Keaton Good

2022-01-25

Can $\mathrm{cos}\left({36}^{\circ }\right)$ be calculated using known cosines like $\mathrm{cos}\left({37}^{\circ }\right)$ and the derivative formula
${f}^{\prime }\left(x\right)\approx \frac{f\left(x+\mathrm{△}x\right)-f\left(x\right)}{\mathrm{△}x}$

Wilson Mitchell

Expert

If we know that $\mathrm{cos}\left({37}^{\circ }\right)\approx 0.8$ and $\mathrm{sin}\left({37}^{\circ }\right)\approx 0.6$ then we can use the fact that
$\frac{d}{dx}\mathrm{cos}\left(x\right)=-\mathrm{sin}\left(x\right)$
to approximate $\mathrm{cos}\left({36}^{\circ }\right)$ as
$\mathrm{cos}\left({36}^{\circ }\right)\approx 0.8+0.6\left(\frac{\pi }{180}\right)=0.810\dots$
Where ${1}^{\circ }=\frac{\pi }{180}$ radians which must be used here.

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