Approximating cos⁡(47∘) Given that cos⁡(45∘)=22 what would cos⁡(47∘) be. Using differential approximation, I get cos⁡(47∘)...

When you write ${\displaystyle \cos {}^{\prime }\left(\theta \right)=-\sin \left(\theta \right)}$ (dropping the minus sign) you need to measure ${\displaystyle \theta }$ in radians. That comes out in your formula in the times 2, which should be times ${\displaystyle 2\cdot \frac{\pi }{180}}$. So
${\displaystyle \cos \left({47}^{\circ }\right)\approx \cos \left({45}^{\circ }\right)-\frac{2\pi }{180}\sin \left({45}^{\circ }\right)\approx \frac{\sqrt{2}}{2}(1-0.035)\approx 0.682}$

Check your units. The general form for differential approximation is
${\displaystyle f\left(x_0\right)+(x-x_0)\cdot \frac{d}{dx}f}$
You convert your 47, which I'm assuming is in degrees to radians by multiplying by ${\displaystyle \frac{\pi }{180}}$. This is fine. But then you use 2 degrees (I'm assuming) as your ${\displaystyle x-x_0}$ 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 ${\displaystyle (\pm )}$