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Prove some properties of the complex sine and cosine functions

by RoRi

The following definitions extend the sine and cosine functions to take arguments z \in \mathbb{C}:

    \[ \cos z = \frac{e^{iz} + e^{-iz}}{2}, \qquad \sin z = \frac{e^{iz} - e^{-iz}}{2i}. \]

Prove the following formulas, where u,v,z are complex numbers and z = x + iy.

  1. \sin (u+v) = \sin u \cos v + \cos u \sin v.
  2. \cos (u+v) = \cos u \cos v - \sin u \sin v.
  3. \sin^2 z + \cos^2 z = 1.
  4. \cos (iy) = \cosh y, \qquad \sin(iy) = i \sinh y.
  5. \cos z = \cos x \cosh y - i \sin x \sinh y.
  6. \sin z = \sin x \cosh y + i \cos x \sinh y.
  1. Proof. Using the given definition of the sine of complex numbers we have

        \begin{align*} \sin (u+v) &= \frac{e^{i(u+v)} - e^{-i(u+v)}}{2i} \\[9pt] &= \frac{e^{iu}e^{iv} - e^{-iu}e^{-iv}}{2i} \\[9pt] &= \frac{2 e^{iu}e^{iv} - 2 e^{-iu}e^{-iv}}{2(2i)} \\[9pt] &= \frac{2e^{iu}e^{iv} + e^{iu}e^{-iv} - e^{iu}e^{-iv} + e^{-iu}e^{iv} - e^{-iu}e^{iv} - 2e^{-iu}e^{-iv}}{2(2i)} \\[9pt] &= \frac{e^{iu}(e^{iv} + e^{-iv}) - e^{-iu}(e^{iv} + e^{-iv}) + e^{iv}(e^{iu} + e^{-iu}) - e^{-iv}(e^{iu} + e^{-iv})}{2(2i)} \\[9pt] &= \frac{(e^{iu} - e^{-iu})(e^{iv} + e^{-iv})}{2(2i)} + \frac{(e^{iv} - e^{-iv})(e^{iu} + e^{-iu})}{2(2i)} \\[9pt] &= \sin u \cos v + \sin v \cos u. \qquad \blacksquare \end{align*}

  2. Proof. Similar to part (a) we compute

        \begin{align*} \cos (u+v) &= \frac{1}{2} \big( e^{i(u+v)} + e^{-i(u+v)} \big) \\[9pt] &= \frac{1}{2} \big( e^{iu} e^{iv} + e^{-iu} e^{-iv} \big) \\[9pt] &= \frac{1}{4} \big( 2 e^{iu} e^{iv} + 2 e^{-iu} e^{-iv} \big) \\[9pt] &= \frac{1}{4} \big( 2e^{iu}e^{iv} + e^{-iu}e^{iv} - e^{-iu}e^{iv} + e^{iu}e^{-iv} - e^{iu}e^{-iv} + 2 e^{-iu}e^{-iv} \big) \\[9pt] &= \frac{1}{4} \big( e^{iu}(e^{iv} + e^{-iv}) + e^{-iu}(e^{iv} + e^{-iv}) + e^{iu}(e^{iv} - e^{-iv}) - e^{-iu}(e^{iv} - e^{-iv}) \big) \\[9pt] &= \frac{1}{4} \big( (e^{iu} + e^{-iu})(e^{iv} + e^{-iv}) \big) + \frac{1}{4} \big( (e^{iu} - e^{-iu})(e^{iv} - e^{-iv})\big) \\[9pt] &= \cos u \cos v - \frac{1}{4i^2} \big( (e^{iu} - e^{-iu})(e^{iv} - e^{-iv})\big)\\[9pt] &= \cos u \cos v - \sin u \sin v. \qquad \blacksquare \end{align*}

  3. Proof. We compute

        \begin{align*} \sin^2 z + \cos^2 z &= \left( \frac{1}{2i} \right)^2 (e^z - e^{-z})^2 + \left( \frac{1}{2} \right)^2 (e^z + e^{-z})^2 \\[9pt] &= -\frac{1}{4} (e^z - e^{-z})^2 + \frac{1}{4}(e^z + e^{-z})^2 \\[9pt] &= \frac{1}{4} \left( (e^z+e^{-z})^2 - (e^z - e^{-z})^2 \right) \\[9pt] &= \frac{1}{4} \left( (e^z + e^{-z} + e^z - e^{-z})(e^z + e^{-z} - e^z + e^{-z}) \right) \\[9pt] &= \frac{1}{4} \left( (2e^z)(2e^{-z}) \right) \\[9pt] &= 1. \qquad \blacksquare \end{align*}

