Question

Find the values of $\sin 67\frac{1^{\circ }}{2}$ and $\cos 67\frac{1^{\circ }}{2}$

Answer 1

$\cos {\left(67\frac{1}{2}\right)}^{\circ }$

$\cos \frac{\theta }{2}=\cos {135}^{\circ }$

$\cos {\left(135\right)}^{\circ }=\cos (-45+180)=-\cos {45}^{\circ }$

we know that

$\cos 2\theta =2{\cos }^2\theta -1$

$\Rightarrow 2{\cos }^2\theta =\cos 2\theta +1$

$\Rightarrow 2{\cos }^2\theta =\frac{-\sqrt{2}}{2}+1=\frac{2-\sqrt{2}}{2}$

$\Rightarrow co{s}^2\theta =\frac{2-\sqrt{2}}{4}$

$\Rightarrow cos\theta =\sqrt{\frac{2-\sqrt{2}}{4}}$

$\Rightarrow \cos 67{\frac{1}{2}}^{\circ }=\frac{\sqrt{2-\sqrt{2}}}{2}$

Now,

by using the half angle formula,

$\sin \frac{\theta }{2}=\sqrt{\frac{1-\cos \theta }{2}}$

$\sin 67{\frac{1}{2}}^{\circ }=\sin {135}^{\circ }=\sin \frac{\theta }{2}$

$\Rightarrow \sin \frac{\theta }{2}=\sqrt{\frac{1-\cos \theta }{2}}$

we take only positive value because it's lies on fist quadrant

$\sin \frac{\theta }{2}=\sqrt{\frac{1-\cos {135}^{\circ }}{2}}$

$\Rightarrow \sqrt{\frac{1-\left(\frac{-\sqrt{2}}{2}\right)}{2}}$

$\Rightarrow \sqrt{\frac{2+\sqrt{2}}{4}}$

$\Rightarrow \frac{\sqrt{2+\sqrt{2}}}{2}$

$\sin 67{\frac{1}{2}}^{\circ }=\frac{\sqrt{2+\sqrt{2}}}{2}$

Hence, $\sin 67{\frac{1}{2}}^{\circ }=\frac{\sqrt{2+\sqrt{2}}}{2}$ and $\cos 67{\frac{1}{2}}^{\circ }=\frac{\sqrt{2-\sqrt{2}}}{2}$