> I am sure you can do better than that
Okay...
sin(5x) = sin(x + x + x + x + x)
ergo, as sin(A + B) = sinAcosB + sinBcosA
sin(5x) = sin4xcosx + sinxcos4x
...//cos(2x) => 1, by the way
sin(5x) = (sin2xcos2x + sin2xcos2x)cosx + sinx(sin(2x)^2 + cos(2x)^2)
sin(5x) = (sinxcosx + sinxcosx + sinxcosx + sinxcosx)cosx + sinx
sin(5x) = 4(sinxcosx)cosx + sinx
sin(5x) = sinx(4cos(x)^2 + 1)
sin(5x) = sinx(4(1 - sin(x)^2) + 1)
sin(5x) = sinx(4 - 4sin(x)^2 + 1)
sin(5x) = -4sinx(sin(x)^2 - 1.25)
sin(5x) = -4sin(x)^3 + 5sin(x)
Yeah... there's no way that's all correct...
So, anyway, my several answers for your question.
~Cheers
What about properties of addition and multiplication?
sin(5 * x) =
sin(1 * 5 * x)
sin(0 + 5 * x)
sin(5 * 1 * (x + 0))
sin(0 + 1 * 5 * 1 + 0) * (0 + 1 * x * 1 + 0))
sin(cos(0) * 5 * x * cos(0))
sin((cos(0) + cos(0) + cos(0) + cos(0) + cos(0)) * x)
Or inverse sin function, arcsin?
sin(arcsin(sin(arcsin(sin(arcsin(sin(arcsin(sin(5 * arcsin(sin(x)))))))))))
Or... Java?
int y = 0;
for(y = 0; y <= 5; y++);
int z = x * y;
result = sin(z)
But... what was your *real* intention? get an efficient implementation of sin(5 * x) or create a crazy math tournament?