Question

Given $\cos {45}^o=\frac{1}{\sqrt{2}}$, then find the value of $\sin 22{\frac{1}{2}}^{o}and\cos 22{\frac{1}{2}}^o$

Answer 1

$\cos {45}^{\circ }=\frac{1}{\sqrt{2}}$

We know that $2{\sin }^{2}A=1-\cos 2A$

So,

$2{\sin }^2{\left(\frac{45}{2}\right)}^{\circ }=1-\cos {45}^{\circ }$

$2{\sin }^2{\left(\frac{45}{2}\right)}^{\circ }=1-\frac{1}{\sqrt{2}}$

$2{\sin }^2{\left(\frac{45}{2}\right)}^{\circ }=1-\frac{\sqrt{2}}{2}$

${\sin }^2{\left(\frac{45}{2}\right)}^{\circ }=\frac{1}{2}-\frac{\sqrt{2}}{4}$

$\sin {\left(\frac{45}{2}\right)}^{\circ }=\sqrt{\frac{1}{2}-\frac{\sqrt{2}}{4}}$

$=\frac{\sqrt{2-\sqrt{2}}}{2}$

Now, $2{\cos }^{2}A=1+\cos 2A$

So,

$2{\cos }^2{\left(\frac{45}{2}\right)}^{\circ }=1+\cos {45}^{\circ }$

$2{\cos }^2{\left(\frac{45}{2}\right)}^{\circ }=1+\frac{1}{\sqrt{2}}$

$2{\cos }^2{\left(\frac{45}{2}\right)}^{\circ }=1+\frac{\sqrt{2}}{2}$

${\cos }^2{\left(\frac{45}{2}\right)}^{\circ }=\frac{1}{2}+\frac{\sqrt{2}}{4}$

$\cos {\left(\frac{45}{2}\right)}^{\circ }=\sqrt{\frac{1}{2}+\frac{\sqrt{2}}{4}}$

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