sinC = (sinA + sinB) / (cosA + cosB), shape of ABC = ?

  

\(\textrm{Let}\quad\overline{BC}=a,\quad\overline{CA}=b,\quad\overline{AB}=c\\ \frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}=2R\quad (\textrm{R=radius of the circumscribed circle of triangle ABC})\\ \therefore sinA=\frac{a}{2R},\quad sinB=\frac{b}{2R},\quad sinC=\frac{c}{2R}\\ cosA=\frac{b^2+c^2-a^2}{2bc},\quad cosB=\frac{c^2+a^2-b^2}{2ca},\quad cosC=\frac{a^2+b^2-c^2}{2ab}\\ \therefore \frac{c}{2R}=\frac{\frac{a}{2R}+\frac{b}{2R}}{\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ca}}\\ \quad c=\frac{2abc(a+b)}{a(b^2+c^2-a^2)+b(c^2+a^2-b^2)}=\frac{2abc(a+b)}{ab^2+c^2a-a^3+bc^2+a^2b-b^3}=\frac{2abc(a+b)}{c^2(a+b)-(a-b)^2(a+b)}=\frac{2abc}{c^2-(a-b)^2}\\ \therefore c^2-(a-b)^2=2ab\\ \quad c^2=2ab+(a-b)^2=a^2+b^2\\ \therefore \textrm{Triangle ABC is a right triangle that has angle C as the right angle.}\)

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The 2R comes from the standard proof of the sine rule.

Draw your triangle ABC with its circumscribing circle, radius R, centre O.

With the usual notation, angle BOC = 2A, so if you drop a perpendicular from O to BC you get a rt-angled triangle with

two sides R and a/2 and an angle A, from which a/2R = sin(A) and a/sin(A) = 2R.

Repeat for B and C and equate the three.

 

Here's an alternative answer to the original question.

 

LHS,

 \(\displaystyle \sin(C)=2\sin(C/2)\cos(C/2)=2\sin(90-(A+B)/2)\cos(90-(A+B)/2)\\ =2\cos((A+B)/2)\sin((A+B)/2) \dots\dots\dots(1)\)

 

RHS

\(\displaystyle \frac{\sin(A)+\sin(B)}{\cos(A)+\cos(B)}=\frac{2\sin((A+B)/2)\cos((A-B)/2)}{2\cos((A+B)/2)\cos((A-B)/2)}=\frac{\sin((A+B)/2)}{\cos((A+B)/2)}\dots\dots(2)\)

 

Equate (1) and (2), cancel and cross multiply,

\(\displaystyle \cos^{2}((A+B)/2)=1/2,\\\text{so }\cos((A+B)/2) = 1/\sqrt{2}\\\text{so }(A+B)/2=45\text{ deg}\\A+B=90\text{ deg}\\C=90\text{ deg}.\)

 

Tiggsy.