Solve $\sin \frac{\alpha - \beta}2 + \sin \frac{\alpha - \gamma}2 + \sin \frac{3\alpha}2 =\frac 3 2$

Question

Solve the following trigonometric eqation where $\alpha, \beta, \gamma$ are angles in a triangle ($\alpha + \beta + \gamma = 180$):

$$\sin \frac{\alpha - \beta}2 + \sin \frac{\alpha - \gamma}2 + \sin \frac{3\alpha}2 =\frac 3 2$$

Transforming it into $2 \sin \frac{3\alpha-180}4 \cos \frac{\beta - \gamma}4 + \sin \frac{3\alpha}2 =\frac{3}2$ and $\cos \frac{\beta - \gamma}4 =\cos \frac{180- \alpha}4 + 2 \sin \frac{\beta}4 \sin \frac{\gamma}4$ is as far as I came.

Answer 1

I took an approach which may be only superficially different from abel's because of the underlying trig identities and follows a lead SoulEater was taking, but along a slightly different path.

We can get rid of $ \ \alpha \ $ , to start off, by writing $ \ \alpha - \beta \ = \ \pi - 2 \beta - \gamma \ $ and, similarly, $ \ \alpha - \gamma \ = \ \pi - \beta - 2\gamma \ $ . From this, we have

$$ \sin \left( \frac{\alpha - \beta}{2} \right) \ = \ \sin \left( \frac{\pi}{2} - [ \ \beta + \frac{\gamma}{2} \ ] \right) \ = \ \cos \left( \ \beta + \frac{\gamma}{2} \ \right) \ \ $$

and $ \ \sin \left( \frac{\alpha - \gamma}{2} \right) \ = \ \cos \left( \ \frac{\beta}{2} + \gamma \ \right) \ $ . [So far, this looks like what abel did, while deriving the variable transformations.] We will also write

$$ \sin \frac{3 \alpha}{2} \ = \ \sin \left( \frac{3}{2} [ \pi - \beta - \gamma ] \right) \ = \ \sin \left( \frac{3 \pi}{2} - \ \frac{3}{2} [\beta + \gamma ] \ \right) \ = \ -\cos \left( \frac{3}{2} [\beta + \gamma ] \ \right) \ \ . $$

The original equation is now

$$ \cos \left( \ \beta + \frac{\gamma}{2} \ \right) \ + \ \cos \left( \ \frac{\beta}{2} + \gamma \ \right) \ - \ \cos \left( \frac{3}{2} [\beta + \gamma ] \ \right) \ = \ \frac{3}{2} \ \ . $$

Applying a "sum-to-product rule" on the first two terms here gives

$$ 2 \ \cos \left( \frac{3}{4} [\beta + \gamma ] \ \right) \cos \left( \frac{1}{4} [\beta - \gamma ] \right) - \ \cos \left( \frac{3}{2} [\beta + \gamma ] \ \right) \ = \ \frac{3}{2} \ \ . $$

This seemingly doesn't appear to have done much of anything useful, but it turns out we've actually finished the "hard part"...

We will label the angles $ \ \Theta \ = \ \frac{3}{4} [\beta + \gamma ] \ $ and $ \ \Phi \ = \ \frac{1}{4} [\beta - \gamma ] \ $ , allowing us to write

$$ 2 \ \cos \Theta \ \cos \Phi \ - \ \cos \ ( 2 \Theta ) \ = \ \frac{3}{2} \ \ \Rightarrow \ \ 2 \ \cos \Theta \ \cos \Phi \ - \ 2 \ \cos^2 \Theta \ + \ 1 \ = \ \frac{3}{2} $$

$$ \Rightarrow \ \ \cos \Theta \ \cos \Phi \ - \ \cos^2 \Theta \ = \ \frac{1}{4} \ \ . $$

Now, if $ \ \beta \ = \ \gamma \ $ , then $ \ \Phi \ = \ 0 \ $ and we just have the quadratic equation in $ \ \cos \Theta \ $ ,

$$ \cos^2 \Theta \ - \ \cos \Theta \ + \ \frac{1}{4} \ = \ \left( \cos \Theta \ - \ \frac{1}{2} \right)^2 \ = \ 0 \ \ . $$

