# Taylor approximation of cos function

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)?

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

$$f(x_0+Δx)≈f(x_0)+f′(x_0)Δx$$
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.