  4. Proof. The two computations are as follows,

        \begin{align*} \cos(iy) &= \frac{1}{2} \big( e^{i(iy)} + e^{-i(iy)} \big) \\[9pt] &= \frac{1}{2}(e^{-y} + e^y) \\[9pt] &= \frac{1}{2}(e^y + e^{-y}) \\[9pt] &= \cosh y. \\[9pt] \sin (iy) &= \frac{1}{2i} \big( e^{i(iy)} - e^{-i(iy)} \big) \\[9pt] &= \frac{1}{2i} (e^{-y} - e^y) \\[9pt] &= -\frac{1}{2i} (e^y - e^{-y}) \\[9pt] &= i \sinh y. \qquad \blacksquare \end{align*}

  5. Proof. We have,

        \[ \cos z = \cos (x+iy) = \cos x \cos (iy) - \sin x \sin (iy) = \cos x \cosh y - i \sin x \sinh y. \qquad \blacksquare \]

  6. Proof. We have,

        \[ \sin z = \sin(x+iy) = \sin x \cos (iy) + \cos x \sin (iy) = \sin x \cosh y + i \cos x \sinh y. \qquad \blacksquare \]

Prove the orthogonality relations for sine and cosine using complex exponentials

by RoRi
  1. Prove the integral formula,

        \[ \int_0^{2 \pi} e^{inx} e^{-imx} \, dx = \begin{cases} 0 & \text{if } m \neq n, \\ 2 \pi & \text{if } m = n. \end{cases} \]

    for integers m and n.

  2. Prove the following orthogonality relations of sine and cosine using the relation in part (a), where m and n are integers with m^2 \neq n^2.

        \[ \int_0^{2 \pi} \sin (nx) \cos (mx) \, dx = \int_0^{2 \pi} \sin (nx) \sin (mx) \, dx = \int_0^{2 \pi} \cos (nx) \cos (mx) \, dx = 0, \]

        \[ \int_0^{2 \pi} \sin^2 (nx) \, dx = \int_0^{2 \pi} \cos^2 (nx) \, dx = \pi \qquad \text{if } n \neq 0. \]

  1. Proof. First, if n = m then we have

        \[ \int_0^{2 \pi} e^{inx} e^{-imx} \, dx = \int_0^{2 \pi} e^{ix(n-m)} \, dx = \int_0^{2 \pi} dx = 2 \pi. \]

    If n \neq m then we have

        \begin{align*} \int_0^{2 \pi} e^{inx} e^{-imx} \, dx &= \int_0^{2 \pi} e^{ix(n-m)} \, dx \\[9pt] &= \int_0^{2\pi} e^{ikx} \, dx &(\text{for some } k \neq 0) \\[9pt] &= \frac{1}{k} \left( e^{ikx} \Bigr \rvert_0^{2\pi} \right) \\[9pt] &= \frac{1}{k} \left( e^{2 \pi i k} - e^0 \right) \\[9pt] &= \frac{1}{k} (1-1) = 0. \qquad \blacksquare \end{align*}

  2. Proof. These are all direct computations using part (a). Here they are,

        \begin{align*} \int_0^{2 \pi} \sin (nx) \cos (mx) \, dx &= \int_0^{2 \pi} \left( \frac{e^{inx} - e^{-inx}}{2i} \right) \left( \frac{e^{imx} + e^{-imx}}{2} \right) \, dx \\[9pt] &= \frac{1}{4i} \int_0^{2 \pi} \big( e^{ix(n+m)} - e^{ix(m-n)} + e^{ix(n-m)} - e^{ix(-n-m)} \big) \, dx \\[9pt] &= 0. \end{align*}

    (The final line follows by part (a) and since m^2 \neq n^2 by hypothesis which implies n \neq m, -n \neq m and -n \neq -m.) Next,

        \begin{align*} \int_0^{2\pi} \sin (nx) \sin (mx) \, dx &= \int_0^{2 \pi} \left( \frac{e^{inx} - e^{-inx}}{2i} \right) \left( \frac{e^{imx} - e^{-imx}}{2i} \right) \, dx \\[9pt] &= -\frac{1}{4} \int_0^{2 \pi} \big( e^{ix(n+m)} - e^{ix(n-m)} - e^{ix(m-n)} + e^{ix(-n-m)} \big) \, dx \\[9pt] &= 0. \end{align*}

