Sine of Twice Angle minus Sum of Sines of Twice Other Two Angles of Triangle

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Theorem

Let $\triangle ABC$ be a triangle.

Then:

$\sin 2 A - \sin 2 B - \sin 2 C = -4 \sin A \cos B \cos C$


Proof

First we note that:

\(\ds A + B + C\) \(=\) \(\ds 180 \degrees\) Sum of Angles of Triangle equals Two Right Angles
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds A + B\) \(=\) \(\ds 180 \degrees - C\)

That is, $C$ is the supplement of $A + B$.


Then:

\(\ds \sin 2 A - \sin 2 B - \sin 2 C\) \(=\) \(\ds 2 \map \cos {A + B} \map \sin {A - B} - \sin 2 C\) Sine minus Sine
\(\ds \) \(=\) \(\ds 2 \map \cos {180 \degrees - C} \map \sin {A - B} - \sin 2 C\) from $(1)$
\(\ds \) \(=\) \(\ds -2 \cos C \map \sin {A - B} - \sin 2 C\) Cosine of Supplementary Angle
\(\ds \) \(=\) \(\ds -2 \cos C \paren {\sin A \cos B - \cos A \sin B} - \sin 2 C\) Sine of Difference
\(\ds \) \(=\) \(\ds -2 \cos C \paren {\sin A \cos B - \cos A \sin B} - 2 \sin C \cos C\) Double Angle Formula for Sine
\(\ds \) \(=\) \(\ds -2 \sin A \cos B \cos C + 2 \paren {\cos A \sin B - \sin C} \cos C\) multiplying out and rearranging
\(\ds \) \(=\) \(\ds -2 \sin A \cos B \cos C + 2 \paren {\cos A \sin B - \map \sin {180 \degrees - \paren {A + B} } } \cos C\) from $(1)$
\(\ds \) \(=\) \(\ds -2 \sin A \cos B \cos C + 2 \paren {\cos A \sin B - \map \sin {A + B} } \cos C\) Sine of Supplementary Angle
\(\ds \) \(=\) \(\ds -2 \sin A \cos B \cos C + 2 \paren {\cos A \sin B - \paren {\sin A \cos B + \cos A \sin B} } \cos C\) Cosine of Sum
\(\ds \) \(=\) \(\ds -2 \sin A \cos B \cos C + 2 \paren {\cos A \sin B - \cos A \sin B} \cos C - 2 \sin A \cos B \cos C\) multiplying out
\(\ds \) \(=\) \(\ds -4 \sin A \cos B \cos C\) simplifying

$\blacksquare$


Sources

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