Past exam of the mathematics course of the University of Cambridge / 2020 / ia / Paper 1 / 9D / b / iii

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- Solution iii

Solution

Differentiation gives

\frac{d}{d x} (s (x)^{2} + c (x)^{2}) = 2 sc - 2 cs = 0.

The initial conditions set the constant to one, so

s (x)^{2} + c (x)^{2} = 1,

the Pythagorean trigonometric identity.

Since

∣ c ∣ \leq 1

, Taylor theorem with Lagrange remainder gives

c (1) = 1 - \frac{1}{2} c (ξ) > 0

for some

0 < ξ < 1

, while

c (2) = 1 - \frac{2 ^{2}}{2} + \frac{2 ^{4}}{4 !} c (η) \leq - 1 + \frac{2}{3} < 0

for some

0 < η < 2

. The intermediate value theorem supplies

k \in (1, 2)

with

c (k) = 0

. The sine addition formula gives

s (2 k) = 2 s (k) c (k) = 0

. The analogous cosine addition formula gives

c (2 k) = c (k)^{2} - s (k)^{2} = - 1

, so

s (x + 2 k) = - s (x), s (x + 4 k) = s (x) .

Thus $s$ is a periodic function.

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