How do you do a sign function?

[box]

I have always used a calculator to find the sin when the angle is given. But how would you figure this out if you didn't have a calculator

 

[arildno]

I have always used a calculator to find the sin when the angle is given. But how would you figure this out if you didn't have a calculator

Look up in a sine table.

 

[krab]

$$\sin x=x-x^3/6+x^5/120-...$$
One can use the symmetries of the function to avoid large values of x (where convergence is slow) and only ever have to calculate with x as large as pi/2. For example, sin(2)=sin(pi-2)=sin(1.14159..). One can even do better by using the expansion of cosine:
$$\cos x=1-x^2/2+x^4/24-...$$
Then using sin^2+cos^2=1, you only need the series up to x=pi/4. The series converges very quickly for such small x.

 

[amcavoy]

...and to a limited extent you can calculate them by hand using the angles you already know (30, 45, 60, 90, etc.) with identities such as the double and half-angle formulas.

 

[HallsofIvy]

You can use the Taylor's series for sin(x)= x- x²/2+ x⁴/4!- x⁶/6!+ ... where the general term is (-1)ⁿx²ⁿ/(2n)! and a few terms should get you pretty good accuracy.

My understanding is that calculators and computers actually use the "CORDIC" method:
http://www.dspguru.com/info/faqs/cordic.htm

 

[VietDao29]

You can use the Taylor's series for sin(x)= x- x²/2+ x⁴/4!- x⁶/6!+ ... where the general...

There's a slight error there... You just mixed up the Taylor expansion for sin(x), and cos(x) .
$$\sin x = x - \frac{x ^ 3}{3!} + \frac{x ^ 5}{5!} - \frac{x ^ 7}{7!} + ... = \sum_{n = 0} ^ {\infty} \frac{(-1) ^ n}{(2n + 1)!} x ^ {2n + 1}$$
$$\cos x = 1 - \frac{x ^ 2}{2!} + \frac{x ^ 4}{4!} - \frac{x ^ 6}{6!} + ... = \sum_{n = 0} ^ {\infty} \frac{(-1) ^ n}{(2n)!} x ^ {2n}$$
Viet Dao,

 

[HallsofIvy]

Oops!

No wonder I keep getting my trig problems wrong!