Which expression represents the Maclaurin series for cos x truncated to degree 2?

Which expression represents the Maclaurin series for cos x truncated to degree 2?

• A. 0.65
• B. 0.75
• C. 0.875
• D. 0.92

Answer: (C)

The idea is to approximate cos x near 0 using its Maclaurin (Taylor around 0) series and stop after the x^2 term. The Maclaurin series for cos x is 1 − x^2/2! + x^4/4! − ..., so truncating to degree 2 gives the quadratic approximation 1 − x^2/2. If you’re evaluating this truncated expression at a specific x, say x = 1/2, you get 1 − (1/2)^2/2 = 1 − 1/8 = 7/8 = 0.875. That matches the given numeric choice. So the expression itself is 1 − x^2/2, and the value 0.875 corresponds to plugging in x = 1/2.

The idea is to approximate cos x near 0 using its Maclaurin (Taylor around 0) series and stop after the x^2 term. The Maclaurin series for cos x is 1 − x^2/2! + x^4/4! − ..., so truncating to degree 2 gives the quadratic approximation 1 − x^2/2.

If you’re evaluating this truncated expression at a specific x, say x = 1/2, you get 1 − (1/2)^2/2 = 1 − 1/8 = 7/8 = 0.875. That matches the given numeric choice. So the expression itself is 1 − x^2/2, and the value 0.875 corresponds to plugging in x = 1/2.