How do you prove ${(cos A + cos B)}^{2} + {(sin A + sin B)}^{2} = 2 [1 + cos (A - B)]$ $(CosA + CosB)^2 + (SinA + SinB)^2 = 2[1 + cos(A - B)]$ ?

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How do you prove ${(cos A + cos B)}^{2} + {(sin A + sin B)}^{2} = 2 [1 + cos (A - B)]$ $(CosA + CosB)^2 + (SinA + SinB)^2 = 2[1 + cos(A - B)]$ ?

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Answer 1 · 2015-06-06T04:05:15

${cos}^{A} + {cos}^{2} B + 2 cos A . cos B + {sin}^{2} A + {sin}^{2} B + 2 sin A . sin B =$ $cos^A + cos^2 B + 2cos A.cos B + sin^2A + sin^2 B + 2sin A.sin B =$

$= 2 + (2 cos A . cos B - 2 sin A . sin B) = 2 [1 + cos (A - B)]$ $= 2 + (2cos A.cos B - 2sin A.sin B) = 2[1 + cos (A - B)]$

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How do you prove ${(cos A + cos B)}^{2} + {(sin A + sin B)}^{2} = 2 [1 + cos (A - B)]$ $(CosA + CosB)^2 + (SinA + SinB)^2 = 2[1 + cos(A - B)]$ ?

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How do you prove (CosA + CosB)^2 + (SinA + SinB)^2 = 2[1 + cos(A - B)](cosA+cosB)2+(sinA+sinB)2=2[1+cos(A−B)](CosA + CosB)^2 + (SinA + SinB)^2 = 2[1 + cos(A - B)]?

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How do you prove ${(cos A + cos B)}^{2} + {(sin A + sin B)}^{2} = 2 [1 + cos (A - B)]$ $(CosA + CosB)^2 + (SinA + SinB)^2 = 2[1 + cos(A - B)]$ ?