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

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

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

Answer: (D)

This question tests truncating a Maclaurin (Taylor at 0) series to a given degree. For sin x, the Maclaurin series is sin x = x - x^3/3! + x^5/5! - ... Keeping terms up to degree 3 means we include the x term and the x^3 term and drop higher powers. Since 3! = 6, the cubic term is -x^3/6, giving sin x ≈ x - x^3/6. The x^2 and all even powers don’t appear because sin is an odd function, so its expansion contains only odd powers. A form with just x would be missing the next correction, and a form with an x^2 term would violate the function’s parity.

This question tests truncating a Maclaurin (Taylor at 0) series to a given degree. For sin x, the Maclaurin series is sin x = x - x^3/3! + x^5/5! - ... Keeping terms up to degree 3 means we include the x term and the x^3 term and drop higher powers. Since 3! = 6, the cubic term is -x^3/6, giving sin x ≈ x - x^3/6.

The x^2 and all even powers don’t appear because sin is an odd function, so its expansion contains only odd powers. A form with just x would be missing the next correction, and a form with an x^2 term would violate the function’s parity.