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