Since we are limiting ourselves to angles of a triangle, we obtain

$$ \cos \left( \frac{3}{4} [\beta + \gamma ] \ \right) \ = \ \frac{1}{2} \ \ \Rightarrow \ \ \frac{3}{4} [\beta + \gamma ] \ = \ \frac{\pi}{3} \ \ \Rightarrow \ \ \beta \ + \ \gamma \ = \ \frac{4 \pi}{9} \ \ . $$

Thus, $ \ \beta \ = \ \gamma \ = \ \frac{2 \pi}{9} \ \ \Rightarrow \ \ \alpha \ = \ \frac{5 \pi}{9} \ $ .

Are other solutions permissible? If we allow $ \ \beta \ \neq \ \gamma \ $ and call $ \ \cos \Phi \ = \ \varphi \ $ , our quadratic equation becomes $ \ \cos^2 \Theta \ - \ \varphi \ \cos \Theta \ + \ \frac{1}{4} = \ 0 \ $ , which has potential solutions

$$ \cos \Theta \ = \ \frac{\varphi \ \pm \ \sqrt{ \ \varphi^2 \ - \ 1 } } {2} \ \ . $$

But real solutions for $ \ \cos \Theta \ $ would only be possible here for $ \ \varphi \ = \ \cos \Phi \ \ge \ 1 \ $ . So the only solution for our original equation is the one we've already found with $ \ \varphi \ = \ 1 \ $ or $ \ \beta \ = \ \gamma \ $ .

Answer 2

When we exchange $\beta$ and $\gamma$ in the equation, we get the same equation back. To me, that means $\beta = \gamma+4k\pi$, where $k$ is an integer, is a third equation that we can use. Of course, only $k=0$ satisfies the constraints that $\alpha+\beta+\gamma=\pi$ and $\alpha, \gamma, \beta>0$, since they are angles in a triangle. I haven't been able to derive it, but $\alpha=5\pi/9$, $\beta=2\pi/9$, and $\gamma=2\pi/9$ does appear to be the solution.

Answer 3

i think the answer is $\alpha = 100^\circ, \beta = 40^\circ$ and $\gamma = 40^\circ$

here is my attempt at the solution. i will use a change of variable $$b = \beta+{1 \over 2}\gamma,\ c = \gamma + {1 \over 2}\beta$$ and the following inverse relations $$\alpha = 180^\circ - {2 \over 3}(b+c), \ \beta = {4 \over 3}(b - {c \over 2})\ , \gamma = {4 \over 3}(c - {b \over 2}) $$ this change of variable transforms $\sin({\alpha - \beta \over 2}) + \sin({\alpha - \gamma \over 2}) + \sin{3\alpha \over 2} = {3 \over 2}$ into a symmetric $2\pi$-periodic equation $$f(b,c)=\cos b + \cos c - \cos(b+c) = {3 \over 2}.$$ observe that $f = 1$ on the boundary and $f(b,b) = 2\cos b - \cos 2b = -2\cos^2 b + 2 \cos b +1$ is maximum $3/2$ at $b = \pi/3 \text{ and } 5\pi/3.$

we will show that the global maximum of $f(b,c)$ on the square $[0,2\pi] \times [0,2\pi]$ is ${3 \over 2}$ at $b = c = \pi/3.$

at a local extremum, the partial derivatives $$f_b = -\sin b + \sin(b+c) = 0,\ f_c = -\sin c + \sin(b+c) = 0$$

so we need to solve $$\sin b = \sin c = \sin(b+c).$$ the solutions are $$c = b, c = \pi - b, c = 3\pi - b $$

only the case $c = b$ is relevant. we have $\sin b = \sin 2b$ an the solution is $\cos b = 1/2, \sin b = 0 $

now we can transform the $$b = c = 60^\circ$$ to the values for $\alpha, \beta$ and $\gamma$ claimed at the top of the post.