Question

Find Maclaurin Series of cos^3 x​

Answer 1

Answer:
`\bold{1-(3)/(2)x^2+(7)/(8)x^4-(61)/(240)x^6+(547)/(13440)x^8+...}`

Explanation:

First, find the derivatives at x = 0:

`f(x)=cos^3(x)\\f(0)=cos^3(0)\\.\qquad =1\\\\\\f'(x)=-3\ cos^2(x)\cdot sin(x)\\f'(0)=-3\ cos^2(0)\cdot sin(0)\\.\qquad =0\\\\\\f''(x)=-3[sin(2x)\cdot sin(x)+cos^3(x)]\\f''(0)=-3[sin(2\cdot 0)\cdot sin(0)+cos^3(0)]\\.\qquad =-3\\\\\\f'''(x)=-3[-2\ cos(2x)\cdot sin(x)-cos(x)\cdot sin(x)\cdot sin(2x)-3\ cos^2(x)\cdot sin(x)]\\f'''(0)=-3[-2\ cos(2\cdot 0)\cdot sin(0)-cos(0)\cdot sin(0)\cdot sin(2\cdot 0)-3\ cos^2(0)\cdot sin(0)]\\.\qquad =0`

`f^(IV)(x)=-3[8\ sin(x)\cdot sin(2x)-4\ cos(2x)\cdot cos(x)-3\ cos^3(x)]\\f^(IV)(0)=-3[8\ sin(0)\cdot sin(2\cdot 0)-4\ cos(2\cdot 0)\cdot cos(0)-3\ cos^3(0)]\\.\qquad =21\\\\\\f^V(x)=-3[16\ sin(2x)\cdot cos(2x)+20\ sin(x)\cdot cos(2x)+9\ cos^2(x)\cdot sin(x)]\\f^V(0)=-3[16\ sin(2\cdot 0)\cdot cos(2\cdot 0)+20\ sin(0)\cdot cos(2\cdot 0)+9\ cos^2(0)\cdot sin(0)]\\.\qquad =0`

`f^(VI)(x)=-3[52\ cos(x)\cdot cos(2x)-65\ sin(2x)\cdot sin(2x)+9\ cos^3(x)\cdot sin(x)]\\f^(VI)(0)=-3[52\ cos(0)\cdot cos(2\cdot 0)-65\ sin(2\cdot 0)\cdot sin(2\cdot 0)+9\ cos^3(0)\cdot sin(0)]\\.\qquad =-183\\\\\\f^(VII)(x)=-3[-182\ cos(2x)\cdot sin(x)-169\ cos(x)\cdot sin(2x)-27\ cos^2(x)\cdot sin(x)]\\f^(VII)(0)=-3[-182\ cos(2\cdot 0)\cdot sin(0)-169\ cos(0)\cdot sin(2\cdot 0)-27\ cos^2(0)\cdot sin(0)]\\.\qquad =0`

`f^(VIII)(x)=-3[560\ sin(x)\cdot sin(2x)-520\ cos(2x)\cdot cos(x)-27\ cos^3(x)]\\f^(VIII)(0)=-3[560\ sin(0)\cdot sin(2\cdot 0)-520\ cos(2\cdot 0)\cdot cos(0)-27\ cos^3(0)]\\.\qquad =1641`

The Maclaurin Series of f(x) at 0 is:

`f(x)=f(0)+(f'(0))/(1!)(x)+(f''(0))/(2!)(x^2)+(f'''(0))/(3!)(x^3)+(f^(IV)(0))/(4!)(x^4)\\\\.\qquad \qquad +(f'(0))/(5!)(x^5)+(f^(VI)(0))/(6!)(x^6)+(f^(VII)(0))/(7!)(x^7)+\frac{f{VIII}(0)}{8!}(x^8)+...\\\\\text{Insert the derivatives into the equation above and simplify:}\\\boxed{f(x) =1-(3)/(2)x^2+(7)/(8)x^4-(61)/(240)x^6+(547)/(13440)x^8+...}`