# Stuck at Two Trigonometric Inequalities

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## Problem

Suppose that $z = x + iy, x,y \in \R$. Then $\abs{\sin z} \ge \abs{\sin x}$ and $\abs{\sin z} \ge \abs{\sinh y}$.

## Forgotten facts

I’ve forgotten many facts about complex variables learnt almost two years ago. For example, the defhinition of sines and cosines. Eight years ago, I can memorize well the trigonometric formulae. Now, I have difficulties in proving them.

\begin{align} \cos z &= \frac{e^{iz}+e^{-iz}}{2} \label{defa} \\ \sin z &= \frac{e^{iz}-e^{-iz}}{2i} \label{defb} \end{align}

## Look at the hint

In the book that I’m reading, there’re two formulae.

\begin{align} \abs{\sin z}^2 &= \sin^2 x + \sinh^2 y \label{hint1a} \\ \abs{\cos z}^2 &= \cos^2 x + \sinh^2 y \label{hint1b} \end{align}

Even though the problem can be solved once these two formulae are accepted, I don’t know why they’re true.

## Look further up

I found another two formulae.

\begin{align} \sin z &= \sin x \cosh y + i \cos x \sinh y \label{hint2a} \\ \cos z &= \cos x \cosh y - i \sin x \sinh y \label{hint2b} \end{align}

I tried to verify \eqref{hint2a} after having read \eqref{defhb}.

## First failed attempt

\begin{equation} \begin{aligned} &\quad\: \sin z \\ &= \frac{e^{iz} - e^{-iz}}{2i} \\ &= \frac{e^{i(x + iy)} - e^{-i(x + iy)}}{2i} \\ &= \frac{e^{ix} e^{-y} - e^{-ix} e^{y}}{2i} \\ &= \frac{e^{ix} - e^{-ix}}{2i} e^{-y} + \frac{e^{-ix - y}}{2i} - \frac{e^{-ix} e^{y}}{2i} \\ &= \sin x \cdot e^{-y} + \frac{e^{-ix}}{i} \cdot \frac{e^{-y} - e^{y}}{2}\\ &= \sin x \cdot e^{-y} + i e^{-ix} \frac{e^{y} - e^{-y}}{2} \\ &= \sin x \cdot e^{-y} + i e^{-ix} \sinh y \end{aligned} \label{fail1} \end{equation}

I was stuck at this point because I didn’t know how to change \eqref{fail1} to \eqref{hint2a}.

## More forgotten facts

In fact, some simple identities involving hyperbolic functions can simplify matters.

\begin{align} \cosh z &= \frac{e^{y}+e^{-y}}{2} \label{defha} \\ \sinh z &= \frac{e^{y}-e^{-y}}{2} \label{defhb} \\ \cos(iy) &= \cosh y \label{idha} \\ \sin(iy) &= i \sinh y \label{idhb} \end{align}

Then I realised that \eqref{hint2a} follows from the compound angle formula.

\begin{equation} \begin{aligned} \sin(x + iy) &= \sin x \cos(iy) + \cos x \sin(iy) \\ &= \sin x \cosh y + i \cos x \sinh y \end{aligned} \label{link} \end{equation}

Thus, I need to prove something more general.

## Compound angle formulae

\begin{align} \cos(z_1 + z_2) &= \cos z_1 \cos z_2 - \sin z_1 \sin z_2 \label{cpda} \\ \sin(z_1 + z_2) &= \sin z_1 \cos z_2 + \cos z_1 \sin z_2 \label{cpdb} \end{align}

I tried proving \eqref{cpdb} in the same way like \eqref{fail1}.

## Second failed attempt

\begin{equation} \begin{aligned} &\quad\: \sin(z_1 + z_2) \\ &= \frac{e^{i(x_1 + iy_1 + x_2 + iy_2)} - e^{-i(x_1 + iy_1 + x_2 + iy_2)}}{2i} \\ &= \frac{1}{2i} [e^{i(x_1 + x_2)} e^{-(y_1 + y_2)} - e^{-i(x_1 + x_2)} e^{y_1 + y_2}] \end{aligned} \label{fail2} \end{equation}

Then I didn’t know how to continue.

