Question

What is the maximum value of the function

$y=3\sin x+4\cos x$

Answer 1

Now,

$\begin{array}{c}f\left(x\right)=3\sin x+4\cos x................\left(1\right)\\ Let,3=a\cos \left(y\right)...............\left(2\right)\\ and4=a\sin \left(y\right).................\left(3\right)\\ Now,{\left(acos\left(y\right)\right)}^2+{\left(a\sin \left(y\right)\right)}^2=3^2+4^2\\ \Rightarrow a^2\left({\cos }^2\left(y\right)\right)+{\sin }^2\left(y\right)=9+16\\ \Rightarrow a^2=25\\ \Rightarrow a=\mathrm{5......}\left(4\right)\\ \Rightarrow 3=5\cos \left(y\right)..........\left(5\right)\\ \Rightarrow 4=5\sin \left(y\right)...........\left(6\right)\\ and\tan \left(y\right)=\frac{4}{3}\\ soeqation\left(1\right)becomes,\\ f\left(x\right)=5\cos \left(y\right)\sin \left(x\right)+\sin \left(y\right)\cos \left(x\right)\\ =5\cos \left(y\right)\sin \left(x\right)+\sin \left(y\right)\cos \left(x\right)\\ =5\sin \left(x+y\right)\\ from\left(7\right)wehave,\\ f\left(x\right)=5\sin \left(x+{\tan }^{-1}\left(\frac{4}{3}\right)\right)........\left(8\right)\\ as,-1\le \sin \left(x+{\tan }^{-1}\left(\frac{4}{3}\right)\right)\le 1\\ \Rightarrow -5\le \sin \left(x\right)+{\tan }^{-1}\left(\frac{4}{3}\right)\le 5\\ somaxf\left(x\right)=5\end{array}$