# Thread: Working with Trig identies to show that a triangle is a right triangle

1. ## Working with Trig identies to show that a triangle is a right triangle

If angles A, B, and C are the angles of a triangle such that sin(A+B)=1/sin(C) and cos(A+B)=cos(C), then show that the triangle is a right triangle

2. Originally Posted by bbanbury

If angles A, B, and C are the angles of a triangle such that sin(A+B)=1/sin(C) and cos(A+B)=cos(C), then show that the triangle is a right triangle
Hmm, interesting! I'm sure there are many ways to prove that ABC is a right triangle, but here's one method:

So let's assume ABC is a right triangle.

$c$ is the hypotenuse of the triangle
$a$ and $b$ are both legs of the triangle
Leg $a$ is the opposite side to angle $A$; leg $a$ isn't adjacent to angle $A$. So the sine of A would be $\frac{a}{c}$.

$cos(A+B) = cosC$ can be expanded to

$cos(A)cos(B) - sin(A)sin(B) = cosC$

$\frac{b}{c} \times \frac{a}{c}-\frac{a}{c} \times \frac{b}{c} = cosC$

$\frac{ab}{c^2} - \frac{ab}{c^2} = cosC$

$\frac{ab}{c^2} - \frac{ab}{c^2} = 0$

$cosC = 0$

The cosine of $C$ is only $0$ if angle $C$ is $90^{\circ}$ or $270^{\circ}.$ Since a triangle cannot have a $270^{\circ}$ angle, angle $C$ must be $90^{\circ}$, and ABC is a right triangle.

3. ## Trigonometry

Hello bbanbury
Originally Posted by bbanbury

If angles A, B, and C are the angles of a triangle such that sin(A+B)=1/sin(C) and cos(A+B)=cos(C), then show that the triangle is a right triangle
This is very straightforward.

We know that, for any angle $\theta,\, \sin\theta = \sin(180-\theta)$. And in a triangle $ABC, A+B = 180 - C$, so $\sin(A+B) = \sin(180-C) = \sin C$ in any triangle.

If, in addition, we know that $\sin(A+B) = \frac{1}{\sin C}$, then

$\sin C = \frac{1}{\sin C}$

$\Rightarrow \sin^2C= 1$

$\sin C = +1$ (since $\sin C = -1$ is impossible if $0)

$\Rightarrow C = 90^o$

So the triangle is right-angled at C.

(Notice that we don't need the extra fact about $\cos(A+B) = \cos C$. But you could equally well use this instead, and say that $\cos C = -\cos(180 - C) = -\cos(A+B)$ in any triangle. So in this case $\cos C = \cos(A+B) \Rightarrow \cos C = -\cos C = 0 \Rightarrow C=90^o$.)