Is there a formula for sin(x/3)?
I do not think it is likely because it is impossible to trisect an angle exactly in geometry. Since you can not do that I do not know how you could make a construction of right triangles that would lead to a triangle with exactly 1/3 of the original angle in it.
I guess you're right. I was looking forward to finding the value of the sin 1 (deg).
There is the infinite series
sin (x/3) = x/3 - (1/2)(x/3)^2
+ (1/6)(x/3)^3 + ...
-(-1)^n *1/n!* (x/3)^n
(n-> infinity)
Damon has made a good argument that there may be no closed form equation for sin (x/3) in terms of trig functions of x.
I tired Googling sin(x/3) and found nothing
For angles that small, the approximation sin x = x is very good. x must be in radians to use it.
Sin 1 degree = sin pi/180 radian
If you use the first term of the infinite series, that gives you pi/180 = 0.0174533...
The exact value is 0.174524...
Yes, there is a formula for sin(x/3). To find the formula, we can use the half-angle identity for the sine function.
The half-angle identity for sine states that sin(x/2) = ±√[(1 - cos(x))/2], where the sign depends on the quadrant in which x/2 lies.
We can rewrite sin(x/3) as sin((x/2) / (2/3)). By substituting x/2 with a new angle, let's call it u, we get sin(u/ (2/3)) = sin((3u)/2).
Now we can apply the half-angle identity to find the formula for sin((3u)/2):
sin((3u)/2) = ±√[(1 - cos(3u))/2]
So, the formula for sin(x/3) is sin(x/3) = ±√[(1 - cos((3x/2)))/2]. The sign of the square root depends on the quadrant in which x/3 lies.