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Prove that sin and cos are the unique functions satisfying given properties

Let f, g be given functions with derivatives f', g' satisfying

    \[ f'(x) = g(x), \quad \text{and} \quad g'(x) = -f(x), \quad \text{and} \quad f(0) = 0, \quad \text{and} \quad g(0) = 1, \]

for all x in some interval J which contains 0. (Note: f(x) = \sin x and g(x) = \cos x satisfy all of these conditions.)

  1. Prove that these functions must satisfy the Pythagorean identity

        \[ f^2(x) + g^2(x) = 1 \qquad \text{for all } x \in J. \]

  2. If F(x) and G(x) are any other pair of functions satisfying all of these conditions prove that

        \[ F(x) = f(x) \qquad \text{and} \qquad G(x) = g(x) \qquad \text{for all } x \in J. \]

  3. What else can we say regarding the functions f and g?

  1. Proof. First, we show that f^2(x) + g^2(x) = c for some constant, and then we will prove that constant must be 1. To show f^2 + g^2 is constant we take the derivative

        \begin{align*}  \left( f^2(x) + g^2(x) \right)' &= \left( f^2(x) \right)' + \left( g^2(x) \right)' \\  &= 2 f(x) f'(x) + 2 g(x)g'(x) \\  &= 2 f(x) g(x) - 2 f(x) g(x) &(f'(x) = g(x), \ g'(x) = -f(x)) \\  &= 0. \end{align*}

    This holds for all x \in J; hence, by the zero derivative theorem (Theorem 4.7 (c) in Apostol) we must have f^2(x) + g^2(x) = c for some constant c for all x \in J. Now, to show this constant c = 1, we know by hypothesis that 0 \in J and we have

        \[ f^2(0) + g^2(0) = 0 + 1 = 1. \]

    Hence, f^2 (x) + g^2(x) = 1 everywhere on J. \qquad \blacksquare

  2. Proof. Let F(x) and G(x) be another pair of functions satisfying the given relations:

        \[ F'(x) = G(x), \qquad G'(x) = -F(x), \qquad F(0) = 0, \qquad G(0) = 1. \]

    Then define

        \[ h(x) = (F(x) - f(x))^2 + (G(x) - g(x))^2. \]

    This implies

        \[ h'(x) = 2(F(x) - f(x))(F'(x) - f'(x)) + 2(G(x) - g(x))(G'(x) - g'(x)). \]

    Using the relations G'(x) = -F(x), \ g'(x) = -f(x), \ G(x) = F'(x), and g(x) = f'(x) we have

        \begin{align*}    h'(x) &= 2(F(x) - f(x))(F'(x) - f'(x)) + 2(F'(x) - f'(x))(f(x) - F(x))\\  &= 2(F(x) - f(x))(F'(x) - f'(x)) - 2(F(x) - f(x))(F'(x) - f'(x)) \\  &= 0   \end{align*}

    for all x \in J. Hence, h(x) = c for some constant c. Then, we evaluate h(0) and use the given relations f(0) = F(0) = 0 and g(0) = G(0) = 1,

        \begin{align*}  h(0) = c &= (F(0) - f(0))^2 + (G(0) - g(0))^2 \\  &= 0 + (1-1)^2 \\  &= 0. \end{align*}

    Hence, h(x) = 0 for all x \in J. Since it is a sum of squares (which must be nonnegative) we have that h(x) = 0 if and only if

        \[ F(x) - f(x) = 0 \qquad \text{and} \qquad G(x) - g(x) = 0 \]

    for all x \in J. Hence we must have

        \[ F(x) = f(x) \qquad \text{and} \qquad G(x) = g(x). \qquad \blacksquare \]

  3. I’m not entirely sure what Apostol wants us to say here with respect to this solution. Since we have established that \sin x and \cos x satisfy these properties and that any functions satisfying these properties are unique, we can conclude that \sin x and \cos x are the unique functions which satisfy the given properties.

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