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

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(C) We know that $1^{\circ} = \alpha$ radians,so $1^{\prime} = \frac{\alpha}{60}$ radians. Using the differential approximation,$\cos(x + \Delta x) \a

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

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(C) We know that $1^{\circ} = \alpha$ radians,so $1^{\prime} = \frac{\alpha}{60}$ radians. Using the differential approximation,$\cos(x + \Delta x) \approx \cos(x) - \sin(x) \Delta x$. Here,$x = 60^{\circ} = \frac{\pi}{3}$ and $\Delta x = 1^{\prime} = \frac{\alpha}{60}$. Thus,$\cos(60^{\circ} 1^{\prime}) \approx \cos(60^{\circ}) - \sin(60^{\circ}) 	imes \frac{\alpha}{60}$. Substituting the values $\cos(60^{\circ}) = \frac{1}{2}$ and $\sin(60^{\circ}) = \frac{\sqrt{3}}{2}$,we get: $\cos(60^{\circ} 1^{\prime}) \approx \frac{1}{2} - \frac{\sqrt{3}}{2} 	imes \frac{\alpha}{60} = \frac{1}{2} - \frac{\alpha \sqrt{3}}{120}$.

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

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