can you provide me links that explain this a little more in-depth? – HERO Nov 09 '19 at 03:51
  • @AmandaMacaurenni Sure, I'll add those in an edit to my post. Just give me a couple minutes. – Will Nov 09 '19 at 04:00
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    Yeah sure, take your time. – HERO Nov 09 '19 at 04:14
  • @AmandaMacaurenni I've added a couple links above. – Will Nov 09 '19 at 04:26
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    I have learned about radians lol. Also, many many thanks for providing those links. They actually seem helpful. – HERO Nov 09 '19 at 06:12