Find the approximate value of $$\cos\bigg(\frac{\pi}{5}+0.07\bigg)-\cos\bigg(\frac{\pi}{5}\bigg)$$ using linear approximation.

My attempt:

The tangent line approximation of $f(x)=\cos(x)$ at $\displaystyle x=\frac{\pi}{5}$ is

$$f(x)\approx f\bigg(\frac{\pi}{5}\bigg)+\bigg(x-\frac{\pi}{5}\bigg)f'\bigg(\frac{\pi}{5}\bigg).$$

Putting $x=\frac{\pi}{5}+0.07$, we get

$$f(x)-f\bigg(\frac{\pi}{5}\bigg)\approx -\sin\bigg(\frac{\pi}{5}\bigg)(0.07)=0.0411$$

Is my solution is right? If not, then how do I solve it?

  • $\begingroup$ Yea this is correct :) $\endgroup$ Aug 12, 2020 at 18:23

2 Answers 2


You're missing the negative sign, otherwise it's correct.

In fact we have $\cos(\frac{\pi}{5}+0.07)-\cos(\frac{\pi}{5})\approx -0.0430926$ which is not too far away from the linear approximation.


Here's another way to approach it. For very a small angle $\theta \approx 0$, we have $\cos\theta \approx 1$ and $\sin\theta \approx \theta$. You can verify that these are, in fact, the linear approximations to $f(\theta)=\cos\theta$ and $f(\theta)=\sin\theta$ near $\theta=0$.

Then, using the angle sum identity for cosine, i.e., $\cos(\alpha+\beta) = \cos\alpha\cos\beta-\sin\alpha\sin\beta$,

$$ \begin{align} \cos\left(\frac\pi5+0.07\right)&=\cos\frac\pi5\cos0.07-\sin\frac\pi5\sin0.07\\ &\approx\cos\frac\pi5-0.07\sin\frac\pi5, \end{align} $$

giving you the same result.

  • $\begingroup$ It ends up being the same as linear approximations, but it's not exactly showing an understanding of linear approximations. $\endgroup$ Aug 13, 2020 at 3:02
  • $\begingroup$ @Acccumulation Presumably the OP did the question as it was intended, so my answer is more of a side comment just to show another way. It uses linear approximations as well; I added a sentence to make that clear. $\endgroup$
    – Théophile
    Aug 13, 2020 at 3:25
  • 1
    $\begingroup$ I think this answer provides more insight than the mechanical calculation with derivatives. Unlike the latter, it connects the approximation directly to the Maclaurin series for the trigonometric functions, which --depending on the text in use-- may be the definitions used. $\endgroup$
    – tobi_s
    Aug 13, 2020 at 4:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .