Question
Using differentials, find the approximate values of the following:
$\cos61^\circ$ it being given that $\sin60^\circ=0.86603$ and $0.01745$ radian
$\cos61^\circ$ it being given that $\sin60^\circ=0.86603$ and $0.01745$ radian
✓
Answer
Consider the function $\text{y}=\text{f} (\text{x})=\cos\text{x}^\circ$ Let:
x = 60° $\text{x}+\triangle \text{x}=61^\circ$ Then,
$\triangle\text{x}=1^ \circ=0.01745$For $\text{x}=60^\circ$
$\text{y}\cos60^\circ- 0.5$Let:
$\text{dx}=\triangle \text{x}=0.01745$ Now, $\text{y}=\cos\text {x}$
$\Rightarrow\frac {\text{dy}}{\text{dx}}=-\sin\text{x}$ $\Rightarrow\Big(\frac {\text{dy}}{\text{dx}}\Big)_{\text{x} =60}=-0.86603$ $\therefore\triangle \text{y}=\text{dy}=\frac{\text{dy}} {\text{dx}}\text{dx}=- 0.86603\times0.01745=-0.01511$ $\Rightarrow\text{y} =0.01511$ $\therefore\cos61^ \circ=\text{y}+\triangle\text{y} =0.48488\approx0.48489$
x = 60° $\text{x}+\triangle \text{x}=61^\circ$ Then,
$\triangle\text{x}=1^ \circ=0.01745$For $\text{x}=60^\circ$
$\text{y}\cos60^\circ- 0.5$Let:
$\text{dx}=\triangle \text{x}=0.01745$ Now, $\text{y}=\cos\text {x}$
$\Rightarrow\frac {\text{dy}}{\text{dx}}=-\sin\text{x}$ $\Rightarrow\Big(\frac {\text{dy}}{\text{dx}}\Big)_{\text{x} =60}=-0.86603$ $\therefore\triangle \text{y}=\text{dy}=\frac{\text{dy}} {\text{dx}}\text{dx}=- 0.86603\times0.01745=-0.01511$ $\Rightarrow\text{y} =0.01511$ $\therefore\cos61^ \circ=\text{y}+\triangle\text{y} =0.48488\approx0.48489$
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