Problem of the Week

Problem 8 solution

(i)

By Pythagoras, $(\sin\theta)^2+(\cos\theta)^2=1^2=1$ . Thus $\sin\theta=\sqrt{1-(\cos\theta)^2}$ since $0 \le \theta \le 180^\circ$ and hence $\sin\theta \ge 0$ .

(ii) By the cosine rule, $a^2 = b^2 + c^2 - 2bc\cos\alpha$ and hence

$\begin{aligned} \cos\alpha &= \frac{b^2+c^2-a^2}{2bc} \\ \sin\alpha &= \sqrt{1-(\cos\alpha)^2} \text{ since $0 \le \theta \le 180^\circ$ by (i) } \\ &= \sqrt{1-\left(\frac{b^2+c^2-a^2}{2bc}\right)^2} \\ &= \sqrt{\frac{4b^2c^2 - (b^2+c^2-a^2)^2}{4b^2c^2}} \\ &= \frac{\sqrt{A}}{2bc} \text{ where $A = 4b^2c^2 - (b^2+c^2-a^2)^2$.} \end{aligned}$

Expanding,

$\begin{aligned} A &= 4b^2c^2 - (a^4 + b^4 + c^4 - 2a^2b^2 - 2a^2c^2 + 2b^2c^2) \\ &= -a^4 - b^4 - c^4 + 2a^2b^2 + 2a^2c^2 + 2b^2c^2 \text{.} \end{aligned}$

By $A$ ‘s symmetry in $a$ , $b$ and $c$ , we also have

$\sin\beta = \frac{\sqrt{A}}{2ac}$

and hence

$\sin\alpha \sin\beta = \frac{A}{4abc^2} \text{.}$

Also by symmetry,

$\cos\beta = \frac{a^2+c^2-b^2}{2ac}$

and hence

$\begin{aligned} \cos\alpha \cos\beta &= \frac{(b^2+c^2-a^2)(a^2+c^2-b^2)}{4abc^2} \\ &= \frac{-a^4 - b^4 + c^4 + 2a^2b^2}{4abc^2} \text{.} \end{aligned}$

Thus

$\begin{aligned} \cos(\alpha+\beta) &= \cos\alpha \cos\beta - \sin\alpha \sin\beta \text{ (identity given in the question)} \\ &= \frac{(-a^4 - b^4 + c^4 + 2a^2b^2) - \overbrace{(-a^4 -b^4 - c^4 + 2a^2b^2 + 2a^2c^2 + 2b^2c^2)}^A}{4abc^2} \\ &= \frac{2c^4 - 2a^2c^2 - 2b^2c^2}{4abc^2} \\ &= \frac{c^2 - a^2 - b^2}{2ab} \\ &= -\frac{a^2 + b^2 - c^2}{2ab} \\ &= -\cos\gamma \text{.} \end{aligned}$

(iii)

$\begin{aligned} \cos(\alpha+\beta) &= -\cos\gamma \text{ by (ii)} \\ &= \cos(180^\circ - \gamma) \\ \end{aligned}$

Since $0 < a \le b \le c$ , $\cos\alpha = \frac{b^2+c^2-a^2}{2bc} \ge 0$ and $\cos\beta = \frac{a^2+c^2-b^2}{2ac} \ge 0$ . Thus $\alpha, \beta \le 90^\circ$ . Then since $0 \le \alpha+\beta \le 180^\circ$ and $0 \le 180^\circ - \gamma \le 180^\circ$ , it follows from

$\cos(\alpha+\beta) = \cos(180^\circ - \gamma)$

that $\alpha+\beta = 180^\circ - \gamma$ , since $\cos\theta$ is always decreasing as $\theta$ increases from 0 to $180^\circ$ , and hence $\alpha+\beta+\gamma = 180^\circ$ .