Approximate $\sin 29^\circ$ using differentials

Question

Using differential find approximate value of $\sqrt[3]{1.02}$

I did this. We know $f(x_0+\Delta x)-f(x_0)=\frac{df}{dx}(x_0)\cdot\Delta x$

$$f(x)=\sqrt[3]{1+x}$$

$$f(0.02)-f(0)=\frac{df}{dx}(0)\cdot0.02$$

And after calculation I solved.

But the problem I don't know how to solve is approximate using differential $\sin29^\circ$.

Answer 1

The value of $\sin (30^\circ) = \sin(\frac\pi6) = 0.5$ is known.

Now you have to convert the angles to radians.

Let $x_0 = \frac\pi6$, $\Delta x = - 1^\circ (\text{in radian}) =- \frac{\pi}{180} $.

And $\frac{df}{dx}(x_0) = \cos(x)_{x_0= \pi/6} $

$\sin(29^\circ)=f(x+x_0) = f(x_0)+\frac{df}{dx}(x_0)\Delta x = \sin\frac\pi6 - (\cos\frac\pi6)\cdot\frac\pi{180} \approx 0.4848$

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P.S. This question has answers for why we should convert degrees to radians in calculus