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Angles in a triangle (Tomato Subjective 116)

Problem: If A, B, C are the angles of a triangle, then show that \(\displaystyle { \sin A + \sin B – \cos C \le \frac {3 \sqrt{3}}{2}}\)

Discussion:

\(\displaystyle { \sin A + \sin B – \cos C } \)
\(\displaystyle { = \sin A + \sin B – \cos (\pi – (A+B)) } \)
\(\displaystyle { = \sin A + \sin B + \sin (A+B) } \)
\(\displaystyle { = 2\sin \frac{(A+B)}{2} \cos \frac{(A-B)}{2} + 2\sin \frac{(A+B)}{2} \cos \frac{(A+B)}{2} } \)
\(\displaystyle { = 2\sin \frac{(A+B)}{2} \left( \cos \frac{(A-B)}{2} + \cos \frac{(A+B)}{2}\right ) } \)
\(\displaystyle { = 2\sin \frac{(A+B)}{2} 2\cos \frac{A}{2} \cos \frac{B}{2} } \)
\(\displaystyle { = 4\sin \frac{(\pi -C)}{2} \cos \frac{A}{2} \cos \frac{B}{2} } \)
\(\displaystyle { = 4\sin \left(\frac{\pi}{2} – \frac{C}{2} \right) \cos \frac{A}{2} \cos \frac{B}{2} } \)
\(\displaystyle { = 4\cos \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} } \)

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November 27, 2015

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