Question

(i) If sin $x=\frac{1}{3}$, find the value of sin $3x$

(ii) If cos $x=\frac{1}{2}$, find the value of cos $3x$

Answer 1

$\left(i\right)\text{sin}3x=(3\text{sin}x-4{\text{sin}}^{3}x)$

=$\{(3\times \frac{1}{3})-4\times {\left(\frac{1}{3}\right)}^3\}=(1-\frac{4}{27})=\frac{23}{27}$

(ii)$\text{cos}3x=(4{\text{cos}}^{3}x-3\text{cos}x)$

=$\{(4\times \frac{1}{8})-(3\times \frac{1}{2})\}=(\frac{1}{2}-\frac{3}{2})=-1$