Question

Prove that:

$cot22{\frac{1}{2}}^{\circ }=\sqrt{2}+1$

Answer 1

We know that,

$sin\frac{A}{2}=\pm \sqrt{\frac{1-cosA}{2}}PutA={45}^{\circ },sin22{\frac{1}{2}}^{\circ }=\sqrt{\frac{1-cos{45}^{\circ }}{2}}\{\text{since}sin22\frac{1}{2},\text{is positive}\}=sqrt\frac{:+\frac{1}{2}}{2}sin22{\frac{1}{2}}^{\circ }=\sqrt{\frac{\sqrt{2}-1}{2\sqrt{2}}}$

Put $A={45}^{\circ }$

$cos22{\frac{1}{2}}^{\circ }=\sqrt{\frac{1+cos{45}^{\circ }}{2}}=\sqrt{\frac{1-\frac{1}{2}}{2}}$

$cos22{\frac{1}{2}}^{\circ }=\sqrt{\frac{\sqrt{2}+1}{2\sqrt{2}}}$

Now,

$cos22{\frac{1}{2}}^{\circ }=\frac{cos22{\frac{1}{2}}^{\circ }}{sin22{\frac{1}{2}}^{\circ }}=\sqrt{\frac{\sqrt{2}+1}{2\sqrt{2}}\times \frac{2\sqrt{2}}{\sqrt{2}-1}}=\sqrt{\frac{\sqrt{2}+1}{\sqrt{2}-1}}$

Rationalizing denominator

$=\sqrt{\frac{\sqrt{2}+1}{\sqrt{2}-1}\times \frac{\sqrt{2}+1}{\sqrt{2}+1}}=\sqrt{\frac{(\sqrt{2}+1{)}^2}{2-1}}\Rightarrow cot22{\frac{1}{2}}^{\circ }=\sqrt{2}+1$