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I know that angles in a triangle can be determined using inverse trigonometric functions and the SOH CAH TOA mnemonic. A blind spot for me is how an angle $\theta$, expressed in degrees, is algebraically related to the ratio.

For example, if $\theta$ is the $30^\circ$ angle in a 30-60-90 triangle with hypotenuse $4$, adjacent side $2$, and opposite side $2\sqrt{3}$, then $$ \sin(30^\circ) = \frac{2}{4} = \frac{1}{2}. $$

I don't see the mathematical use of the degree value $30$ in the trigonometric function, only an intuitive relationship to the ratio.

How do I get degrees from the ratio? How do I get $30^\circ$ from $1/2$? If I wanted to calculate sine on my own, what would the steps be? How do I mathematically use the value $30^\circ$ in those steps?

  • Suggest a re-check of sides of 30-60-90 triangle. They aren't all rationals as you wrote for H,A,O which I assume mean hypotenuse, adjacent, opposite. – coffeemath Feb 13 '21 at 21:14
  • Thanks, but that doesn't answer it. Nothing shows in the above link how the theta value is applied in the calculation of trigonometric function, only how it is related to a length ratio. – Ray Walker Feb 13 '21 at 21:26
  • And log tables don't really address this issue, either, because theta values can be continuous. For example, sin(30.22212345°) is a valid angle. So, discrete theta values, as they would be in a table, cannot address continuity. – Ray Walker Feb 13 '21 at 21:34
  • The trigonometric functions are transcendental (i.e. they are not algebraic which roughly speaking, means that they cannot be expressed in terms of powers, roots, multiplication/division, and addition/subtraction). Evaluating transcendental functions at arbitrary points typically requires tools from analysis (limits, Taylor series approximations, Newton's method, etc). These techniques generally don't give exact answers, but approximations with arbitrary precision. – Xander Henderson Feb 13 '21 at 21:40
  • I have made significant edits to your question to (in my mind) improve readability and notation. If you are unhappy with my edits, please revert them. – Xander Henderson Feb 13 '21 at 21:45
  • Got it. Thank you so much for your time. I see I overlooked the use of Taylor Serious in the link provided by David K. That works for me. – Ray Walker Feb 13 '21 at 21:45
  • Xander Henderson, not at all. Much appreciated. – Ray Walker Feb 13 '21 at 21:45
  • As you have indicated that the link David K provided solved your problem, I have closed this question as a duplicate. – Xander Henderson Feb 13 '21 at 22:00

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