What type of triangle satisfies the equation $\cos(A)-\cos(B)+\sin(C)=0$? A triangle with angle $A,B,C$ satisfies the equation $\cos(A)-\cos(B)+\sin(C)=0$.
What type of triangle is this? Regular, acute, right, obtuse etc.
I tried using sine and cosine rule, but no result.
 A: Since $A+B+C=\pi$ we have
$$\cos B-\cos A=\sin(A+B)\ .$$
Writing
$$\phi=\frac{A+B}{2}\ ,\quad \psi=\frac{A-B}{2}$$
gives
$$\sin2\phi=\cos(\phi-\psi)-\cos(\phi+\psi)$$
and hence
$$2\sin\phi\cos\phi=2\sin\phi\sin\psi\ .$$
We may assume that $0<A+B<\pi$, so $0<A+B<\frac{\pi}{2}$.  Therefore $\sin\phi\ne0$, so
$$\eqalign{\cos\phi=\sin\psi\quad
  &\Rightarrow\quad 2\cos^2\phi=2\sin^2\psi\cr
  &\Rightarrow\quad 1+\cos(A+B)=1-\cos(A-B)\cr
  &\Rightarrow\quad \cos(A+B)+\cos(A-B)=0\cr
  &\Rightarrow\quad 2\cos A\cos B=0\ .\cr}$$
So the triangle is right-angled.  And in fact the right angle is at $A$, because $\cos A+\sin C=0$ is impossible when $A+C=\frac{\pi}{2}$.
Conversely, it is easy to check that any triangle with a right angle at $A$ satisfies the given condition.
A: Using Prosthaphaeresis Formulas,
$$\sin C=\cos B-\cos A=2\sin\dfrac{A+B}2\sin\dfrac{A-B}2$$
Now, $\displaystyle \sin\dfrac{A+B}2=\sin\dfrac{\pi-C}2=\cos\dfrac C2$
Using double angle formula the given relation becomes, $$2\sin\dfrac C2\cos\dfrac C2=2\cos\dfrac C2\sin\dfrac{A-B}2$$  
which implies 
$(1)$ either $\displaystyle\cos\dfrac C2=0\iff\frac C2=(2n+1)\frac\pi2\iff C=(2n+1)\pi$ where $n$ is any integer 
But it is impossible as $0<C<\pi$
$(2)$ or $\displaystyle\sin\dfrac C2 =\sin\dfrac{A-B}2$
$\displaystyle\implies \dfrac C2=m\pi+(-1)^m\dfrac{A-B}2\iff C=2m\pi+(-1)^m(A-B)$ where $m$ is any integer 
If $m$ is even $=2r$(say), $\displaystyle C=4r\pi+(A-B)\iff B+C=4r\pi+A$
$\displaystyle\implies \pi-A=4r\pi+A\iff A=\dfrac{(1-4r)\pi}2$
As $\displaystyle0<A<\pi, 0<\dfrac{(1-4r)\pi}2<\pi\iff 0<1-4r<2\iff0>4r-1>-2$
$\displaystyle\implies r=0\implies A=\dfrac{(1-4\cdot0)\pi}2$
Similarly check for odd $m=2r+1$(say)
