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Find the maximum value of the expression $E=\sin\theta+\cos\theta+\sin2\theta$.

My approach is as follow ,let $E=\sin\theta+\cos\theta+\sin2\theta$, solving we get

$E^2=1+\sin^22\theta+\sin2\theta+2\sin2\theta(\sin\theta+\cos\theta)$ not able to approach from here.

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    $\begingroup$ Maybe try to find the roots of the derivative? $\endgroup$
    – lisyarus
    Jul 13, 2020 at 13:46
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    $\begingroup$ Let $x:=\sin(\theta)+\cos(\theta)$. Show that $E=x^2+x-1$ and the set of all possible values of $x$ is $[-\sqrt2,+\sqrt2]$. $\endgroup$ Jul 13, 2020 at 13:48
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    $\begingroup$ Even simpler, note that $\sin(\theta)+\cos(\theta)\leq \sqrt{2}$ and $\sin(2\theta)\leq 1$. (However, if you want to find the minimum value of $E$, my first comment is a better way.) $\endgroup$ Jul 13, 2020 at 13:56

4 Answers 4

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Following @Batominovski's hint $$\left(\sin x+\cos x+\frac12\right)^2-\frac54=2\sin x\cos x+\sin x+\cos x$$ and the maximum value of $\sin x+\cos x$ is $\sqrt2$. Hence

$$\left(\sqrt2+\frac12\right)^2-\frac54=\sqrt2+1.$$

(The minimum is $-\dfrac54$ because the squared expression can vanish. There is also a local maximum with value $\left(-\sqrt2+\dfrac12\right)^2-\dfrac54=-\sqrt2+1$.)

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  • $\begingroup$ The maximum value of $sinx+cosx$ is $\sqrt2$ and the max value of $sinx.cosx$ is $\frac{1}{2}$ and also occurs simultaneously. So, should this not be more straightforward to get $\sqrt 2+1$, unless we have to also find the minima? $\endgroup$
    – Math Lover
    Jul 13, 2020 at 14:22
  • $\begingroup$ It is a coincidence that this happens. In the domain $0 ≤ x ≤ \infty$, the minimum of $x$ is $0$, and the minimum of $\frac{1}{x}$ approaches $0$. Yet the minimum of $x + \frac{1}{x}$ is not $0 + 0$, but $2$. $\endgroup$
    – Toby Mak
    Jul 13, 2020 at 14:24
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    $\begingroup$ @MathLover: the second insight of Batominovski is indeed an excellent one. In fact, there are two coincident stationary points, giving two stationary points of the sum, one being the global maximum. $\endgroup$
    – user65203
    Jul 13, 2020 at 14:34
  • $\begingroup$ @TobyMak - it is not a coincidence. It is due to the nature of the sine and cos curve. I am not able to relate to your analogy as those are happening for two different values of x. $\endgroup$
    – Math Lover
    Jul 13, 2020 at 14:41
  • $\begingroup$ @YvesDaoust, saw your edit. It is definitely interesting. Will study this further. Thanks. $\endgroup$
    – Math Lover
    Jul 13, 2020 at 14:42
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$E = sin(\theta) + cos(\theta) + sin(2\theta) $ $=$ $ \sqrt{2} sin(\theta + \frac{\pi}{4}) + sin(2\theta)$

Replace $ \theta$ by $\theta -\frac{\pi}{4} $,

Then $E(\theta - \frac{\pi}{4}) = \sqrt{2} sin(\theta) - cos(2\theta) $ = $\sqrt{2}x - 1 + 2x^2 $ where $ x = sin(\theta)$

Now you can find the maximum of the quadratic function in the domain $[-1,1]$ which will be the maximum of the function $E$.

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$$E=\sin (\theta )+\cos (\theta )+\sin (2 \theta )$$ $$E'=\cos (\theta )-\sin (\theta )+2 \cos (2 \theta )$$ Use the tangent half-angle formula $$\theta=2 \tan ^{-1}(x) \implies E'=\frac{x^4-2 x^3-12 x^2-2 x+3 } {(1+x^2)^2 }$$ $$x^4-2 x^3-12 x^2-2 x+3=\left(x^2-4 x-3\right) \left(x^2+2 x-1\right)$$

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Let $$ \sin \theta = s, \cos \theta = c $$ We can use Lagrange Multiplier method to maximize/minimize

$$ c+s + 2 s c $$

subject to trig constraint

$$ c^2+s^2 =1 $$

$$\dfrac{1+2 s }{1+2c}=\dfrac{2c}{2s}$$

which simplifies to $$ (s-c) (1+2s+2c)=0\;$$

$$s=c,\; s+c=-\dfrac12$$

These are two conditions one each for maximum and minimum evaluation:

First case maximum

$$ s=c; s^2+c^2=1\rightarrow s=c=1/\sqrt{2};\;$$

Maximum value $$ 1/\sqrt{2}+ 1/\sqrt{2}+1 = 1+\sqrt{2}$$

Second case minimum $$ s+c=-\frac12,\; s^2+c^2=1\;$$

$$ s= \dfrac{\sqrt7-1}{4};\;c= -\dfrac{\sqrt7+1}{4};$$

Minimum value $$ s+c+ 2sc = =\dfrac{-5}{4}. $$

Both solutions verify on plot of $f( \theta)= s+c+sc\;$.

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