The Indian mathematician Bhaskara I (600-680AD) found the following approximation for sin(x):

sin(x) ≈ b(x) = 16x(π − x)/ (5π^2 − 4x(π − x))

a. Find the errors for this approximation on [0, π] and on [−π, 0]:

A sin(x),b(x) (0, π) =

A sin(x),b(x) (−π, 0) =.

b. Modify Bhaskara’s approximation to get a good approximation c(x) for sin(x) on [−π, 0]. c(x) =

c. How good is your approximation? At −π/6, compute: sin(−π/6) = c(−π/6) =

Esin(x),c(x) (−π/6) =

Do the same thing for −π/3. sin(−π/3) = c(−π/3) = Esin(x),c(x) (−π/3)

Answer rating: 100% (QA)

a Given sin x bx 16 x x 5 2 4x x b x 0 16 0 0 5 2 4 0 0 0 0 b x 0 b x 0 16 ... View the full answer