Which expression represents the Maclaurin series for sin x truncated to degree 3?

Which expression represents the Maclaurin series for sin x truncated to degree 3?

• A. sin x ≈ x + x^3/6
• B. sin x ≈ x - x^3/6
• C. sin x ≈ x
• D. sin x ≈ 1 - x^2/2

Answer: (B)

The Maclaurin series for sin x is an infinite alternating power series: sin x = x − x^3/3! + x^5/5! − ... Truncating to degree 3 means keeping terms up to x^3, since 3 is the highest power included. So sin x ≈ x − x^3/6. This matches the given expression because 3! = 6 and the sign of the x^3 term is negative. The other options either miss the x^3 term, use the wrong sign, or correspond to a different function (cosine) or a lower-degree truncation.

The Maclaurin series for sin x is an infinite alternating power series: sin x = x − x^3/3! + x^5/5! − ... Truncating to degree 3 means keeping terms up to x^3, since 3 is the highest power included. So sin x ≈ x − x^3/6. This matches the given expression because 3! = 6 and the sign of the x^3 term is negative. The other options either miss the x^3 term, use the wrong sign, or correspond to a different function (cosine) or a lower-degree truncation.