If 1^{circ} = alpha radians,then the approxim | vedclass

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(C) We know that $60^{\circ} 1^{\prime} = 60^{\circ} + 1^{\prime} = 60^{\circ} + \left(\frac{1}{60}\right)^{\circ}$.Since $1^{\circ} = \alpha$ radians

Vedclass · Accepted Answer

$\frac{1}{2} - \frac{\alpha \sqrt{3}}{120}$

Vedclass · Answer

(C) We know that $60^{\circ} 1^{\prime} = 60^{\circ} + 1^{\prime} = 60^{\circ} + \left(\frac{1}{60}\right)^{\circ}$.Since $1^{\circ} = \alpha$ radians,then $1^{\prime} = \left(\frac{1}{60}\right)^{\circ} = \frac{\alpha}{60}$ radians.Using the approximation $\cos(A + B) \approx \cos A - B \sin A$ for small $B$ (where $B$ is in radians):$\cos(60^{\circ} + 1^{\prime}) \approx \cos(60^{\circ}) - \left(\frac{\alpha}{60}\right) \sin(60^{\circ})$.Substituting the values $\cos(60^{\circ}) = \frac{1}{2}$ and $\sin(60^{\circ}) = \frac{\sqrt{3}}{2}$:$\cos(60^{\circ} 1^{\prime}) \approx \frac{1}{2} - \left(\frac{\alpha}{60}\right) \left(\frac{\sqrt{3}}{2}\right) = \frac{1}{2} - \frac{\alpha \sqrt{3}}{120}$.

If $1^{\circ} = \alpha$ radians,then the approximate value of $\cos(60^{\circ} 1^{\prime})$ is

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