Math · Calculus II · Worked example

Find the Maclaurin series of sin x

Find the Maclaurin series of sin x, and use its first two terms to approximate sin 0.5.

Derivatives at 0

The derivatives of sin x repeat every four steps.

Derivatives of sin x at 0
k	0	1	2	3	4	5
f⁽ᵏ⁾(x)	sin x	cos x	−sin x	−cos x	sin x	cos x
f⁽ᵏ⁾(0)	0	1	0	−1	0	1

Build the series

Only odd powers survive, with alternating signs.

sin x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \dots

Approximate sin 0.5

Keep the first two terms.

0.5 - \frac{{0.5}^{3}}{6} = \frac{23}{48}

Bound the error

At x = 0.5 the series alternates with shrinking terms, so the error is less than the next term, 0.5⁵/5! ≈ 0.00026. The true value, sin 0.5 = 0.479426, is 0.000259 away.

\frac{{0.5}^{5}}{120} = \frac{1}{3840}

Result

sin x = x − x³/3! + x⁵/5! − ⋯; sin 0.5 ≈ 0.479167, with error below 0.00026.

Your turn

Find the first three nonzero terms of the Maclaurin series of cos x.

Show the answer and explanation

1 − x²/2 + x⁴/24.

The derivatives of cos x at 0 cycle 1, 0, −1, 0, so cos x = 1 − x²/2! + x⁴/4! − ⋯. It is also the derivative of the sine series, term by term.

Keep exploring

In Taylor Approximation, raise the degree from 3 to 5. The sampled error on [0, 0.5] falls from 0.00026 to 0.0000015.

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