Question

Find the value of
$\sin {120}^o$
$\cos (-{600}^o)$

Answer 1

$sin\left({120}^{\circ }\right)=2sin\left({60}^{\circ }\right)cos\left({60}^{\circ }\right)$ $sin2\theta =2sin\theta cos\theta$

$\Rightarrow sin\left({120}^{\circ }\right)=2\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right)$

$\Rightarrow \left[sin\right({120}^{\circ })=(\frac{\sqrt{3}}{2}\left)\right]$

$cos(-{600}^{\circ })=cos\left({600}^{\circ }\right)$

$cos(-{600}^{\circ })=cos[2\times ({360}^{\circ })-{120}^{\circ }]$

$cos(-{600}^{\circ })=+cos{120}^{\circ }$ $cos(2nn-\theta )=-cos\theta$

$cos(-{600}^{\circ })=+[2co{s}^2{60}^{\circ }-1]$ $co{s}^2\theta =2co{s}^2\theta$

$cos(-{600}^{\circ })=+\left[2\right(1/2{)}^2-1]$

$\left[cos\right(-{600}^{\circ })=-(\frac{1}{2}\left)\right]$