Trigonometry Formulae - Part 6

$\begin{aligned} \cos A \color{#3D99F6}\sin C & = \cos A \color{#3D99F6} \sin (150^\circ - A) & \small \color{#3D99F6} \text{For }B = 30^\circ \\ & = \cos A (\sin 150^\circ \cos A - \cos 150^\circ \sin A) \\ & = \frac 12 \cos^2 A + \frac {\sqrt 3}2 \sin A \cos A \\ & = \frac 14 + \frac 14 \cos 2A + \frac {\sqrt 3}4 \sin 2A \\ & = \frac 14 + \frac 12 \sin (2A + 30^\circ) \end{aligned}$

This implies that $\cos A \sin C$ is minimum and maximum when $\sin (2A+30^\circ)$ is minimum and maximum. For $A \in (0^\circ, 150^\circ)$ ,

$\begin{cases} \min (\cos A \sin C) = \dfrac 14 + \dfrac 12 (-1) = - \dfrac 14 & \text{when }A = 120^\circ \\ \max (\cos A \sin C) = \dfrac 14 + \dfrac 12 (1) = \dfrac 34 & \text{when }A = 30^\circ \end{cases}$

Therefore, the range of $\cos A \sin C$ is $\boxed{\left[-\dfrac 14, \dfrac 34 \right]}$ .

Since $\angle B = 30°$ , then by the angle sum of a triangle, $C = 150° - A$ . Therefore, using some trigonometric identities,

$\cos A \sin C$

$= \cos A \sin (150° - A)$

$= \cos A (\sin 150° \cos A - \sin A \cos 150°)$

$= \sin 150° \cos^2 A - \cos 150° \sin A \cos A$

$= \sin 150° \cos^2 A - \frac{1}{2}\sin 150° - \cos 150° \sin A \cos A + \frac{1}{2} \sin 150°$

$= \frac{1}{2}(\sin 150° (2 \cos^2 A - 1) - \cos 150° (2 \sin A \cos A)) + \frac{1}{2} \sin 150°$

$= \frac{1}{2}(\sin 150° \cos 2A - \cos 150° \sin 2A) + \frac{1}{2} \sin 150°$

$= \frac{1}{2}\sin (150° - 2A) + \frac{1}{2} \sin 150°$

$= \frac{1}{2}\sin (150° - 2A) + \frac{1}{4}$

Therefore, $\cos A \sin C$ is equivalent to a sine curve with an amplitude of $\frac{1}{2}$ and shifted up $\frac{1}{4}$ , so its range is $[-\frac{1}{4}, \frac{3}{4}]$ .

CosASinC=CosA( $\dfrac{1}{2}$ CosA+ $\dfrac{√3}{2}$ SinA) = $\dfrac{1}{4}$ [1+2Sin(2A+α)], where Tanα= $\dfrac{1}{√3}$ . Since the range of Sin(2A+α) is [-1,1], therefore range of CosASinC is [- $\dfrac{1}{4}$ , $\dfrac{3}{4}$ ]