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I have the following problem:

Knowing the linear approximation of the Taylor approximation of the form(1):

$$ f(x_{0} + \Delta x) \approx f(x_{0}) + f'(x_{0}) \Delta x $$

I have to determine the linear approximation(2) $$\cos (32) = \cos(30^{\circ} + 2^{\circ}) = cos(\frac{\pi}{6}+\frac{2\pi}{180})$$ with the help of (3)

$$\cos(30^{\circ}) = \cos( \frac {\pi}{6}) = \frac{\sqrt{3}}{2}$$

The answer is: (4)

$$\cos(32^{\circ}) = cos(\frac{\pi}{6}+\frac{2\pi}{180})$$ (5) $$\approx \cos(\frac{\pi}{6}) - \sin(\frac{\pi}{6})\frac{\pi}{90}$$

(6) $$= \frac{\sqrt{3}}{2}-\frac{2\pi}{180}$$

Which (trigonometric?) rule allows us to pass from (4) to (5)?

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  • $\begingroup$ I believe they used the small-angle approximation to go from (4) to (5). When $\alpha$ and $\beta$ are small, $\cos(\alpha+\beta)\approx\cos(\alpha)-\beta\sin(\alpha)$. $\endgroup$ – Sriram Gopalakrishnan Sep 28 '18 at 22:22
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    $\begingroup$ $(4)$ to $(5)$ would be $f(x_{0} + \Delta x) \approx f(x_{0}) + f'(x_{0}) \Delta x$ that you mentioned above. $\endgroup$ – robjohn Sep 28 '18 at 22:29
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Your answer is in your question. In the beginning of your question, you provide

$f(x_0+Δx)≈f(x_0)+f′(x_0)Δx$

All you need to do is plug $x_0=\frac{\pi}{6}$, $Δx=\frac{\pi}{90}$, and $f(x)=\cos(x)$ into $f(x_0+Δx)≈f(x_0)+f′(x_0)Δx$, and you arrive at your answer.

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