How to find the minimum value of the function $f=\sin\theta_1 + \sin\theta_2$, where $\theta_1 + \theta_2 + \phi = π.$

Question

How to find the minimum value of the function $f=\sin\theta_1 +$ $\sin\theta_2$, where $\theta_1 + \theta_2 + \phi = π,$ where $\phi$ is a variable such that $0 < \phi < π$

we can easily do the substitution $\theta_1 =π - \theta_2 - \phi$, and take the derivative of the $f$ and find the critical value of $f$, but when we are doing that, we have to also need to know that $\dfrac {\mathrm d \phi}{\mathrm d \theta_1}$, so how can we solve this problem ?

Edit:Basically $\theta_1$, $\theta_2$ and $\phi$ are the inner angle of a triangle which is created by connecting a point on an ellipse to the focus of that ellipse, and $\phi$ is the angle that sees the major axis of the ellipse, and I want to find the minimum value of $sin\theta_1 + sin\theta_2$.

Answer 1

$$ f=\sin(\theta_1) + \sin(\theta_2) $$

Constrained to $\theta_1 + \theta_2 + \phi = \pi$ which also implies that $\theta_2=\pi-\phi-\theta_1$

$$ f(\theta_1)=\sin(\theta_1)+\sin(\pi-\phi-\theta_1)$$

$$ f(\theta_1)=\sin(\theta_1)+\sin(\pi-(\phi+\theta_1))$$

$$ f(\theta_1)=\sin(\theta_1)+\ \sin(\pi)\cos(\phi+\theta_1)-\cos(\pi)\sin(\phi+\theta_1)$$

$$ f(\theta_1)= \sin(\theta_1)+\sin(\phi+\theta_1)$$

$$ \frac{df}{d(\theta_1)} = \cos(\theta_1)+\cos(\phi+\theta_1)$$

For maxima/minima $ \frac{df}{d(\theta_1)}=0$

$$ \cos(\theta_1)+\cos(\phi+\theta_1) =0 $$

$$ 2\cos\left(\frac{2\theta_1+\phi}{2}\right)\cos\left(\frac{\phi}{2} \right)=0$$

Clearly $ 2\cos\left(\frac{\phi}{2}\right)$ is a constant so

$$\cos\left(\frac{2\theta_1+\phi}{2}\right)=0$$

$$\left(\frac{2\theta_1+\phi}{2}\right) = \pm \frac{\pi}{2} + 2\pi k $$

$$ 2\theta_1 +\phi = \pm \pi + 4\pi k $$

$$ \theta_1 = \frac{1}{2}\left(\pm \pi + 4\pi k - \phi \right)$$

Let $k=0$ then $\theta_1= \frac{1}{2}(\pi - \phi)$

Subbing this back into $f$

$$ f(\theta_1) = \sin(\theta_1)+\sin(\phi+\theta_1)$$

$$ f(\theta_1)= \sin\left( \frac{1}{2}(\pi - \phi)\right) + \sin \left( \frac{1}{2}(\pi - \phi)+\phi \right) $$

$$ f(\theta_1)= \sin \left( \frac{\pi}{2} - \frac{\phi}{2}\right) + \sin \left(\frac{\pi}{2}+\frac{\phi}{2} \right) $$

$$ f(\theta_1)= 2\sin\left(\frac{\pi}{2} \right) \cos \left(\frac{-\phi}{2} \right) $$

$$ f(\theta_1)_{\text{max}} = 2\cos\left(\frac{\phi}{2}\right) $$

Note that you do not need to know $\dfrac {\mathrm d \phi}{\mathrm d \theta_1}$ to solve the problem since $\phi$ is a constant and not a function.

Answer 2

you can do this geometrically. i will take $x = \cos t, y = \sin t$ on the unit circle $x^2 + y^2 = 1$ with $t$ the arc length. the constraint $\theta_1 + \theta_2 + \phi = \pi$ can be turne into $$\frac{\theta_1 + \theta_2}2 = \pi/2 - \phi/2 \tag 1$$ look at the radial line $l: t = \pi/2 - \phi/2,$ the constraint (1) implies that the midpoint of the secant line $(\cos \theta_1, \sin \theta_1) \mbox{ and } (\cos \theta_2, \sin \theta_2)$ is on $l.$ the lowest of $l$ has coordinates $(-\sin(\phi/2), -\cos(\phi/2)).$ the minimum of $-sin \theta_1 + \sin \theta_2$ is twice the $y$ coordinate of the lowest point; that is $-2 \cos (\phi/2).$

Answer 3

$\forall x,y,x+y\in (0,\pi)$

\begin{align*} f(x,y) &= \sin x+\sin (\pi-x-y) \\ &= \sin x+\sin(x+y) \\ &= 2\cos \frac{y}{2} \sin \left( \frac{x}{2}+\frac{x+y}{2} \right) \\ &\ge 0 \end{align*}

Case I

$$\cos \dfrac{y}{2}=0 \implies y=\pi$$ which is beyond the domain.

Case II

$$\sin \left( x+\dfrac{y}{2} \right)=0 \implies \dfrac{x}{2}+\dfrac{x+y}{2}=0 \quad \text{or} \quad \pi$$ which is also beyond the domain.

Can only approach to zero when $(\theta_1,\phi)=(x,y)=(0^{+},0^{+})$, $(0^{+},\pi^{-})$ or $(\pi^{-},0^{+})$