Question

Let
$f\left(x\right)=cosx(sinx+\sqrt{si{n}^{2}x+si{n}^2\theta })$,
$\theta$ is a given constant,
then max value of $f\left(x\right)$is

Answer 1

Given that
$f\left(x\right)=cosx(sinx+\sqrt{si{n}^{2}x+si{n}^2\theta })f\left(x\right)=\frac{1}{secx}(sinx+\sqrt{si{n}^{2}x+si{n}^2\theta })\Rightarrow secx.f\left(x\right)=sinx+\sqrt{si{n}^2+si{n}^2\theta }\Rightarrow secx.f\left(x\right)-sinx=\sqrt{si{n}^2+si{n}^2\theta }\Rightarrow (secx.f(x)-sinx{)}^2=si{n}^{2}x+si{n}^2\theta \Rightarrow se{c}^{2}x{f}^2(x)+si{n}^{2}x-2sinxsecxf(x)=si{n}^{2}x+si{n}^2\theta \Rightarrow f^2(x\left)\right[1+ta{n}^{2}x]-2f(x)tanx=si{n}^2\theta \Rightarrow f^2(x)ta{n}^{2}x-2f(x)tanx+f^2(x)-si{n}^2\theta =0\text{because values of}\tan x\text{are real and it is represented by a quadratic equation}\text{So Discriminant D of this should be}D\ge 0\Rightarrow 4f^2(x)-4f^2(x\left)\right(f^2\left(x\right)-si{n}^2\theta )\ge 0\Rightarrow 4f^2(x)-4f^4(x)+4f^2(x)si{n}^2\theta \ge 0\Rightarrow 4f^4(x)-4f^2(x)si{n}^2\theta -4f^2(x)\le 0\Rightarrow 4f^2(x\left)(f^2\right(x)-si{n}^2\theta -1)\le 0So\Rightarrow (f^2(x)-si{n}^2\theta -1)\le 0\Rightarrow f^2(x)\le 1+si{n}^2\theta \Rightarrow |f(x)|\ge \sqrt{1+si{n}^2\theta }$

So max value of $f\left(x\right)$ is=$\sqrt{1+si{n}^2\theta }$