sin(A + B) + sin(B + C) + cos(A + C) = 3/2. Find each angle.

Question

I'm given the fact that in $\triangle ABC$, $\sin(A + B) + \sin(B + C) + \cos(A + C) = \frac{3}{2}$. I'm asked to get the angles $A, B, C$. So far what I've done is I've substituted $A+B = 180-C$ and so on in order to get

$\sin(180 - C) + \sin(180 - A) + \cos(A + C) = \frac{3}{2}$.

From here I've used the sin addition identity to get

$\sin(C) + \sin(A) + \cos(A + C) = \frac{3}{2}$

From here I'm stuck. Help!

Answer 1

$\sin(A+B)=\frac{1}{2}$ , $\sin(B+C)=\frac{1}{2}$ , $\cos(A+C)=\frac{1}{2}$. Therefore, $A+B=\frac{\pi}{6}$ , $B+C=\frac{\pi}{6}$ , $A+C=\frac{\pi}{3}$. Therefore after solving we get the value of $A=\frac{\pi}{6}$ , $B=0$ , $C=\frac{\pi}{6}$. This is the required solution.

Answer 2

Let consider

$$f(x,y)=\sin x + \sin y + \cos (x+y)$$

then by

• $f_x=\cos x -\sin(x+y)=0$
• $f_y=\cos y -\sin(x+y)=0$

we obtain as possible solutions (critical points) $x=y=\frac \pi 2, \frac \pi 6$, $\frac 5 6\pi$ and only $x=y=\frac \pi 6$ is possible which leads to $A=\frac \pi 6$, $B=\frac {2\pi} 3$, $C=\frac \pi 6$.

Answer 3

Hint:

As $A+B+C=\pi,$

$$\dfrac32=\sin C+\sin A-\cos B$$ $$=2\sin\dfrac{C+A}2\cos\dfrac{C-A}2-\left(2\cos^2\dfrac B2-1\right)$$

$$=2\cos\dfrac B2\cos\dfrac{C-A}2-2\cos^2\dfrac B2+1$$

$$\iff2\cos^2\dfrac B2-2\cos\dfrac B2\cos\dfrac{C-A}2+\dfrac12=0$$

Use

$ \cos {A} \cos {B} \cos {C} \leq \frac{1}{8} $

OR

Solve equation $\cos (a)\cos(b)\cos(a+b) = -\frac18$ over $0 < a,b < \frac\pi2$