Track / Stage 6 / 214 of 400 #1214 of 2000
Problem 1214 National olympiad, first round Difficulty 6.3 Prove it
3.205. sin α + sin β + sin γ − sin ( α + β ) cos γ − cos ( α + β ) sin γ = \sin \alpha+\sin \beta+\sin \gamma-\sin (\alpha+\beta) \cos \gamma-\cos (\alpha+\beta) \sin \gamma= sin α + sin β + sin γ − sin ( α + β ) cos γ − cos ( α + β ) sin γ = = 4 sin α + β 2 sin β + γ 2 sin γ + α 2 =4 \sin \frac{\alpha+\beta}{2} \sin \frac{\beta+\gamma}{2} \sin \frac{\gamma+\alpha}{2} = 4 sin 2 α + β sin 2 β + γ sin 2 γ + α
This one wants a proof. Work it on paper, read the official solution, then mark
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I solved it I didn't Skip
Official solution ## Solution.
sin α + sin β + sin γ − sin ( α + β ) cos γ − cos ( α + β ) sin γ = = ( sin α + sin β ) + sin γ − ( sin ( α + β ) cos γ + cos ( α + β ) sin γ ) = = [ sin x + sin y = 2 sin x + y 2 cos x − y 2 ; sin x cos y + cos x sin y = sin ( x + y ) ] = = 2 sin α + β 2 cos α − β 2 + sin γ − sin ( α + β + γ ) = = 2 sin α + β 2 cos α − β 2 + ( sin γ − sin ( α + β + γ ) ) = = [ sin x − sin y = 2 cos x + y 2 sin x − y 2 ] =
\begin{aligned}
& \sin \alpha + \sin \beta + \sin \gamma - \sin (\alpha + \beta) \cos \gamma - \cos (\alpha + \beta) \sin \gamma = \\
& = (\sin \alpha + \sin \beta) + \sin \gamma - (\sin (\alpha + \beta) \cos \gamma + \cos (\alpha + \beta) \sin \gamma) = \\
& = \left[\sin x + \sin y = 2 \sin \frac{x + y}{2} \cos \frac{x - y}{2} ; \sin x \cos y + \cos x \sin y = \sin (x + y)\right] = \\
& = 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} + \sin \gamma - \sin (\alpha + \beta + \gamma) = \\
& = 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} + (\sin \gamma - \sin (\alpha + \beta + \gamma)) = \\
& = \left[\sin x - \sin y = 2 \cos \frac{x + y}{2} \sin \frac{x - y}{2}\right] =
\end{aligned}
sin α + sin β + sin γ − sin ( α + β ) cos γ − cos ( α + β ) sin γ = = ( sin α + sin β ) + sin γ − ( sin ( α + β ) cos γ + cos ( α + β ) sin γ ) = = [ sin x + sin y = 2 sin 2 x + y cos 2 x − y ; sin x cos y + cos x sin y = sin ( x + y ) ] = = 2 sin 2 α + β cos 2 α − β + sin γ − sin ( α + β + γ ) = = 2 sin 2 α + β cos 2 α − β + ( sin γ − sin ( α + β + γ )) = = [ sin x − sin y = 2 cos 2 x + y sin 2 x − y ] =
= 2 sin α + β 2 cos α − β 2 + 2 cos α + β + 2 γ 2 sin ( − α + β 2 ) = = 2 sin α + β 2 cos α − β 2 − 2 cos α + β + 2 γ 2 sin α + β 2 = = 2 sin α + β 2 ( cos α − β 2 − cos α + β + 2 γ 2 ) = = [ cos x − cos y = − 2 sin x + y 2 sin x − y 2 ] = = 2 sin α + β 2 ( − 2 sin α + γ 2 sin ( − β + γ 2 ) ) = 4 sin α + β 2 sin β + γ 2 sin γ + α 2
\begin{aligned}
& = 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} + 2 \cos \frac{\alpha + \beta + 2 \gamma}{2} \sin \left(-\frac{\alpha + \beta}{2}\right) = \\
& = 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} - 2 \cos \frac{\alpha + \beta + 2 \gamma}{2} \sin \frac{\alpha + \beta}{2} = \\
& = 2 \sin \frac{\alpha + \beta}{2} \left(\cos \frac{\alpha - \beta}{2} - \cos \frac{\alpha + \beta + 2 \gamma}{2}\right) = \\
& = \left[\cos x - \cos y = -2 \sin \frac{x + y}{2} \sin \frac{x - y}{2}\right] = \\
& = 2 \sin \frac{\alpha + \beta}{2} \left(-2 \sin \frac{\alpha + \gamma}{2} \sin \left(-\frac{\beta + \gamma}{2}\right)\right) = 4 \sin \frac{\alpha + \beta}{2} \sin \frac{\beta + \gamma}{2} \sin \frac{\gamma + \alpha}{2}
\end{aligned}
= 2 sin 2 α + β cos 2 α − β + 2 cos 2 α + β + 2 γ sin ( − 2 α + β ) = = 2 sin 2 α + β cos 2 α − β − 2 cos 2 α + β + 2 γ sin 2 α + β = = 2 sin 2 α + β ( cos 2 α − β − cos 2 α + β + 2 γ ) = = [ cos x − cos y = − 2 sin 2 x + y sin 2 x − y ] = = 2 sin 2 α + β ( − 2 sin 2 α + γ sin ( − 2 β + γ ) ) = 4 sin 2 α + β sin 2 β + γ sin 2 γ + α
The identity is proven.
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