1. Fill in the coefficients in Taylor expansion sin(32) = sin(pi/6 + x) as A + Bx +
Cx^2.
Answer: Using Taylor series expansion, value of a function at a point x+Δx is
given as
F(x+Δx)= F(x) + F’(x)* Δx + F’’(x)* Δx2 +F’’’(x)* Δx3/3! + …….. and so on.
In the given question, value of sin(32) needs to be determined.
sin(pi/6 + x) = sin(pi/6) + sin’(pi/6)*x +sin’’(pi/6)*x2/2.
By comparing the co-efficients, A= sin(pi/6) B=sin’(pi/6) and C=sin’’(pi/6)*1/2.
A=0.5.
B=sin’(pi/6)=cos(pi/6)
B=0.866.
C= sin’’(pi/6)*1/2.
C= -sin(pi/6)*1/2
C=-0.250.
2. Evaluate the expression x = 32 - pi/6 in radians to 3dp.
Answer:32 in radians is = 0.559.
x=0.035.
3. Evaluate the error sin(32) - A + Bx + Cx^2 for the x above.
Answer:Substituting the value of A, B, C and x found above, we get
A + Bx + Cx^2=0.5 + 0.866 * 0.035 -0.25*(0.035)2.
A + Bx + Cx^2=0.530 upto 3 decimal points.
So, using Taylor series, sin(32) = 0.530
Actual Value of sin(32)=0.529919
Approximating the value upto 3 decimal places, sin(32)=0.530.
Error = sin(32)- A + Bx + Cx^2
=0.530 -0.530
=0.
Hence, Error = 0.