    The third formula,

        \begin{align*} \int_0^{2 \pi} \cos (nx) \cos(mx) \, dx &= \int_0^{2 \pi} \left( \frac{e^{inx} + e^{-inx}}{2} \right) \left( \frac{e^{imx} + e^{-imx}}{2} \right) \, dx \\[9pt] &= \frac{1}{4} \int_0^{2 \pi} \big( e^{ix(n+m)} + e^{ix(n-m)} + e^{ix(m-n)} + e^{ix(-n-m)} \big) \, dx \\[9pt] &= 0. \end{align*}

    For the next one we use the identities for \cos^2 \theta and \sin^2 \theta derived in this exercise (Section 9.10, Exercise #4(b)).

        \begin{align*} \int_0^{2 \pi} \sin^2 (nx) \, dx &= \int_0^{2 \pi} \frac{1}{2}(1 - \cos(2nx)) \, dx \\[9pt] &= \pi - \int_0^{2 \pi} \cos (2nx) \, dx \\ &= \pi. \end{align*}

    Finally,

        \begin{align*} \int_0^{2 \pi} \cos^2 (nx) \, dx &= \frac{1}{2} \int_0^{2 \pi} (1 - \sin^2 nx)) \, dx \\[9pt] &= 2 \pi - \pi = \pi. \qquad \blacksquare \end{align*}

Prove DeMoivre’s theorem using complex numbers

by RoRi
  1. Prove DeMoivre’s theorem,

        \[ (\cos \theta + i \sin \theta)^n = \cos (n \theta) + i \sin (n \theta), \]

    for all \theta \in \mathbb{R} and all n \in \mathbb{Z}_{>0}.

  2. Prove the triple angle formulas for sine and cosine,

        \[ \sin (3 \theta) = 3 \cos^2 \theta \sin \theta - \sin^3 \theta, \qquad \cos (3 \theta) = \cos^3 \theta - 3 \cos \theta \sin^2 \theta, \]

    by letting n = 3 in part (a).

  1. Proof. Since \cos \theta + i \sin \theta = e^{i \theta} we have

        \[ (\cos \theta + i \sin \theta)^n = (e^{i \theta})^n = e^{ni \theta} = \cos (n \theta) + i \sin (n \theta). \qquad \blacksquare \]

  2. Letting n = 3, we first apply DeMoivre’s theorem to get

        \[ (\cos \theta + i \sin \theta)^3 = \cos (3 \theta) + i \sin (3 \theta). \]

    On the other hand, we can expand the product,

        \[ (\cos \theta + i \sin \theta)^3 = \cos^3 \theta + 3 i \cos^2 \theta \sin \theta - 3 \cos \theta \sin^2 \theta - i \sin^3 \theta \]

    Equating real and imaginary parts from the two expressions we obtain the requested identities:

        \[ \cos (3 \theta) = \cos^3 \theta - 3 \cos \theta \sin^2 \theta \quad \text{and} \quad \sin (3 \theta) = 3 \cos^2 \theta \sin \theta - \sin^3 \theta. \]

Prove formula relating trig functions of real numbers to the complex exponential

by RoRi
  1. Prove that for \theta \in \mathbb{R} we have the following formulas,

        \[ \cos \theta = \frac{e^{i \theta} + e^{-i \theta}}{2}, \qquad \sin \theta = \frac{e^{i \theta} - e^{-i\theta}}{2i}. \]

  2. Using part (a) prove that

        \[ \cos^2 \theta = \frac{1}{2}(1 + \cos (2 \theta)), \qquad \sin^2 \theta = \frac{1}{2} (1 - \cos (2 \theta)). \]

  1. Proof. We compute, using the definition of the complex exponential, e^{i \theta} = \cos \theta + i \sin \theta:

        \begin{align*} \cos \theta &= \frac{1}{2} (2 \cos \theta) \\ &= \frac{1}{2} \left( \cos \theta + i \sin \theta + \cos (\theta) - i \sin (\theta) ) \\ &= \frac{1}{2} \left( \cos \theta + i \sin \theta + \cos (-\theta) + i \sin (-\theta)) \\ &= \frac{e^{i \theta} + e^{-i \theta}}{2}. \end{align*}

