Question

Taking $\theta ={30}^{\circ }$, verify that:

$\left(i\right)sin3\theta =3sin\theta -4si{n}^3\theta$

$\left(ii\right)cos3\theta =4co{s}^3\theta -3cos\theta$ [2 MARKS]

Answer 1

Solution: 1 Mark each

When $\theta ={30}^{\circ }$, we have $3\theta ={90}^{\circ }$

$\left(i\right)sin3\theta =sin{90}^{\circ }=1$

$3sin\theta -4si{n}^3\theta$

$=3sin{30}^{\circ }-4si{n}^3{30}^{\circ }$

$=[3\times \frac{1}{2}-4\times {\left(\frac{1}{2}\right)}^3]=(\frac{3}{2}-4\times \frac{1}{8})$

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

$\therefore sin3\theta =3sin\theta -4si{n}^3\theta$


$\left(ii\right)cos3\theta =cos{90}^{\circ }=0$

$4co{s}^3\theta -3cos\theta =4co{s}^3{30}^{\circ }-3cos{30}^{\circ }$

$=[4\times {\left(\frac{\sqrt{3}}{2}\right)}^3-3\times \frac{\sqrt{3}}{2}]$

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

$=(\frac{3\sqrt{3}}{2}-\frac{3\sqrt{3}}{2})=0$

$\therefore cos3\theta =4co{s}^3\theta -3cos\theta$