# Small angle approximation on cosine

The problem is

Using the small angle approximation of cosine, show that $$3-2\cos(x)+4\cos^2(x)\approx 5-kx^2$$ where k is a positive constant

I did solve it by using $$\cos^2(x)=1-\sin^2(x)$$ on the $$\cos^2(x)$$, by plugging $$\sin^2(x)\overset{x\to 0}{\approx}x^2$$ and $$\cos(x)\overset{x\to 0}{\approx}1-\frac{x^2}{2}$$ to get $$3-2(1-\frac{x^2}{2})+4(1-x^2)=5-3x^2$$ hence $$k=3$$. But why does using $$\cos^2(x)\overset{x\to 0}{\approx}(1-\frac{x^2}{2})^2$$ doesn't work out? I originally tried plugging that into the $$\cos^2(x)$$ but got another complete thing. why?

• What "complete other thing" did you get? The approximation you suggest should have worked. We can't show you your mistake if you don't show us what you did. Commented Nov 29, 2020 at 15:57
• You can also use $4\cos ^2 x = 2 + 2\cos (2x)$ to get rid of squaring.
– Gary
Commented Nov 29, 2020 at 15:58
• let me post my work, $$3-2(1-\frac{x^2}{2})+4(1-\frac{x^2}{2})^2$$ $$3-2+x^2+4(1-x^2+\frac{x^4}{4})$$ $$1+x^2+4-4x^2+x^4$$ $$x^4-3x^2+5$$ Commented Nov 29, 2020 at 16:00
• "got another complete thing": you must tell us which.
– user65203
Commented Nov 29, 2020 at 16:09
• the one i just posted? @YvesDaoust Commented Nov 29, 2020 at 16:11

## 3 Answers

Directly subbing $$\cos^2x=(1-x^2/2+\cdots)^2$$ should work out, provided you expand properly: $$\cos^2x=1-2(x^2/2)+\dots=1-x^2+\cdots$$ $$1-\sin^2x=1-(x-\cdots)^2=1-x^2+\cdots$$

• what are the dots for? can you show how it should work, because i didn't get to the answer by that method at all, i can post my unsuccessful working if you want. Commented Nov 29, 2020 at 15:56
• @Xetrez Dots indicate $x^3$ or higher terms, which we don't need. If you had thrown away the $x^4$ - because the question only required expanding to $x^2$ - you would have got it. Commented Nov 29, 2020 at 16:04
• so is it because this is an approximation, and x^4 when x tends to 0, are smaller and negligible than for example an x^2? Commented Nov 29, 2020 at 16:12
• @Xetrez Yes.${}$ Commented Nov 29, 2020 at 16:12

I think you simply forgot to plug $$cos\theta \approx 1 - \frac{x^2}{2}$$ in for both $$cos\theta$$ and $$cos\theta^2$$ in the expression.

• i posted my working at a post above this one on the comment section, you may want to take a look Commented Nov 29, 2020 at 16:03

By Taylor, as the function is even the development to second order is $$f(0)+\dfrac{f''(0)}2x^2$$ or

$$(3-2\cos(0)+4\cos^2(0))+(2\cos(0)+8\sin^2(0)-8\cos^2(0))\frac{x^2}2.$$

• I'm just wondering; would it be more appropiate to say "By Macluarin" or is "By Taylor" also fine as obviously it's a generalisation of Maclaurin series? Commented Nov 29, 2020 at 21:18
• @A-LevelStudent: on this site Taylor seems to be the preferred name in all cases, though Maclaurin is indeed more specific to this case. As you can trade one for the other by a simple shift of the variable, the distinction is unimportant.
– user65203
Commented Nov 30, 2020 at 7:32
• Ok, thanks for the clarification. Commented Nov 30, 2020 at 16:46