# Approximating $\cos(47^{\circ})$

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?

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 $$\cos(47^\circ)\approx \cos(45^\circ)-\frac {2\pi}{180}\sin(45^\circ)\approx\frac {\sqrt 2}2(1-.035)\approx 0.682$$
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 ($$\pm$$).
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}