Trigonometry! #50

Geometry Level 3

If α + β − γ = π \alpha+\beta-\gamma=\pi then sin ⁡ 2 α + sin ⁡ 2 β − sin ⁡ 2 γ \sin^{2} \alpha + \sin^{2} \beta - \sin^{2} \gamma is equal to which of the following?

This problem is part of the set Trigonometry .

sin ⁡ α sin ⁡ β cos ⁡ γ \sin \alpha \sin \beta \cos \gamma 2 sin ⁡ α sin ⁡ β cos ⁡ γ 2\sin \alpha \sin \beta \cos \gamma 2 cos ⁡ α cos ⁡ β sin ⁡ γ 2\cos \alpha \cos \beta \sin \gamma cos ⁡ α cos ⁡ β sin ⁡ γ \cos \alpha \cos \beta \sin \gamma

Omkar Kulkarni by Omkar Kulkarni , 22, India

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1 solution

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sin ⁡ 2 θ cos ⁡ 2 ϕ = sin ⁡ 2 θ − sin ⁡ 2 θ sin ⁡ 2 ϕ . . . . . . ( 1 ) γ = ( α + β ) − π cos ⁡ γ = cos ⁡ ( α + β − π ) = cos ⁡ { ( α + β ) ( − 1 ) + 0 } ∴ cos ⁡ γ = − cos ⁡ ( α + β ) . . . . . . . . . . ( 2 ) sin ⁡ γ = sin ⁡ { ( α + β ) − π } = sin ⁡ ( α + β ) ( cos ⁡ π ) − 0 ∴ sin ⁡ γ = − { sin ⁡ α cos ⁡ β + cos ⁡ α sin ⁡ β } ∴ sin ⁡ 2 γ { u s i n g ( 1 ) f o r sin ⁡ 2 α cos ⁡ 2 β a n d sin ⁡ 2 α cos ⁡ 2 β } = sin ⁡ 2 α cos ⁡ 2 β + cos ⁡ 2 α sin ⁡ 2 β + 2 sin ⁡ α sin ⁡ β ( cos ⁡ α cos ⁡ β ) = sin ⁡ 2 α + sin ⁡ 2 β − 2 sin ⁡ 2 α sin ⁡ 2 β + 2 sin ⁡ α sin ⁡ β cos ⁡ α cos ⁡ β = sin ⁡ 2 α + sin ⁡ 2 β − 2 sin ⁡ α sin ⁡ β ( sin ⁡ α sin ⁡ β − cos ⁡ α cos ⁡ β ) = sin ⁡ 2 α + sin ⁡ 2 β + 2 sin ⁡ α sin ⁡ β cos ⁡ ( α + β ) = sin ⁡ 2 α + sin ⁡ 2 β − 2 sin ⁡ α sin ⁡ β cos ⁡ γ . . . . . . f r o m ( 2 ) ∴ sin ⁡ 2 α + sin ⁡ 2 β − sin ⁡ 2 γ = sin ⁡ 2 α + sin ⁡ 2 β − { sin ⁡ 2 α + sin ⁡ 2 β − 2 sin ⁡ α sin ⁡ β cos ⁡ γ } 2 sin ⁡ α sin ⁡ β cos ⁡ γ \sin^2\theta\cos^2\phi =\sin^2\theta-\sin^2\theta\sin^2\phi......(1)\\ \gamma=(\alpha+\beta)-\pi \\ \cos \gamma= \cos( \alpha+\beta~~-~~\pi) =\cos\{(\alpha+\beta)(-1) +0\} \\\therefore~\cos \gamma=-\cos(\alpha+\beta) ..........(2) \\ \sin \gamma= \sin\{(\alpha+\beta)-\pi \} =\sin(\alpha+\beta)(\cos \pi)-0 \\ \therefore~\sin \gamma=-\{\sin\alpha\cos\beta +\cos\alpha\sin\beta\} \\ \therefore~\sin^2 \gamma ~~\{using ~~(1)~for~\sin^2\alpha\cos^2\beta ~and~\sin^2\alpha\cos^2\beta \}\\~~~~= \sin^2\alpha\cos^2\beta +\cos^2\alpha\sin^2\beta + 2\sin\alpha\sin\beta(\cos\alpha\cos\beta)\\ ~~~~=\sin^2\alpha+\sin^2\beta-2 \sin^2\alpha\sin^2\beta+2\sin\alpha\sin\beta\cos\alpha\cos\beta \\~~~~=\sin^2\alpha+\sin^2\beta -2\sin\alpha\sin\beta(\sin\alpha\sin\beta-\cos\alpha\cos\beta)\\~~~~ =\sin^2\alpha+\sin^2\beta+2\sin\alpha\sin\beta\cos(\alpha+\beta) \\~~~~=\sin^2\alpha+\sin^2\beta-2\sin\alpha\sin\beta\cos\gamma......from (2)\\ \color{#20A900}{\therefore~\sin^{2} \alpha + \sin^{2} \beta - \sin^{2} \gamma\\~~~~=\sin^{2} \alpha + \sin^{2} \beta - \{\sin^{2} \alpha + \sin^{2} \beta - 2\sin\alpha\sin\beta\cos\gamma\} }\\ ~~~~~~~~\boxed {\color{#D61F06}{2\sin\alpha\sin\beta\cos\gamma} }

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