Review question

# Can we prove $\sin a \sin b \leq \sin^2\frac{1}{2}(a+b)$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8525

## Solution

Prove that, for all values of $a$ and $b$, $\sin a \sin b \le \sin^2\frac{1}{2}(a+b).$

We know that $\cos (x+y) = \cos x\cos y-\sin x\sin y \quad \text{and} \quad \cos (x-y) = \cos x\cos y+\sin x\sin y.$

Thus $\sin x\sin y = \dfrac{1}{2}(\cos (x-y)-\cos (x+y)).$

This means $\sin \left(\dfrac{a+b}{2}\right)\sin \left(\dfrac{a+b}{2}\right) = \dfrac{1}{2}(1-\cos (a+b)).$

We also have $\sin a\sin b = \dfrac{1}{2}(\cos (a-b)-\cos (a+b)).$

Since $\cos(a-b) \leq 1$, on comparing the last two equations we see $\sin a\sin b \leq \sin^2 \left(\dfrac{a+b}{2}\right)$.

Show further that, if $a$, $b$, $c$ and $d$ all lie between $0$ and $\pi$, then $\sin a \sin b \sin c \sin d \leq \left( \sin\frac{1}{4}(a+b+c+d)\right)^4;$

If $0 \leq a, b, c, d \leq \pi,$ then

$0 \leq \sin a, \sin b, \sin c, \sin d \leq 1.$

Therefore

$0 \leq \sin a \sin b \leq \sin^2{\frac{1}{2}(a+b)}$

and

$0 \leq \sin c \sin d \leq \sin^2{\frac{1}{2}(a+b)}.$

Because the left hand side of each inequality is greater than or equal to zero, we can multiply the inequalities to get

$$$\sin a \sin b \sin c \sin d \leq \sin^2{\frac{1}{2}(a+b)} \sin^2{\frac{1}{2}(c+d)}. \label{eq:1}$$$ Using the inequality from a), we also have \begin{align} \sin{\frac{1}{2}(a+b)} \sin{\frac{1}{2}(c+d)} &\leq \sin^2{\frac{1}{2}\bigl[\frac{1}{2}(a+b)+\frac{1}{2}(c+d)\bigr]} \notag \\ \quad &= \sin^2{\frac{1}{4}(a + b + c +d)}. \label{eq:2} \end{align}

We know

$0 \leq \frac{1}{2}(a+ b), \frac{1}{2}(c+d) \leq \pi,$

and so

$0 \leq \sin \frac{1}{2}(a+b) \sin \frac{1}{2}(c+d),$

which means we can square $\eqref{eq:2}$ to get

$$$\sin^2{\frac{1}{2}(a+b)} \sin^2 \frac{1}{2}(c+d) \leq \sin^4{\frac{1}{4}(a+b+c+d)}. \label{eq:3}$$$

Combining $\eqref{eq:1}$ and $\eqref{eq:3}$ we get

\begin{align*} \sin a \sin b \sin c \sin d &\leq \sin^4\frac{1}{4}(a+b+c+d) \\ & \quad = \bigl(\sin \frac{1}{4}(a+b+c+d)\bigr)^4, \end{align*}

as required.

… and, by writing $d=\frac{1}{3}(a+b+c)$, deduce that $\sin a \sin b \sin c \le \left( \sin\frac{1}{3}(a+b+c)\right)^3.$

Setting $d = \frac{1}{3}(a+b+c),$ we see that

$a + b + c + d = \frac{4}{3}(a+b+c),$

and we can deduce that

$$$\sin a \sin b \sin c \sin \frac{1}{3}(a+b+c) \leq \bigl(\sin \frac{1}{3}(a+b+c)\bigr)^4. \label{eq:4}$$$

From $0 \leq a, b, c \leq \pi$, we deduce that $0 \leq \frac{1}{3}(a+b+c) \leq \pi$ and, except in the cases $a,b,c=0$ and $a,b,c=\pi$, we have the strict inequality $0 < \frac{1}{3}(a+b+c) < \pi$, and so

$\sin \frac{1}{3}(a+b+c) > 0.$

Given this, we can divide $\eqref{eq:4}$ by $\sin \frac{1}{3}(a+b+c)$ to get

$\sin a \sin b \sin c \leq \bigl(\sin \frac{1}{3}(a+b+c)\bigr)^3,$

as required.

The final task is to consider the cases $a = b = c = 0$ and $a = b = c = \pi.$ In both of these cases, both sides of the inequality are $0$ (as $\sin 0 = \sin \pi = 0$), and so the inequality still holds.