Question

Prove that: $\left|sin\theta sin\right(60-\theta \left)sin\right(60+\theta \left)\right|\le \frac{1}{4}$ for all values of $\theta$

Answer 1

$\left|sin\theta sin\right(60-\theta \left)sin\right(60+\theta \left)\right|=\left|sin\theta \right(si{n}^{2}60-si{n}^2\theta \left)\right|\left\{\text{Since}sin\right(A+B\left)sin\right(A-B)=si{n}^{2}A-si{n}^{2}B\}=\left|sin\theta (\frac{3}{4}-si{n}^2\theta )\right|=\left|\frac{1}{4}sin\theta \right(3-4si{n}^2\theta \left)\right|=\left|\frac{1}{4}sin3\theta \right|=\frac{1}{4}|sin3\theta |\le \frac{1}{4}\{sin\theta |sin3\theta |\le ||\}$

So,

$|sin\theta sin({60}^{\circ }-\theta )sin({60}^{\circ }+\theta )|\le \frac{1}{4}$