Question

Let $y=\sin x.\sin ({60}^{\circ }+x).\sin ({60}^{\circ }-x).$ Then for all real $x$

• A. the minimum value of $y$ is $-\frac{1}{4}$
• B. the maximum value of $y$ is $1$
• C. $y\le \frac{1}{4}$
• D. $y\ge -1$

Answer 1

Answer: (A), (C)

$y\le \frac{1}{4}$
the minimum value of $y$ is $-\frac{1}{4}$

$y=\sin x.\sin ({60}^{\circ }+x).\sin ({60}^{\circ }-x)$
$=\sin x.(\sin {60}^{\circ }.\cos x+\cos {60}^{\circ }.\sin x).(\sin {60}^{\circ }.\cos x-\cos {60}^{\circ }.\sin x)$
$=\sin x.({\sin }^2{60}^{\circ }.{\cos }^{2}x-{\cos }^2{60}^{\circ }.{\sin }^{2}x)$
$=\sin x.(\frac{3}{4}{\cos }^{2}x-\frac{1}{4}{\sin }^{2}x)$ $=\frac{1}{4}\sin x.(3{\cos }^{2}x-{\sin }^{2}x)$
$=\frac{1}{4}\sin x.(3-4{\sin }^{2}x)$ $=\frac{1}{4}(3\sin x-4{\sin }^{3}x)$ $=\frac{\sin 3x}{4}$
So range of $y$ is $[-\frac{1}{4},\frac{1}{4}]$