Question
53 If \( \triangle A B C \) is a scalene triangle, show that: \( \cos (A+B)=-\cos C \)
Ask by Hodges Barker. in South Africa
Mar 11,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
In any triangle, the sum of the angles is \( \pi \). Therefore, \( A + B = \pi - C \). Using the cosine identity for supplementary angles, \( \cos(\pi - C) = -\cos C \). So, \( \cos(A+B) = -\cos C \).
Solution
1. In any triangle, the sum of the angles is \( \pi \). Therefore,
\[
A + B + C = \pi.
\]
2. Solving for \( A+B \), we get
\[
A + B = \pi - C.
\]
3. By the trigonometric identity for the cosine of a supplementary angle,
\[
\cos(\pi - C) = -\cos C.
\]
4. Thus, substituting back, we have
\[
\cos(A+B) = \cos(\pi - C) = -\cos C.
\]
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Simplify this solution Beyond the Answer
To prove that \( \cos(A + B) = -\cos C \) for a scalene triangle \( \triangle ABC \), we can use the fact that the sum of angles in a triangle is \( 180^\circ \). Thus, we have: \[ A + B + C = 180^\circ \] From this, we can express \( C \) in terms of \( A \) and \( B \): \[ C = 180^\circ - (A + B) \] Now, we apply the cosine function: \[ \cos C = \cos(180^\circ - (A + B)) \] Using the cosine identity \( \cos(180^\circ - x) = -\cos x \), we can write: \[ \cos C = -\cos(A + B) \] Rearranging this gives us: \[ \cos(A + B) = -\cos C \] Thus, we have shown that \( \cos(A + B) = -\cos C \), completing the proof.
