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19 August, 20:25

Write cos (3x) - cos x as a product

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  1. 19 August, 20:40
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    -2sin (x) * sin (2x)

    or any equivalent form, such as 4cos (x) [cos² (x) - 1]

    Step-by-step explanation:

    simplify cos (3x) first:

    cos (3x) = cos (2x+x) = cos (2x) cos (x) - sin (2x) sin (x)

    using trig identities

    = [2cos² (x) - 1]cos (x) - [2sin (x) cos (x) ]sin (x)

    = 2cos³ (x) - cos (x) - 2sin² (x) cos (x)

    substituting using trig identity sin² (x) + cos² (x) = 1

    2cos³ (x) - cos (x) - 2[1-cos² (x) ]cos (x)

    2cos³ (x) - cos (x) - 2cos (x) + 2cos³ (x)

    4cos³ (x) - 3cos (x)

    remember this cos (3x), we still have to subtract cos (x)

    4cos³ (x) - 3cos (x) - cos (x) = 4cos³ (x) - 4cos (x)

    we can factor 4cos (x) to write this as a product of:

    4cos (x) [cos² (x) - 1]

    further simplification if you want

    trig identity sin² (x) + cos² (x) = 1

    simplifying: sin² (x) = 1-cos² (x)

    simplifying: - sin² (x) = cos² (x) - 1

    4cos (x) [cos² (x) - 1]

    4cos (x) [-sin² (x) ]

    -4cos (x) sin² (x)

    trig identity: sin (2a) = 2cos (a) sin (a)

    -2sin (x) * 2cos (x) sin (x)

    -2sin (x) * sin (2x)
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