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

Table of contents
- i b
  - Solution i
- ii b
  - Solution ii
- iii b
  - Solution iii

i

Solution

Differentiate

F (x) = s (x) c (a - x) + c (x) s (a - x) .

Using

s^{'} = c

and

c^{'} = - s

gives

F^{'} (x) = c (x) c (a - x) + s (x) s (a - x) - s (x) s (a - x) - c (x) c (a - x) = 0.

By the zero derivative on a connected open set,

F

is independent of

x

.

ii

Solution

The constant from part (i) is found by setting

x = 0

:

F (0) = s (a) .

Taking

a = x + y

therefore gives the sine addition formula

s (x + y) = s (x) c (y) + c (x) s (y) .

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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