Question

expand cosx in powers of (x+π) using taylors theorem​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

f(x)=cosx   here x = -π

f(x) = cosx      f(-π) = 1

f'(x)= -sinx     f'(-π) = 0

f''(x)= -cosx     f"(-π) =  1

f'''(x) = sinx      f"'(-π) = 0

f''''(x) = -cosx      f""(-π) = 1

Taylor series : f(x) = f(a)+(x-a)f'(x)+(x-a)^2/2! f"(x)+(x-a)^3/3!f"'(x)+------------(1)

substitute above value in 1

cosx = 1+(x+π)(0)+(x+π)^2/2!(1)+(x+π)^3/3!(0)+(x+π)^4/4!(1)+-----

-->cosx = 1+(x+π)^2/2!+(x+π)^4/4!+-----