Question

The minimum value of sinx. sin(60-x). sin(60 + x) is ....​

Answer 1

$\: \: \: \: \: \: sinx.sin(60 + x).sin(60 - x) \\ = sinx.(sin60cosx + cos60sinx).(sin60cosx - cos60sinx) \\ = sinx.( \frac{ \sqrt{3} }{2} cosx + \frac{1}{2} sinx).( \frac{ \sqrt{3} }{2} cosx - \frac{1}{2} sinx) \\ = sinx.( \frac{3}{4} {cos}^{2} x - \frac{1}{4} {sin}^{2} x) \\ = sinx.( \frac{3 {cos}^{2} x - {sin}^{2}x }{4} ) \\ = sinx.( \frac{3(1 - {sin}^{2}x) - {sin}^{2} x }{4} ) \\ = sinx.( \frac{3 - 3 {sin}^{2} x - {sin}^{2}x }{4} ) \\ = sinx( \frac{3 - 4 {sin}^{2}x }{4} ) \\ = \frac{3sinx - 4 {sin}^{3}x }{4} \\ = \frac{1}{4} sin3x$

if the derivation is asked, you have to write this or this actually is a direct formula i.e.

sinx.sin(60 + x).sin(60 - x) = (1/4) (sin3x)

now, we know that minimum value of sin3x = -1

(because sine of any angle cannot preceed -1)

hence, minimum value of (1/4)(sin3x) = (1/4)(-1) = -1/4

Hence, minimum value of sinx.sin(60 + x).sin(60 - x) is -1/4 or -0.25.