Can CSI and AM-GM Inequalities Solve Trigonometric Equation?

[anemone]

Solve the equation $\sin a \cos b+ \sin b \cos c+ \sin c \cos a=\dfrac{3}{2}$

 

[Ackbach]

Spoiler

$a=b=c=\dfrac{ \pi}{4}$ solves the equation by inspection. There may be other solutions.

 

[anemone]

Spoiler

$a=b=c=\dfrac{ \pi}{4}$ solves the equation by inspection. There may be other solutions.

Thanks for participating, Ackbach:)...but...

Spoiler

Since the answers to this particular challenge are infinite, I would give you $\dfrac{1}{\infty}$ mark, hehehe...:p

 

[Ackbach]

Spoiler

Fine. Be that way! How about $a,b,c \in \left\{ \dfrac{ \pi}{4} \pm n \pi \bigg| n \in \mathbb{Z} \right \}?$

 

[anemone]

Spoiler

Fine. Be that way! How about $a,b,c \in \left\{ \dfrac{ \pi}{4} \pm n \pi \bigg| n \in \mathbb{Z} \right \}?$

Sorry Ackbach, that is still not correct...

 

[Opalg]

[sp]Using first the Cauchy–Schwarz inequality and then the AM-GM inequality, $$\begin{aligned}\sin a \cos b+ \sin b \cos c+ \sin c \cos a &\leqslant \sqrt{(\sin^2a + \sin^2 b + \sin^2c)(\cos^2a + \cos^2b + \cos^2c)} \\ &\leqslant \tfrac12(\sin^2a + \sin^2 b + \sin^2c + \cos^2a + \cos^2b + \cos^2c) = \tfrac32.\end{aligned}$$ Equality in the Cauchy–Schwarz inequality only occurs if the vectors are scalar multiples of each other. In this case, that means that there exists a scalar $\lambda$ such that $$\sin a = \lambda\cos b, \qquad \sin b = \lambda\cos c, \qquad \sin c = \lambda\cos a.$$ Equality in the AM-GM inequality only occurs if the numbers are equal. In this case, that means that $\lambda^2(\cos^2a + \cos^2b + \cos^2c) = (\cos^2a + \cos^2b + \cos^2c)$, so $\lambda^2 = 1$. It then follows that $$\sin^2a = \cos^2b = 1-\sin^2b = 1-\cos^2c = \sin^2c = \cos^2a.$$ Thus $\cos a = \pm\sin a$ and similarly for $b$ and $c$. So each of $a$, $b$, $c$ is an odd multiple of $\pi/4$. Each of the three products $\sin a \cos b$, $\sin b \cos c$ and $\sin c \cos a$ must be $\pm1/2$, and in fact they must all be $+1/2$ if their sum is to be $3/2$. So $\cos b$ must have the same sign as $\sin a$, and similarly for the other two products. There are thus eight solutions in any interval of length $2\pi$, given by the possible choices of sign in the equations $$\sin a = \cos b = \pm\tfrac1{\sqrt2},\qquad \sin b = \cos c = \pm\tfrac1{\sqrt2},\qquad \sin c = \cos a = \pm\tfrac1{\sqrt2}.$$ In the interval $[-\pi,\pi]$ the solutions are: $$\begin{array}{c|c|c} a&b&c \\ \hline \pi/4 & \pi/4 & \pi/4 \\ 3\pi/4 & \pi/4 & -\pi/4 \\ \pi/4 & -\pi/4 & 3\pi/4 \\ 3\pi/4 & -\pi/4 & -3\pi/4 \\ -\pi/4 & 3\pi/4 & \pi/4 \\ -\pi/4 & -3\pi/4 & 3\pi/4 \\ -3\pi/4 & 3\pi/4 & -\pi/4 \\ -3\pi/4 & -3\pi/4 & -3\pi/4 \\ \end{array}$$[/sp]

 

[I like Serena]

[sp]Using first the Cauchy–Schwarz inequality and then the AM-GM inequality, $$\begin{aligned}\sin a \cos b+ \sin b \cos c+ \sin c \cos a &\leqslant \sqrt{(\sin^2a + \sin^2 b + \sin^2c)(\cos^2a + \cos^2b + \cos^2c)} \\ &\leqslant \tfrac12(\sin^2a + \sin^2 b + \sin^2c + \cos^2a + \cos^2b + \cos^2c) = \tfrac32.\end{aligned}$$ Equality in the Cauchy–Schwarz inequality only occurs if the vectors are scalar multiples of each other. In this case, that means that there exists a scalar $\lambda$ such that $$\sin a = \lambda\cos b, \qquad \sin b = \lambda\cos c, \qquad \sin c = \lambda\cos a.$$ Equality in the AM-GM inequality only occurs if the numbers are equal. In this case, that means that $\lambda^2(\cos^2a + \cos^2b + \cos^2c) = (\cos^2a + \cos^2b + \cos^2c)$, so $\lambda^2 = 1$. It then follows that $$\sin^2a = \cos^2b = 1-\sin^2b = 1-\cos^2c = \sin^2c = \cos^2a.$$ Thus $\cos a = \pm\sin a$ and similarly for $b$ and $c$. So each of $a$, $b$, $c$ is an odd multiple of $\pi/4$. Each of the three products $\sin a \cos b$, $\sin b \cos c$ and $\sin c \cos a$ must be $\pm1/2$, and in fact they must all be $+1/2$ if their sum is to be $3/2$. So $\cos b$ must have the same sign as $\sin a$, and similarly for the other two products. There are thus eight solutions in any interval of length $2\pi$, given by the possible choices of sign in the equations $$\sin a = \cos b = \pm\tfrac1{\sqrt2},\qquad \sin b = \cos c = \pm\tfrac1{\sqrt2},\qquad \sin c = \cos a = \pm\tfrac1{\sqrt2}.$$ In the interval $[-\pi,\pi]$ the solutions are: $$\begin{array}{c|c|c} a&b&c \\ \hline \pi/4 & \pi/4 & \pi/4 \\ 3\pi/4 & \pi/4 & -\pi/4 \\ \pi/4 & -\pi/4 & 3\pi/4 \\ 3\pi/4 & -\pi/4 & -3\pi/4 \\ -\pi/4 & 3\pi/4 & \pi/4 \\ -\pi/4 & -3\pi/4 & 3\pi/4 \\ -3\pi/4 & 3\pi/4 & -\pi/4 \\ -3\pi/4 & -3\pi/4 & -3\pi/4 \\ \end{array}$$[/sp]

Nice!
I hadn't thought of combining Cauchy–Schwarz and AM-GM yet.

 

[anemone]

Awesome, Opalg! I have never used the technique to apply the CSI and AM-GM inequality one after another, you have taught me a great lesson today! Thank you so much!