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Given that $\cos(45^{\circ}) = \frac{\sqrt{2}}{2}$, what would $\cos(47^{\circ})$ be.

Using differential approximation, I get $\cos(47^{\circ})$ is about $\cos\left( \frac{45\pi}{180}\right)-2\sin\left(\frac{47\pi}{180}\right)= -0.755600622$ which is of course not right as $\cos(47^{\circ}) = 0.68199836.$

Where am I going wrong in my calculation?

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When you write $\cos'(\theta)=-\sin(\theta)$ (dropping the minus sign) you need to measure $\theta$ in radians. That comes out in your formula in the times $2$, which should be times $2 \cdot \frac \pi{180}$. So $$\cos(47^\circ)\approx \cos(45^\circ)-\frac {2\pi}{180}\sin(45^\circ)\approx\frac {\sqrt 2}2(1-.035)\approx 0.682$$

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Check your units. The general form for differential approximation is $$f(x_0) + (x-x_0) \cdot \frac{d}{dx}f.$$

You convert your $47$, which I'm assuming is in degrees to radians by multiplying by $\pi/180$. This is fine. But then you use $2$ degrees (I'm assuming) as your $x-x_0$ term. You need this to be in radians.

Try your computation again, using the same method you've already used to convert from degrees to radians for 2 degrees. The answer you then get is off by about $0.004$. But I'll leave it to you to check which way it is off ($\pm$).

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Using the formula of linear approximation $f(x_0+\Delta x)\approx f(x_0)+f'(x_0)\cdot \Delta x$: $$\begin{align}f(45^\circ +2^\circ)&\approx f(45^\circ)+f'(45^\circ)\cdot 2^\circ=\\ &=\cos 45^\circ+(-\sin 45^\circ)\cdot 2^\circ=\\ &=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\cdot \left(2\cdot \frac{\pi}{180^\circ}\right)\approx \\ &\approx \frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\cdot \frac{6.283185}{180^\circ}\approx \\ &\approx \frac{\sqrt{2}}{2}\cdot \left(1-0.0349\right)\approx\\ &\approx 0.7071\cdot 0.9651\approx \\ &\approx 0.6824.\end{align}$$

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