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

## Suggestion

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

Can we find an identity for $\sin a \sin b$?

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 $0 \leq a \leq \pi$, what does that tell us about $\sin a$?

Handy fact: if $0 \leq a \leq b$, and $0 \leq c \leq d$, then $0 \leq ac \leq bd$.

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

Let’s think about the case $0 < a, b, c, d < \pi$ separately to the cases where $a = b = c = 0$ and $a = b = c = \pi.$

For the first case, what do you know about $\sin \frac{1}{3}(a+b+c)$?