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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!

Napoleon Bonaparte
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StrugglinStudent
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3 Answers3

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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 $A=\frac \pi 6$, $B=\frac {2\pi} 3$, $C=\frac \pi 6$.

user
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  • Sorry, I'm not familiar with multi-dimensional function analysis. Are there any online resources to help me better understand what critical points are and how those other two functions were made? – StrugglinStudent Jan 16 '24 at 07:20
  • Refer for example here https://mathstat.slu.edu/~may/ExcelCalculus/sec-6-3-CriticalPointsExtrema.html – user Jan 16 '24 at 07:22
  • The idea is that $3/2$ is a maximum for the function, therefore the given solution is unique (a part renaming the angles). – user Jan 16 '24 at 08:00
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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 $0 < a,b < \frac\pi2$

lab bhattacharjee
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  • I'm not sure if ou are indicating others way by the given links. From here $2\cos^2\dfrac B2-2\cos\dfrac B2\cos\dfrac{C-A}2+\dfrac12=0$ we can conclude that $C-A=0$ and $B=\frac23 \pi$ indeed $p(x)=2x^2-2kx-\frac12$ has minimum equal to $\frac12-\frac{k^2}2$ at $x=\frac k2$ and therefore $p(x)=0\iff k=1$. – user Jan 17 '24 at 20:31
  • @user, can you please re-phrase query/confusion – lab bhattacharjee Jan 18 '24 at 11:18
  • I'm not sure how tha latter identity $2\cos^2\dfrac B2-2\cos\dfrac B2\cos\dfrac{C-A}2+\dfrac12=0$ is related to the given links which follow. In the previuos comment I've indicated a way to conclude that $A=C$ but maybe you are indicating a different way. – user Jan 18 '24 at 12:04
  • Thank you for your solution! I was originally confused about why cos(C+A/2) = cos(B/2), but after a little bit of substitution, I was able to get it. Very clever. – StrugglinStudent Jan 19 '24 at 04:36
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$\sin(A+B)=\frac{1}{2}$ , $\sin(B+C)=\frac{1}{2}$ , $\cos(A+C)=\frac{1}{2}$. 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.

DEB SANKAR ROY
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