nana1banana nana1banana
  • 08-04-2019
  • Mathematics
contestada

(sin Θ − cos Θ)^2 − (sin Θ + cos Θ)^2

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LammettHash
LammettHash LammettHash
  • 08-04-2019

More generally, we have

[tex](a-b)^2-(a+b)^2=(a^2-2ab+b^2)-(a^2+2ab+b^2)=-4ab[/tex]

so that

[tex](\sin\theta-\cos\theta)^2-(\sin\theta+\cos\theta)^2=-4\sin\theta\cos\theta=-2\sin2\theta[/tex]

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