1. ## trig

can anyone tell me a method of finding sin and cos of arbitrary angles like 53 , 57 AND 63 DEGREE

2. Originally Posted by abhishek arora
can anyone tell me a method of finding sin and cos of arbitrary angles like 53 , 57 AND 63 DEGREE
Use the Taylor series expansions.

$\sin{x} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} - \frac{x^{11}}{11!} +\dots$

$\cos{x} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \frac{x^{10}}{10!} + \dots$

The further you go along the series, the more accurate your answer will be.

3. Originally Posted by Prove It
Use the Taylor series expansions.

$\sin{x} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} - \frac{x^{11}}{11!} +\dots$

$\cos{x} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \frac{x^{10}}{10!} + \dots$

The further you go along the series, the more accurate your answer will be.
I should add a warning that the Taylor series expansions are for angles in radians not degrees so you'd have to convert your angles first.

As a quick reference 1 degree = $\frac{\pi}{180}$radians

4. i stumbled upon this exact trigonometric values, very good!

Exact trigonometric constants - Wikipedia, the free encyclopedia

 3°: 60-sided polygon
 6°: 30-sided polygon

 9°: 20-sided polygon

 12°: 15-sided polygon

 15°: dodecagon

 18°: decagon

 21°: sum 9° + 12°

 22.5°: octagon

 24°: sum 12° + 12°

 27°: sum 12° + 15°

 30°: hexagon

 33°: sum 15° + 18°

 36°: pentagon

 39°: sum 18° + 21°

 42°: sum 21° + 21°

 45°: square

 60°: triangle

5. Sin[57°] = Cos[33°] ≈ 0.838670567945 =