Biến đổi thành tổng: $A=4sin3asin2asina$ $B=4sin3asin2acosa$

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Biến đổi thành tổng:
$A=4sin3asin2asina$
$B=4sin3asin2acosa$

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Piper 1 giờ 2021-09-07T11:48:30+00:00 1 Answers 0 views 0

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    2021-09-07T11:50:01+00:00

    $A=4\sin2a(\sin3a.\sin a)$

    $=4.\sin2a.\dfrac{-1}{2}\Big(\cos4a-\cos2a\Big)$

    $=-2\sin2a.(\cos4a-\cos2a)$

    $=-2\sin2a.\cos4a+2\sin2a.\cos2a$

    $=-2.\dfrac{1}{2}\Big(\sin6a+\sin(-2a)\Big)+\sin4a$

    $=-\sin6a+\sin2a+\sin4a$

    $=\sin2a+\sin4a-\sin6a$

    $B=4\sin2a.(\sin3a.\cos a)$

    $=4\sin2a.\dfrac{1}{2}\Big(\sin4a+\sin2a\Big)$

    $=2\sin4a.\sin2a+2\sin^22a$

    $=-(\cos6a-\cos2a)+2.\dfrac{1-\cos4a}{2}$

    $=-\cos6a+\cos2a+2-2\cos4a$

    $=\cos2a-2\cos4a-\cos6a+2$

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