ITPEC FE Subject B April 2025 Question 14
From the answer group below, select the correct combination of answers to be inserted into A and B in the program.
The function cosine( returns an approximate value of the cosine value of z degrees. The function uses Maclaurin series expansion for the cosine shown below and can be used to determine cos(y) for values of y radians.
cos(y) = 1 − y22! + y44! − y66! + ⋯
Answer group
Explanation
A: term and B: squared — each series term is the previous one times -y2 ÷ (2n(2n - 1)), and the loop stops once the latest term is negligible.
The n-th term of the series is (-1)n y2n ÷ (2n)!. Dividing it by term n − 1 leaves -y2 ÷ (2n(2n - 1)): one more factor y2 and two more factorial factors. Tracing the program with B = squared, starting from term = 1:
- •n = 1: 1 × (-y2) ÷ (2 × 1) = -y2 ÷ 2!
- •n = 2: (-y2 ÷ 2) × (-y2) ÷ (4 × 3) = y4 ÷ 4!
- •n = 3: (y4 ÷ 24) × (-y2) ÷ (6 × 5) = -y6 ÷ 6!
Blank A is the stopping test. The factorial outgrows the power, so the terms shrink toward 0; once the absolute value of term is no more than 0.000000001, further terms no longer change cosy meaningfully.
Answer (f)
Why not others:
- •(a), (b) and (c) — A =
cosytests the running total, which approaches cos(y), not 0; for most angles it never falls below the threshold, so the loop never ends - •(d) — y to the power 2n multiplies in too much, because
termalready carries y2(n-1): at n = 2 it produces a y6 term where y4 belongs - •(e) — y to the power 4 is wrong from the first step: n = 1 gives -y4 ÷ 2 instead of -y2 ÷ 2
Key rule: In an iterative series, update the term by the ratio between consecutive terms instead of recomputing powers, and stop on the size of the latest term, not of the sum.
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