What is the taylor series formula for f(x)= cosx at x= pi/4.

Taylor series

cos(x) = f (a)(x-a)^0/0! + f ' (a)(x-a)^1/1! + f '' (a)(x-a)^2/2! + f ''' (a)(x-a)^3/3! + f '''' (a)(x-a)^4/4! + .............

where a = π/4

f (a) = cosx f ' (a) = -sinx f '' (a) = -cosx f ''' (a) = sinx f '''' (a) = cosx

f (π/4) = 1/√2 f ' (π/4) = -1/√2 f '' (π/4) = -1/√2 f ''' (π/4) = 1/√2 f '''' (π/4) = 1/√2

now put the values in Taylor series .

cos(x) = 1/√2 - 1/√2(x-π/4) - 1/√2(x-π/4)^2/2 + 1/√2(x-π/4)^3/6 + 1/√2(x-π/4)^4/24 + ....

any doubt let me know

Thanks a lot.

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