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V1 - 2 sin cos 0 = cos 0 - sin​

V1 - 2 sin cos 0 = cos 0 - sin​
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V1 - 2 sin cos 0 = cos 0 - sin​

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User Jordan Daniels
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1 Answer

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√(1 - 2sin θ \cosθ ) = \sqrt{ {sin}^(2) θ + {cos}^(2)θ - 2 sinθcosθ } = \sqrt{ {(cosθ - sinθ) }^(2) } = cosθ - sinθ

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User Gyurisc
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