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

Question

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.

Answer 1

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 $\left(-\sqrt2+\dfrac12\right)^2-\dfrac54=-\sqrt2+1$.)

Answer 2

$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 $E$.

Answer 3

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

Answer 4

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 $f( \theta)= s+c+sc\;$.