## Reasons for the above failures

• \eqref{fail1} and \eqref{fail2} involve fractions, which are more difficult to deal with than those without fractions.
• In order to change \eqref{fail1} to \eqref{hint2a}, some sort of factorisation is needed, and this requires creating a term that fits the puzzle. Again, this isn’t so easy.

## Just use ‘+’, ‘-‘ and ‘×’, no ‘÷’

From \eqref{defa} and \eqref{defb}, one should know that cosines, sines and exponents are related. For any $z_1, z_2 \in \C$, the formula for $\sin(z_1 + z_2)$ and $\cos(z_1 + z_2)$ aren’t so easy, but the one for $e^{z_1 + z_2}$ is simple. Making use of the Euler’s formula for complex numbers, one has

\begin{align} e^{iz_1} &= \cos z_1 + i \sin z_1 \label{eulerz1} \\ e^{iz_2} &= \cos z_2 + i \sin z_2 \label{eulerz2} \end{align}

Multiply \eqref{eulerz1} by \eqref{eulerz2}.

\begin{equation} \begin{aligned} &\quad\: e^{i(z_1 + z_2)} \\ &= (\cos z_1 \cos z_2 - \sin z_1 \sin z_2) \\ &\quad\: + i(\cos z_1 \sin z_2 + \sin z_1 \cos z_2) \end{aligned} \label{eiz12a} \end{equation}

Remembering Euler’s formula

$e^{i(z_1 + z_2)} = \cos(z_1 + z_2) + i \sin(z_1 + z_2),$

can we say immediately that \eqref{hint2a} and \eqref{hint2b} are true? No, since there’s no guarantee that $\cos(z_1 + z_2)$ and $\sin(z_1 + z_2)$ are real. However, I kept working on \eqref{eiz12a} so that \eqref{hint2a} and \eqref{hint2b} were proved.

From \eqref{defa} and \eqref{defb}, it’s trivial that

$\sin(-z) = -\sin z \text{ and } \cos(-z) = \cos z.$

Therefore, we write an analog of \eqref{eiz12a} for $e^{-i(z_1 + z_2)}$.

\begin{equation} \begin{aligned} &\quad\: e^{-i(z_1 + z_2)} \\ &= [\cos (-z_1) \cos (-z_2) - \sin (-z_1) \sin (-z_2)] \\ &\quad\: + i[\cos (-z_1) \sin (-z_2) + \sin (-z_1) \cos (-z_2)] \\ &= (\cos z_1 \cos z_2 - \sin z_1 \sin z_2) \\ &\quad\: - i(\cos z_1 \sin z_2 + \sin z_1 \cos z_2) \end{aligned} \label{eiz12b} \end{equation}
\begin{aligned} &\quad\: \cos(z_1 + z_2) \quad (\text{by } \frac{\eqref{eiz12a} + \eqref{eiz12b}}{2}) \\ &= \frac{e^{i(z_1 + z_2) - e^{-i(z_1 + z_2)}}}{2} \\ &= \cos z_1 \cos z_2 - \sin z_1 \sin z_2 \end{aligned}

\eqref{hint2a} can be derived in a similar way.

## Remaining work

\eqref{hint1b} can be derived from \eqref{hint2b} by using $\sin^2 x + \cos^2 x = 1$ and $\cosh^2 x - \sinh^2 x = 1$.

\begin{aligned} &\quad\, \abs{\cos z}^2 \\ &= \cos^2 x \cosh^2 y + \sin^2 x \sinh^2 y \\ &= \cos^2 x (1 + \sinh^2 y) + (1 - \cos^2 x) \sinh^2 y \\ &= \cos^2 x + \sinh^2 y \end{aligned}

The derivation of \eqref{hint1a} from \eqref{hint2a} is left as exercise.