Toán Biến đổi thành tổng: $A=4sin3asin2asina$ $B=4sin3asin2acosa$ 07/09/2021 By Piper Biến đổi thành tổng: $A=4sin3asin2asina$ $B=4sin3asin2acosa$
$A=4\sin2a(\sin3a.\sin a)$ $=4.\sin2a.\dfrac{-1}{2}\Big(\cos4a-\cos2a\Big)$ $=-2\sin2a.(\cos4a-\cos2a)$ $=-2\sin2a.\cos4a+2\sin2a.\cos2a$ $=-2.\dfrac{1}{2}\Big(\sin6a+\sin(-2a)\Big)+\sin4a$ $=-\sin6a+\sin2a+\sin4a$ $=\sin2a+\sin4a-\sin6a$ $B=4\sin2a.(\sin3a.\cos a)$ $=4\sin2a.\dfrac{1}{2}\Big(\sin4a+\sin2a\Big)$ $=2\sin4a.\sin2a+2\sin^22a$ $=-(\cos6a-\cos2a)+2.\dfrac{1-\cos4a}{2}$ $=-\cos6a+\cos2a+2-2\cos4a$ $=\cos2a-2\cos4a-\cos6a+2$ Trả lời
$A=4\sin2a(\sin3a.\sin a)$
$=4.\sin2a.\dfrac{-1}{2}\Big(\cos4a-\cos2a\Big)$
$=-2\sin2a.(\cos4a-\cos2a)$
$=-2\sin2a.\cos4a+2\sin2a.\cos2a$
$=-2.\dfrac{1}{2}\Big(\sin6a+\sin(-2a)\Big)+\sin4a$
$=-\sin6a+\sin2a+\sin4a$
$=\sin2a+\sin4a-\sin6a$
$B=4\sin2a.(\sin3a.\cos a)$
$=4\sin2a.\dfrac{1}{2}\Big(\sin4a+\sin2a\Big)$
$=2\sin4a.\sin2a+2\sin^22a$
$=-(\cos6a-\cos2a)+2.\dfrac{1-\cos4a}{2}$
$=-\cos6a+\cos2a+2-2\cos4a$
$=\cos2a-2\cos4a-\cos6a+2$