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3 November, 18:19

Express the complex number sqrt6 (cos315+isin325) in standard form

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  1. 3 November, 18:44
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    I think there is a mistake in the equation √6 (cos315+isin325)

    where sin 325 should be sin 315, if so

    1) the angle 315 is in the 4th quadrant, where cos is positive & sin, negative

    cos 315 = cos (315-360) = - 45 and cos (-45) = √2/2 & sin (-45) = - √2/2

    Then z = √6 (√2 / 2 - i√2 / 2)

    z=√6 x √2 / 2 - √6 x √2 / 2) = √12 / 2 - i√12 / 2)

    Simplify: z = √3 - i√3
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