Math If A+B+C=pi then prove that cos3A.cos3B+cos3B.cos3C+cos3C.cos3A=1

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Math

If A+B+C=pi then prove that cos3A.cos3B+cos3B.cos3C+cos3C.cos3A=1

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Sadie 1 month 2021-10-13T17:57:09+00:00 1 Answer 0 views 0

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    2021-10-13T17:58:19+00:00

    Step-by-step explanation:

    Given:

    A+B+C= π

    <=> 3A+3B+3C = 3π

    <=> cos(3A+3B) = – cos3C

    <=> cos3A.cos3B-sin3A.sin3B = – cos3C

    <=> cos3A.cos3B = sin3A.sin3B  – cos3C (1)

    similarly  apply for the other two angles, we have:

    • cos3B.cos3C = sin3B.sin3C  – cos3A  (2)
    • cos3C.cos3A = sin3C.sin3A  – cos3B  (3)

    Grouping three equations, (1) + (2) + (3), we have:

    <=> cos3A.cos3B+cos3B.cos3C+cos3C.cos3A = sin3A.sin3B + sin3B.sin3C + sin3C.sin3A – ( cos3A  + cos3B + cos3C )

    = 1  

    Hope it can find you well.

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