Question

For $A\in R$, then prove that
i) $\sin A\sin (60+A)\sin (60-A)=\frac{1}{4}\sin 3A$
ii) $\cos A.\cos (60+A).\cos (60-A)=\frac{1}{4}\cos 3A$
iii) $\sin {20}^o.\sin {40}^o.\sin {60}^o.\sin {80}^o=\frac{3}{16}$
iv) $\cos \frac{\pi }{9}.\cos \frac{2\pi }{9}.\cos \frac{3\pi }{9}.\cos \frac{4\pi }{9}=\frac{1}{16}$

Answer 1

i)

$\sin A\sin (60+A)\sin (60-A)$

$=\sin A[{\sin }^{2}60-{\sin }^{2}A]$

$=\sin A[\frac{3}{4}-{\sin }^{2}A]$

$=\frac{3}{4}\sin A-{\sin }^{3}A$

$=\frac{3}{4}\sin A-[\frac{3}{4}\sin A-\frac{1}{4}\sin 3A]$

$=\frac{1}{4}\sin 3A$

iii)

$\sin {20}^{\circ }\cdot \sin {40}^{\circ }\cdot \sin {60}^{\circ }\cdot \sin {80}^{\circ }\cdot$

$=\frac{\sqrt{3}}{2}\left[\sin 20\sin \right(60-20\left)\sin \right(60+20\left)\right]$

$=\frac{\sqrt{3}}{2}\left[\frac{1}{4}\sin 3\right(20\left)\right]$

$=\frac{\sqrt{3}}{2}\left[\frac{1}{4}\sin 60\right]$

$=\frac{\sqrt{3}}{2}\times \frac{1}{4}\times \frac{\sqrt{3}}{2}$

$=\frac{3}{16}$

ii)

$\cos A\cos (60+A)\cos (60-A)$

$=\cos A(\cos A\cos 60-\sin A\sin 60)(\cos A\cos 60+\sin A\sin 60)$

$=\cos A({\cos }^{2}A{\cos }^{2}60-{\sin }^{2}A{\sin }^{2}60)$

$=\cos A(\frac{1}{4}{\cos }^{2}A-\frac{3}{4}{\sin }^{2}A)$

$=\frac{1}{4}\cos A({\cos }^{2}A-3(1-{\cos }^{2}A\left)\right)$

$=\frac{1}{4}\cos A(4{\cos }^{2}A-3)$

$=\frac{1}{4}(4{\cos }^{3}A-3\cos A)$

$=\frac{1}{4}\cos 3A$

iv)

$\cos \frac{\pi }{9}\cdot \cos \frac{2\pi }{9}\cdot \cos \frac{3\pi }{9}\cdot \cos \frac{4\pi }{9}$

$=\frac{1}{2}\cos \frac{\pi }{9}\cos \frac{2\pi }{9}\cdot \cos \frac{4\pi }{9}$

$=\frac{1}{2}\cdot \frac{1}{4}\cos \frac{3\pi }{9}$

$=\frac{1}{2}\cdot \frac{1}{4}\cdot \frac{1}{2}$

$=\frac{1}{16}$