Question : In right angled triangle ABC, if $\angle \mathrm{A}=30^{\circ}$, find the value of $3 \sin \mathrm{A}-4 \sin ^3 \mathrm{~A}$.
Option 1: 2
Option 2: –1
Option 3: –2
Option 4: 1
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Correct Answer: 1
Solution :
Given that $\angle A = 30^{\circ}$
We know that, $\sin A = \sin 30°= \frac{1}{2}$
Now, Substitute $\sin A = \frac{1}{2}$ into the expression $3 \sin A - 4 \sin^3 A$:
$3 \sin A - 4 \sin^3 A = 3 \times \frac{1}{2} - 4 \times \left(\frac{1}{2}\right)^3 = \frac{3}{2} - \frac{4}{8} = \frac{3}{2} - \frac{1}{2} = 1$
Hence, the correct answer is 1.
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