We are tasked with proving that if $\sin C = \cos B + \cos A$ in a triangle $ABC$, then the triangle is a right triangle. Let's break this down step by step.

Step 1: Use the identity for the angles in a triangle

In any triangle $ABC$, the sum of the interior angles is:

$A + B + C = 180^\circ$

Thus, we can express $C$ as:

$C = 180^\circ - (A + B)$

Step 2: Use trigonometric identities

We are given the equation:

$\sin C = \cos B + \cos A$

Using the fact that $C = 180^\circ - (A + B)$, we apply the identity for the sine of an angle sum:

$\sin(180^\circ - (A + B)) = \sin(A + B)$

Thus, the equation becomes:

$\sin(A + B) = \cos B + \cos A$

Step 3: Expand $\sin(A + B)$

Using the sum identity for sine:

$\sin(A + B) = \sin A \cos B + \cos A \sin B$

So the equation now is:

$\sin A \cos B + \cos A \sin B = \cos B + \cos A$

Step 4: Rearrange the equation

Bring all terms involving $\cos B$ and $\cos A$ to one side:

$\sin A \cos B + \cos A \sin B - \cos B - \cos A = 0$

Factor the terms:

$\cos B (\sin A - 1) + \cos A (\sin B - 1) = 0$

This equation suggests that one of the following must be true:

1. $\cos B = 0$
2. $\cos A = 0$

Step 5: Analyze the cases

1. If $\cos B = 0$, then $B = 90^\circ$.
2. If $\cos A = 0$, then $A = 90^\circ$.

Step 6: Conclusion

If either $A = 90^\circ$ or $B = 90^\circ$, the triangle is a right triangle. Therefore, we have shown that if $\sin C = \cos B + \cos A$, then the triangle $ABC$ must be a right triangle.

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Would you like further explanation on any of the steps?

Here are some related questions to help you dive deeper:

1. What are the other trigonometric identities that apply to the angles of a triangle?
2. How does the Pythagorean Theorem relate to right triangles?
3. Can the equation $\sin C = \cos B + \cos A$ hold for non-right triangles?
4. How do you prove that a triangle is a right triangle using the law of cosines?
5. What are some other methods for proving that a triangle is right-angled?

Tip: When working with angle sum identities, it's helpful to remember that $\sin(180^\circ - \theta) = \sin \theta$ and $\cos(180^\circ - \theta) = -\cos \theta$.