$\cos x + \cos 2x = 1$. Maclaurin series.

sequences-and-seriestaylor expansion

I am required to approximate $\cos (x) + \cos (2x) = 1$ using the first three non-zero terms in the respective Maclaurin series.

I have found the first three non-zero terms for both $\cos (x)$ and $\cos(2x)$:

$[\cos (x)] + [\cos (2x)] \implies [1-(1/2!)x^2+(1/4!)x^4] + [1+(-4/2!)x^2+(16/4!)x^4]$

$\therefore\space 2-(5/2)x^2+(17/24)x^4 = 1 $

When I solve the quartic equation:

$ x = +-1.75…$

$ x=+-0.69… $

When I put either value of $x$ into the original equation, I get $\approx 1.99$. Is there a mistake in my calculations or is this a conceptual error?

Thank you.

Edit: I was using degrees rather than radians. When I use $0.69$ I get $0.96$ which is $\approx 1$. Why is the other value present as a solution but does not $=1$?

Best Answer

You have an extra minus sign in the second term of the $\cos 2x$ series, but you fix it the next line.

When I put $x=0.69$ into the original equation I get about $0.96$, very close to $1$. When I plug in $1.75$ I get $-1.11$ but the MacLaurin series is not very accurate this far out.

Related Question