Kaydence Huff

Answered

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 $\mathrm{cos}\left({47}^{\circ}\right)=0.68199836$

Given that

Using differential approximation, I get

Answer & Explanation

Eleanor Shaffer

Expert

2022-01-28Added 16 answers

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

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

Eleanor Shaffer

Expert

2022-01-29Added 16 answers

Check your units. The general form for differential approximation is

$f\left({x}_{0}\right)+(x-{x}_{0})\cdot \frac{d}{dx}f$

You convert your 47, which I'm assuming is in degrees to radians by multiplying by$\frac{\pi}{180}$ . 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$(\pm )$

You convert your 47, which I'm assuming is in degrees to radians by multiplying by

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

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