Biến đổi tích thành tổng biểu thức C = 2 × sinx × sin3x × sin5x 21/08/2021 Bởi Melanie Biến đổi tích thành tổng biểu thức C = 2 × sinx × sin3x × sin5x
$C= 2.sinx.sin3x.sin5x$ $= 2sin3x.\dfrac{1}{2}(cos4x – cos6x)$ $= sin3x.cos4x – sin3x.cos6x$ $= \dfrac{1}{2}(sin7x – sinx) – \dfrac{1}{2}(sin9x – sin3x)$ $= -\dfrac{1}{2}sin9x + \dfrac{1}{2}sin7x + \dfrac{1}{2}sin3x – \dfrac{1}{2}sinx$ Bình luận
$C=2\sin x.\sin 3x.\sin 5x$ $=2.\dfrac{1}{2}.(\sin 6x-\sin 4x).\sin 3x$ $= \sin 6x.\sin 3x-\sin 4x.\sin 3x$ $=-\dfrac{1}{2}(\cos 9x-\cos 6x)+\dfrac{1}{2}(\cos 7x-\cos x)$ $=\dfrac{-1}{2}\cos 9x +\dfrac{1}{2}\cos 6x+\dfrac{1}{2}\cos 7x -\dfrac{1}{2}\cos x$ Bình luận
$C= 2.sinx.sin3x.sin5x$
$= 2sin3x.\dfrac{1}{2}(cos4x – cos6x)$
$= sin3x.cos4x – sin3x.cos6x$
$= \dfrac{1}{2}(sin7x – sinx) – \dfrac{1}{2}(sin9x – sin3x)$
$= -\dfrac{1}{2}sin9x + \dfrac{1}{2}sin7x + \dfrac{1}{2}sin3x – \dfrac{1}{2}sinx$
$C=2\sin x.\sin 3x.\sin 5x$
$=2.\dfrac{1}{2}.(\sin 6x-\sin 4x).\sin 3x$
$= \sin 6x.\sin 3x-\sin 4x.\sin 3x$
$=-\dfrac{1}{2}(\cos 9x-\cos 6x)+\dfrac{1}{2}(\cos 7x-\cos x)$
$=\dfrac{-1}{2}\cos 9x +\dfrac{1}{2}\cos 6x+\dfrac{1}{2}\cos 7x -\dfrac{1}{2}\cos x$