It can be proven that the maximum value of $\sin(x)\cos(y)$ is 0.5 when $x=y$, so can it be simply concluded that the maximum value of $\sin(x)\cos(y) + \sin(y)\cos(z) + \sin(z)\cos(x)$ is 1.5 when $x=y=z$?
Maximum of $\sin(x)\cos(y) + \sin(y)\cos(z) + \sin(z)\cos(x)$
657 Views Asked by user120625 https://math.techqa.club/user/user120625/detail AtThere are 3 best solutions below
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By AM-GM $$\sum_{cyc}\sin{x}\cos{y}\leq\sum_{cyc}|\sin{x}||\cos{y}|\leq\frac{1}{2}\sum_{cyc}(\sin^2x+\cos^2y)=\frac{3}{2}.$$ As you said the equality occurs for $x=y=z=\frac{\pi}{4},$ which says that we got a maximal value.
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"...can it be simply concluded..."
Yes and No
Yes
You have three terms, each of which has a max value of $(1/2)$.
Then the max value of the sum must be less than or equal to
$3 \times (1/2) = 1.5$
However, when considering the three variables, superficially, there is no guarantee that you will be able to find a specific set of values for $x$, $y$, and $z$ such that each term (e.g. $\sin x \cos y$) will achieve its maximum value.
In this case, given that you know that the sum must be
less than or equal to $1.5$,
you can point to the specific example of
$x = y = z = \pi/4$ and demonstrate manually
that the desired sum achieves the maximum possible value with
this group of values for $x,y$, and $z$.
By Cauchy-Schwarz inequality, we get $$ \begin{align} &\ \sin x \cos y + \sin y \cos z + \sin z \cos x \\ &\le \sqrt{\sin^2x+\sin^2y+\sin^2z}~\sqrt{\cos^2x+\cos^2y+\cos^2z} \\ &= \sqrt{\sin^2x+\sin^2y+\sin^2z}~\sqrt{3-(\sin^2x+\sin^2y+\sin^2z)} \end{align} $$
First approach to continue:
Now by taking $t := \sin^2x+\sin^2y+\sin^2z$, the problem comes down to find the maximum value of $$f(t) = \sqrt{t}~\sqrt{3-t}= \sqrt{3t-t^2}$$ But $$f(t) = \sqrt{3t-t^2}=\sqrt{\frac{9}{4}-\left(t-\frac{3}{2}\right)^2} \le \frac{3}{2}$$ with $f(3/2)=3/2$.
Second approach to continue:
Thanks and Credits to @MichaelRozenberg :
By AM-GM: $$ \begin{align} &\ \sqrt{\sin^2x+\sin^2y+\sin^2z}~\sqrt{\cos^2x+\cos^2y+\cos^2z} \\ &\le \frac{(\sin^2x+\sin^2y+\sin^2z) + (\cos^2x+\cos^2y+\cos^2z)}{2} \\ &= \frac{(\sin^2x+\cos^2x)+(\sin^2y + \cos^2y) + (\sin^2z + \cos^2z)}{2} \\ &= \frac{3}{2}. \end{align} $$