Kaydence Huff

2022-01-27

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

Eleanor Shaffer

Expert

When you write $\mathrm{cos}{}^{\prime }\left(\theta \right)=-\mathrm{sin}\left(\theta \right)$ (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
$\mathrm{cos}\left({47}^{\circ }\right)\approx \mathrm{cos}\left({45}^{\circ }\right)-\frac{2\pi }{180}\mathrm{sin}\left({45}^{\circ }\right)\approx \frac{\sqrt{2}}{2}\left(1-0.035\right)\approx 0.682$

Eleanor Shaffer

Expert

Check your units. The general form for differential approximation is
$f\left({x}_{0}\right)+\left(x-{x}_{0}\right)\cdot \frac{d}{dx}f$
You convert your 47, which I'm assuming is in degrees to radians by multiplying by $\frac{\pi }{180}$. 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 $\left(±\right)$

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