Calculus inequality involving sine and cosine

community! I saw the following inequality in a calculus assignment, which I thought was harder to prove than I expected: For every $x\in\mathbb{R},$$(\sin x + a \cos x)(\sin x + b \cos x)\leq 1+(\frac{a+b}{2})^2.$$ By means of a graphing device, I noticed that when$a=b$,$f(x)=(\sin x + a\cos x)(\sin x + b \cos x)$has$1+(\frac{a+b}{2})^2$as its maximum, but in other cases the maximum of$f$is strictly less (and sometimes far less) than this number. I'd appreciate any kind of help :) • I think it will be$1+\frac{a^2+b^2}{2}$– Sachchidanand Prasad Oct 27 '17 at 1:08 3 Answers $$|\sin x+a\cos x|=\left|\sqrt{1+a^2}\left( \frac{1}{\sqrt{1+a^2}}\sin x+\frac{a}{\sqrt{1+a^2}}\cos x \right)\right|=|\sqrt{1+a^2}(\cos \theta \sin x+\sin \theta \cos x)|=|\sqrt{1+a^2}( \sin(x+\theta) )|\le \sqrt{1+a^2}.$$ So the maximum value of$\sin x+a\cos x=\sqrt{1+a^2}.\begin{align*} |(\sin x+a\cos x)(\sin x+b\cos x)|& = |\sin x+a\cos x||\sin x+b\cos x|\\ & \le \sqrt{1+a^2}\sqrt{1+b^2}=\sqrt{(1+a^2)(1+b^2)}\\ & \le \frac{(1+a^2)+(1+b^2)}{2} \tag{1}\\ &=1+\frac{a^2+b^2}{2} \end{align*} In(1)$, I have used AM-GM Inequality. Thank you all. Today I was looking at some old stuff and found this exercise, which was given to me by a student I was tutoring. Now with a fresher view I simply noticed that completing the square we have $$f(x)=(\sin x +a\cos x)(\sin x+b\cos x)=(\sin x+(\frac{a+b}{2})\cos x)^2-(\frac{a-b}{2})^2\cos^2x,$$ and thus$f(x)\leq (\sin x+(\frac{a+b}{2})\cos x)^2=(\sqrt{1+(\frac{a+b}{2})^2}\sin(x+\phi))^2\leq 1+(\frac{a+b}{2})^2.$For any trig0nometric expression of the form $$A\sin x+B\cos x=\sqrt{A^2+B^2}\left(\dfrac{A}{\sqrt{A^2+B^2}}\sin x+\dfrac{B}{\sqrt{A^2+B^2}}\cos x\right)$$ can be written in the form$\sqrt{A^2+B^2}\sin(x+\phi)$with$\tan\phi=B/A$and hence$\vert A\sin x+B\cos x|\le\sqrt{A^2+B^2}$for any$A,B\in\Bbb{R}.\$
In fact these bounds are sharp.

Here I will use above fact to give a pure trig-solution (without calculus).
First observe that

\begin{align} f(x)& = (\sin x+a \cos x)(\sin x+b \cos x) \\ & = \sin^2x+(a+b)\sin x\cos x+ab\cos^2x\\ & =\dfrac{1}{2}\left((1-\cos 2x)+(a+b)\sin 2x+ab(1+\cos 2x)\right) \\ & =\left(\dfrac{a+b}{2}\right)\sin2x+\left(\dfrac{ab-1}{2}\right)\cos 2x+\left(\dfrac{ab+1}{2}\right).\\ \end{align}

Hence $$-\sqrt{\left(\dfrac{a+b}{2}\right)^2+\left(\dfrac{ab-1}{2}\right)^2}\le f(x)-\left(\dfrac{ab+1}{2}\right)\le\sqrt{\left(\dfrac{a+b}{2}\right)^2+\left(\dfrac{ab-1}{2}\right)^2}.$$ Therefore the function has the maximum value $$\left(\dfrac{ab+1}{2}\right)+\sqrt{\left(\dfrac{a+b}{2}\right)^2+\left(\dfrac{ab-1}{2}\right)^2}$$ and minimum value $$\left(\dfrac{ab+1}{2}\right)-\sqrt{\left(\dfrac{a+b}{2}\right)^2+\left(\dfrac{ab-1}{2}\right)^2}.$$