Small angle approximation on cosine

Question

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?

Answer 1

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

Answer 2

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.

Answer 3

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

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