Cos 2 alpha into cos 2 beta + sin square alpha minus beta minus sin square alpha + beta is equal to cos lambda alpha + beta; find lambda
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Answer:
Answer:To Prove: [tex]cos\,2\alpha\;cos\,2\beta+sin^2\,(\alpha-\beta)-sin^2\,(\alpha+\beta)=cos\,2(\
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Identity
sin
2
A−sin
2
B
=sin(A+B)sin(A−B)
=(sin(A)+sin(B))(sin(A)−sin(B))
=2sin(
2
A+B
)cos(
2
A−B
).2cos(
2
A+B
)sin(
2
A+B
)
=2sin(
2
A+B
)cos(
2
A+B
).2sin(
2
A−B
)cos(
2
A−B
)
=sin(A+B)sin(A−B)
Given equation
sin
2
α+[sin
2
β−sin
2
γ]
=sin
2
α+sin(β−γ)sin(β+γ)
=sin
2
α+sin(π−α)sin(β+γ)
=sin
2
α+sinαsin(β+γ)
=sinα[sinα+sin(β+γ)]
=sinα[sin(π−(β−γ))+sin(β+γ)]
=sinα[sin(β−γ)+sin(β+γ)]
=sinα[2sinβcosγ]
=2sinαsinβcosγ
∴λ=2
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