Evaluate: sin(22 frac{1}{2}^{circ}) | vedclass

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Question

(C) We know the half-angle formula: $\sin^2(\frac{A}{2}) = \frac{1 - \cos A}{2}$.Let $A = 45^{\circ}$. Then $\frac{A}{2} = 22 \frac{1}{2}^{\circ}$.Sub

Vedclass · Accepted Answer

$\sqrt{\frac{2-\sqrt{2}}{4}}$

Vedclass · Answer

(C) We know the half-angle formula: $\sin^2(\frac{A}{2}) = \frac{1 - \cos A}{2}$.Let $A = 45^{\circ}$. Then $\frac{A}{2} = 22 \frac{1}{2}^{\circ}$.Substituting the value of $A$:$\sin^2(22 \frac{1}{2}^{\circ}) = \frac{1 - \cos 45^{\circ}}{2}$$= \frac{1 - \frac{1}{\sqrt{2}}}{2}$$= \frac{\frac{\sqrt{2}-1}{\sqrt{2}}}{2} = \frac{\sqrt{2}-1}{2\sqrt{2}}$To rationalize the denominator,multiply numerator and denominator by $\sqrt{2}$:$= \frac{\sqrt{2}(\sqrt{2}-1)}{2\sqrt{2} 	imes \sqrt{2}} = \frac{2-\sqrt{2}}{4}$Therefore,$\sin(22 \frac{1}{2}^{\circ}) = \sqrt{\frac{2-\sqrt{2}}{4}}$.

Evaluate: $\sin(22 \frac{1}{2}^{\circ})$

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