[Toán 10] Lượng giác.

L

l0v3_sweet_381

P

pe_lun_hp

Bài 1:

$B =\dfrac{1}{4} sin^2x - 2sinx.cos\dfrac{2\pi}{3} + \dfrac{3}{4}.cos^2 + \dfrac{1}{4} sin^2x + 2sinx.cos\dfrac{2\pi}{3} + \dfrac{3}{4}.cos^2 + sin^2x$

$B = \dfrac{1}{2}sin^2x + sin^2x + \dfrac{3}{2}cos^2x$

$B= \dfrac{3}{2}$ (đpcm) :)
 
N

noinhobinhyen

bài 2.

$sinx.sin(\dfrac{\pi}{3}-x).sin(\dfrac{\pi}{3}+x)$

$=sinx.\dfrac{-1}{2}.[cos\dfrac{2\pi}{3}-cos2x]$

$=sinx.\dfrac{-1}{2}.(\dfrac{-3}{2}+2.sin^2x)$

$=\dfrac{3}{4}.sinx-sin^3x = \dfrac{1}{4}.sin3x$

:( :)|
 
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