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If $\theta_1,\theta_2(0\leq\theta_1,\theta_2<2\pi)$ are two solutions of $\sin(\theta+\phi)=\frac{1}{2}\sin(2\phi)$, prove that

$$\frac{\sin(\theta_1)+ \sin(\theta_2) }{ \cos(\theta_1)+ \cos(\theta_2)} =\cot\phi$$

I have tried with the following process:

Since $\theta_1,\theta_2(0\leq\theta_1,\theta_2<2\pi)$ are two solutions of $\sin(\theta+\phi)=\frac{1}{2}\sin(2\phi)$, we have $\sin(\theta_i+\phi)=\frac{1}{2}\sin(2\phi)$, I=1,2.

This gives $\sin\theta_i\cos\phi+ \cos\theta_i \sin\phi= \frac{1}{2}\sin(2\phi)$, I=1,2.

I don't know what will be the next process.

MKS
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5 Answers5

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There is probably some particular case to see.

The two equations $$\sin(\theta_1)\cos(\phi)+\sin(\phi)\cos(\theta_1)=\sin(\phi)\cos(\phi)$$

$$\sin(\theta_2)\cos(\phi)+\sin(\phi)\cos(\theta_2)=\sin(\phi)\cos(\phi)$$

can be considered as a system of two linear equations with unknown $\cos(\phi)$ and $\sin(\phi)$. The determinant of the system is then $D=\sin(\theta_1)\cos(\theta_2)-\sin(\theta_2)\cos(\theta_1)=\sin(\theta_2-\theta_1)$. If $D\not =0$, we can solve by Cramer's formula, and we get $$D\cos(\phi)=\sin(\phi)\cos(\phi)(\cos(\theta_2)-\cos(\theta_1))$$ and $$D\sin(\phi)=\sin(\phi)\cos(\phi)(\sin(\theta_1)-\sin(\theta_2))$$

and if all these quantities are non zero, we get $$\cot(\phi)=\frac{\cos(\theta_2)-\cos(\theta_1)}{\sin(\theta_1)-\sin(\theta_2)}=\frac{\sin(\theta_1)+\sin(\theta_2)}{\cos(\theta_1)+\cos(\theta_2)}$$

Kelenner
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  • 's, how do you get the last line can you explain please. – MKS Jul 21 '16 at 15:33
  • If you multiply you have to verify that $(\cos(\theta_2)-\cos(\theta_1))(\cos(\theta_2)+\cos(\theta_1))=(\sin(\theta_1)-\sin(\theta_2))(\sin(\theta_1)+\sin(\theta_2))$ and you simply use that $\sin(x)^2+\cos(x)^2=1$ – Kelenner Jul 21 '16 at 15:37
  • thank you for your response – MKS Jul 21 '16 at 16:45
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$$\sin{\theta}\cos{\phi}+\cos{\theta}\sin{\phi}=\sin{\phi}\cos{\phi} \Longleftrightarrow \cos{\phi}(\sin{\theta}-\sin{\phi})=-\cos{\theta}\sin{\phi}$$ Squaring both sides we get, $$\begin{align}\cos^2{\phi}(\sin^2{\theta}+\sin^2{\phi}-2\sin{\theta}\sin{\theta})=\cos^2{\theta}\sin^2{\phi}\\ \Longleftrightarrow \sin^2{\theta}-2\sin{\phi}\cos^2{\phi}\sin{\theta}-\sin^4{\phi}=0\tag{Why?}\\ \end{align}$$

We can treat this as a quadratic in $\sin{\theta}$ with roots as $\sin{\theta_1}$ and $\sin{\theta_2}$ as $\phi$ is a constant.

So, $$\sin{\theta_1}+\sin{\theta_2}=2\sin{\phi}cos^2{\phi}\tag{1}$$

Similarly, $$\cos{\theta_1}+\cos{\theta_2}=2\cos{\phi}sin^2{\phi}\tag{2}$$

Dividing $(1)$ by $(2)$, we get the desired result.

Roby5
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$$\sin(\theta+\phi)=\frac{1}{2}\sin(2\phi)$$ $$\sin\theta \cos \phi + \sin\phi \cos \theta = \sin \phi \cos\phi$$ $$\sin\theta \cot \phi + \cos \theta = \cos\phi$$ $$\sin\theta \cot \phi - \cos \phi = -\cos\theta$$

Squaring both sides, we get $$\sin^2\theta \cot^2 \phi + \cos^2 \phi - 2\sin\theta \cot \phi \cos \phi= \cos^2\theta=1-\sin^2\theta \tag1$$ $$\sin^2\theta (1+\cot^2 \phi) + \cos^2 \phi - 2\sin\theta \cot \phi \cos \phi -1=0$$ $$\sin^2\theta \csc^2 \phi - 2\sin\theta \cot \phi \cos \phi - \sin^2 \phi =0$$ $$\sin^2\theta - 2\sin\theta \sin \phi \cos^2 \phi - \sin^4 \phi =0$$

Hence $\sin \theta_1+\sin \theta_2=2 \sin \phi \cos^2 \phi$ and accordingly, if instead of converting the $\cos$ to $\sin$ in $(1)$, you convert the $\sin$ to $\cos$ , you will get $\cos \theta_1+\cos \theta_2=2 \sin^2 \phi \cos \phi$

The answer will follow. Hope this is clear.

SchrodingersCat
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Using Prosthaphaeresis Formulas

$$\frac{\sin(\theta_1)+ \sin(\theta_2) }{ \cos(\theta_1)+ \cos(\theta_2)}=\tan\left(\dfrac{\theta_1}2+\dfrac{\theta_2}2\right)$$

Now using Weierstrass Substitution and $\sin2\phi=2\sin\phi\cos\phi$ and setting $\tan\dfrac\theta2=t$

$$t^2\sin\phi(1+\cos\phi)-2t\cos\phi+\sin\phi(1-\cos\phi)=0$$ whose roots are $\tan\dfrac{\theta_1}2,\tan\dfrac{\theta_2}2$

Finally use $\tan(A+B)$ formula.

lab bhattacharjee
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We shall establish that $\dfrac{\sin2\phi}2$ can be replaced with any other value.

$$\sin(\theta_1+\phi)=\dfrac{\sin2\phi}2=\sin(\theta_2+\phi)$$ Method$\#1:$

$$-\sin\phi(\cos\theta_1-\cos\theta_2)=\cos\phi(\sin\theta_1-\sin\theta_2)$$

$$\dfrac{\cos\phi}{\sin\phi}=-\dfrac{\cos\theta_1-\cos\theta_2}{\sin\theta_1-\sin\theta_2}=\tan\left(\dfrac{\theta_1}2+\dfrac{\theta_2}2\right)$$

Method$\#2:$ $\theta_1+\phi=n\pi+(-1)^n(\theta_2+\phi)$ where $n$ is any integer

If $n$ is even, $=2m$(say) $$\theta_1=2m\pi+\theta_2\iff\theta_1\equiv\theta_2\pmod{2\pi}$$ (How to discard this?)

If $n$ is odd, $=2m+1$(say) $$\theta_1+\phi=(2m+1)\pi-(\theta_2+\phi)\implies\dfrac{\theta_1+\theta_2}2=m\pi+\dfrac\pi2-\phi$$

$$\implies\frac{\sin(\theta_1)+ \sin(\theta_2) }{ \cos(\theta_1)+ \cos(\theta_2)}=\tan\left(\dfrac{\theta_1}2+\dfrac{\theta_2}2\right)=?$$

lab bhattacharjee
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