Question
If $2 \sin A-1=0$, show that: $\sin 3 A=3 \sin A-4 \sin ^3 A$
✓
Answer
$2 sinA − 1 = 0$
$\Rightarrow sinA = 1/2$
We know $\sin 30^{\circ}=1 / 2$
So, $A = 30^\circ$
$LHS = \sin 3A = sin90^\circ = 1$
$RHS = 3sinA - 4sin^3A$
$=3 sin30^\circ - 4sin^330^\circ$
$=3\left(\frac{1}{2}\right)-4\left(\frac{1}{2}\right)^3$
$=\frac{3}{2}-\frac{1}{2}=1$
LHS = RHS
$\Rightarrow sinA = 1/2$
We know $\sin 30^{\circ}=1 / 2$
So, $A = 30^\circ$
$LHS = \sin 3A = sin90^\circ = 1$
$RHS = 3sinA - 4sin^3A$
$=3 sin30^\circ - 4sin^330^\circ$
$=3\left(\frac{1}{2}\right)-4\left(\frac{1}{2}\right)^3$
$=\frac{3}{2}-\frac{1}{2}=1$
LHS = RHS
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