    (Where in the second to last line we used that cosine is an even function and sine is odd, i.e., \cos \theta = \cos (-\theta) and \sin \theta = -\sin \theta.)
    For the second formula we compute similarly,

        \begin{align*} \sin \theta &= \frac{1}{2i} (2i \sin \theta) \\ &= \frac{1}{2i} \left( \cos \theta + i \sin \theta - \cos (-\theta) - i \sin (-\theta)) \\ &= \frac{e^{i \theta} - e^{-i \theta}}{2i}. \qquad \blacksquare \end{align*}

  2. Proof. We can compute these directly using the expressions we obtained in part (a),

        \begin{align*} \cos^2 \theta &= \left( \frac{e^{i \theta} + e^{-i \theta}}{2} \right)^2 \\ &= \frac{e^{2i \theta} + e^{-2i \theta} + 2}{4} \\ &= \frac{ \cos (2 \theta) + 2}{2} \\ &= \frac{1}{2} (1 + \cos (2 \theta)). \\ \sin^2 \theta &= \left( \frac{e^{i \theta} - e^{-i \theta}}{2i} \right)^2 \\ &= \frac{e^{2 i \theta} + e^{-2i \theta} - 2}{-4} \\ &= \frac{2 - \cos (2 \theta)}{2} \\ &= \frac{1}{2} (1 - \cos (2 \theta)). \qquad \blacksquare \end{align*}

Find the limit as x goes to 0 of (ax – asin x) / x3

by RoRi

Evaluate the limit.

    \[ \lim_{x \to 0} \frac{a^x - a^{\sin x}}{x^3}. \]

First, we want to get expansions for a^x and a^{\sin x} as x \to 0. For a^x we write a^x = e^{x \log a} and use the expansion (page 287 of Apostol) of e^x. This gives us

    \[ a^x = e^{x \log a} = 1 + (x \log a) + \frac{(\log a)^2}{2} x^2 + \frac{(\log a)^3}{6} x^3 + o(x^3). \]

Next, for a^{\sin x}, again we write a^{\sin x} = e^{\sin x \log a} and then use the expansion for e^x we have

    \[ a^{\sin x} = e^{\sin x \log a} = 1 + (\log a \sin x) + \frac{(\log a)^2}{2} (\sin x)^2 + \frac{(\log a)^3}{6} (\sin x)^3 + o((\sin x)^3). \]

Now, we need use the expansion for \sin x (again, page 287 of Apostol)

    \[ \sin x = x - \frac{x^3}{6} + o(x^4) \]

and substitute this into our expansion of a^{\sin x},

    \begin{align*} a^{\sin x} &= 1 + (\log a \sin x) + \frac{(\log a)^2}{2} (\sin x)^2 + \frac{(\log a)^3}{6} (\sin x)^3 + o((\sin x)^3) \\[9pt] &= 1 + (\log a) \left( x - \frac{x^3}{6} + o(x^4) \right) + \frac{(\log a)^2}{2} \left( x - \frac{x^3}{6} + o(x^4) \right)^2 \\ & \qquad + \frac{(\log a)^3}{6} \left( x - \frac{x^3}{6} + o(x^4) \right)^3 + o\left( \left( x - \frac{x^3}{6} + o(x^4) \right)^3 \right) \\[9pt] &= 1 + (\log a) x + \frac{(\log a)^2}{2} x^2 + \left( -\frac{\log a}{6} + \frac{(\log a)^3}{6} \right) x^3 + o(x^3). \end{align*}

(Again, this is the really nice part of little o-notation. We had lots of terms in powers of x greater than 3, but they all get absorbed into o(x^3), so we don’t actually have to multiply out and figure out what they all were. We only need to figure out the terms for the powers of x up to 3. Of course, the 3 could be any number depending on the situation; we chose 3 in this case because we know that’s what we will want in the limit we are trying to evaluate.)

So, now we have expansions for a^x and a^{\sin x} (in which most of the terms cancel when we subtract) and we can evaluate the limit.

    \begin{align*} \lim_{x \to 0} \frac{a^x - a^{\sin x}}{x^3} &= \lim_{x \to 0} \frac{\frac{\log a}{6} x^3 + o(x^3)}{x^3} \\[9pt] &= \lim_{x \to 0} \left(\frac{\log a}{6} + \frac{o(x^3)}{x^3}\right)\\[9pt] &= \frac{\log a}{6}. \end{align*}