What is maclaurin sin(2x)+cos(2x) ?

The answer to maclaurin sin(2x)+cos(2x) is 1+2x-2x^2-4/3 x^3+2/3 x^4+\ldots

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maclaurin sin(2x)+cos(2x)

maclaurin

sin (2 x) + cos (2 x)

1 + 2 x - 2 x^{2} - \frac{4}{3} x^{3} + \frac{2}{3} x^{4} + \dots

Solution steps

Apply the Maclaurin Formula

= 1 + \frac{\frac{d}{d x} ( sin ( 2 x ) + cos ( 2 x ) ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( sin ( 2 x ) + cos ( 2 x ) ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( sin ( 2 x ) + cos ( 2 x ) ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 1 + \frac{2}{1 !} x + \frac{- 4}{2 !} x^{2} + \frac{- 8}{3 !} x^{3} + \frac{16}{4 !} x^{4} + \dots

Refine

= 1 + 2 x - 2 x^{2} - \frac{4}{3} x^{3} + \frac{2}{3} x^{4} + \dots

\frac{\partial}{\partial x} (e^{(\frac{y}{x})})

sin (\frac{4}{x})

\int x^{2} e^{- 4 x} d x

x = y^{2} + 3, x = y, y = - 1, y = 1

n = 9 \sum \infty \frac{1}{n ( n - 1 )}

What is maclaurin sin(2x)+cos(2x) ?
The answer to maclaurin sin(2x)+cos(2x) is 1+2x-2x^2-4/3 x^3+2/3 x^4+\ldots