Question

if we are about to approximate sin 2.2 with taylor polynomial of sin 2x about x=1, its sin2(1.1)? just substitute x=1.1 into the polynomial?

Answer 1

f(x)=sin 2x
f'(x)=2cos 2x
f''(x)=-2(2) sin 2x
f'''(x)=-2(2)(2) cos 2x

f(1)=0.035
f'(1)=1.999
f''(1)=-4(0.035)
f'''(1)=-4(1.999)

so my 3rd degree taylor series for sin 2x is
0.035 + 1.999(x-1) - 2(0.035)(x-1)^2 -(2/3)(1.999)(x-1)^3 ?

Answer 2

this is correct...then you plug in x=1.1 to get approximation

however its more efficient to use a=0 instead of 1 since sin(0)=0 and cos(0) =1
this way you are not finding sin(2) to approximate sin(2.2)...its redundant since you could just as easily plugged sin(2.2) into your calculator just as you did to find sin(2)

anyway, using a=0, the taylor polynomial will look like this
\[\sin(2x) = \frac{2x}{1!} - \frac{8x^{3}}{3!}+\frac{32x^{5}}{5!}-\frac{128x^{7}}{7!} ...\]
the even exponent terms are 0 because they all include sin(2a) which =0

Answer 3

oh fyi this will evaluate the function in radians NOT degrees
so if you need sin(2.2 degrees) you have to adjust the x value you plug in

Answer 4

owh ic. thx. the x=1 is from the question, cant change anytg lol.