The approximate value of cos left(59^{circ} 3 | vedclass

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Question

(A) Let $f(x) = \cos x$. We need to find the value of $\cos(59^{\circ} 30^{\prime})$.We can write $59^{\circ} 30^{\prime}$ as $60^{\circ} - 30^{\prime

Vedclass · Accepted Answer

$0.5076$

Vedclass · Answer

(A) Let $f(x) = \cos x$. We need to find the value of $\cos(59^{\circ} 30^{\prime})$.We can write $59^{\circ} 30^{\prime}$ as $60^{\circ} - 30^{\prime} = 60^{\circ} - 0.5^{\circ}$.Let $x = 60^{\circ} = \frac{\pi}{3}$ radians and $\Delta x = -0.5^{\circ} = -0.5 	imes 0.0175^{c} = -0.00875^{c}$.Using the differential formula,$f(x + \Delta x) \approx f(x) + f'(x) \Delta x$.Here,$f(x) = \cos x$,so $f'(x) = -\sin x$.$f(60^{\circ} - 0.5^{\circ}) \approx \cos(60^{\circ}) - \sin(60^{\circ}) 	imes \Delta x$.Given $\cos(60^{\circ}) = 0.5$ and $\sin(60^{\circ}) = 0.8660$.$f(59.5^{\circ}) \approx 0.5 - (0.8660) 	imes (-0.00875)$.$f(59.5^{\circ}) \approx 0.5 + 0.0075775 = 0.5075775$.Rounding to four decimal places,we get $0.5076$.

The approximate value of $\cos \left(59^{\circ} 30^{\prime}\right)$ is (given $1^{\circ} = 0.0175^{c}, \sin 60^{\circ} = 0.8660$)

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