## anonymous 5 years ago Find the linearization, say L(x)=ax+b, of fx=sin^(2)x at the point x=pi/4

1. anonymous

Find L(x)

2. anonymous

Hi Lifesaver, You have to take a Taylor series expansion about pi/4 for sin^2(x) and truncate after the first two terms. So, $f(x)=\sum_{k=0}^{\infty}[f^k(x_0)/k!](x-x_0)^k$ Set $x_0=\pi/4$ and truncate after k=1 (i.e. sum up only the first two terms, k=0 and k=1). In this instance, $\sin^2(x)=\sin^2(x_0)+2\sin(x)\cos(x)(x-x_0)+...$ so that, $\sin^2(x)=\sin^2(\pi/4)+2\sin(\pi/4)\cos(\pi/4)(x-\pi/4)+...$ After truncating the above (i.e. dropping everything after the term 2sin(x)cos(x)(x-x_0), and finding values for sin(pi/4), cos(pi/4), etc., you have what you're looking for, $L(x)=x+(1/2-\pi/4)$ as a linear approximation to sin^2(x) about the point pi/4. I hope I've interpreted your question correctly.