Question
If $A =30^o,$ then prove that :$\sin 3A = 3 \sin A - 4 \sin^3A.$
✓
Answer
Given $A = 30^\circ$
$\sin 3A = \sin 3(30^\circ )$
$= \sin 90^\circ$
$=1$
$3 \sin A – 4 \sin^3A = 3 \sin 30^\circ – 4 \sin^330^\circ$
$=3\left(\frac{1}{2}\right)-4\left(\frac{1}{2}\right)^3$
$=\frac{3}{2}-\frac{1}{2}$
$=1$
$\therefore \sin 3 A=3 \sin A-4 \sin ^3 A$
$\sin 3A = \sin 3(30^\circ )$
$= \sin 90^\circ$
$=1$
$3 \sin A – 4 \sin^3A = 3 \sin 30^\circ – 4 \sin^330^\circ$
$=3\left(\frac{1}{2}\right)-4\left(\frac{1}{2}\right)^3$
$=\frac{3}{2}-\frac{1}{2}$
$=1$
$\therefore \sin 3 A=3 \sin A-4 \sin ^3 A$
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