Question

how to prove this that (acostheta+bsintheta)^2is less than or equal to a^2+b^2 by cauchy schwarz inequality . this is linear algebra problem

Answer 1

Let\[u = (a,b)\]\[v = (\cos\theta,\sin\theta)\]by the Cauchy-Schwarz inequality:\[\langle u, v\rangle \leq \|u\|\|v\| \Leftrightarrow a\cos\theta+b\sin\theta \leq \sqrt{a^2+b^2}\sqrt{\cos^2\theta+\sin^2\theta}=\sqrt{a^2+b^2}\]squaring both sides of the inequality,\[(a\cos\theta+b\sin\theta)^2 \leq a^2 + b^2.\]