Bài tập 3 trang 19 SBT Toán 11 Tập 1 Chân trời sáng tạo - CTST

a) Ta có: sinxcos⁵x ‒ cosxsin⁵x

 = sinxcosx(cos⁴x ‒ sin⁴x)

$=\frac{1}{2}\text{sin}2x\left({\text{cos}}^{2}x-{\text{sin}}^{2}x\right)\left({\text{cos}}^{2}x+{\text{sin}}^{2}x\right)$

$=\frac{1}{2}\text{sin}2x\text{cos}2x$

$=\frac{1}{4}\text{sin}4x.$

b) Ta có: $\frac{\text{sin}3x\text{cos}2x+\text{sin}x\text{cos}6x}{\text{sin}4x}$

$=\frac{\frac{1}{2}\left(\text{sin}x+\text{sin}5x\right)+\frac{1}{2}\left[\text{sin}\left(-5x\right)+\text{sin}7x\right]}{\text{sin}4x}$

$=\frac{\text{sin}x+\text{sin}5x-\text{sin}5x+\text{sin}7x}{2\text{sin}4x}$

$=\frac{\text{sin}x+\text{sin}7x}{2\text{sin}4x}$

$=\frac{2\text{sin}4x\text{cos}3x}{2\text{sin}4x}$

$=\text{cos}3x.$

$\text{c) Ta có: }\frac{\text{cos}x-\text{cos}2x+\text{cos}3x}{\text{sin}x-\text{sin}2x+\text{sin}3x}$

$=\frac{\left(\text{cos}x+\text{cos}3x\right)-\text{cos}2x}{\left(\text{sin}x+\text{sin}3x\right)-\text{sin}2x}$

$=\frac{2\text{cos}2x\text{cos}x-\text{cos}2x}{2\text{sin}2x\text{cos}x-\text{sin}2x}$

$=\frac{\text{cos}2x\left(2\text{cos}x-1\right)}{\text{sin}2x\left(2\text{cos}x-1\right)}$

$=\frac{\text{cos}2x}{\text{sin}2x}$

$=\text{cot}2x.$

d) Ta có: $\frac{2\text{sin}\left(x+y\right)}{\text{cos}\left(x+y\right)+\text{cos}\left(x-y\right)}-\text{tan}y$

$=\frac{2\left(\text{sin}x\text{cos}y+\text{cos}x\text{sin}y\right)}{2\text{cos}x\text{cos}y}-\text{tan}y$

$=\frac{\text{sin}x}{\text{cos}x}+\frac{\text{sin}y}{\text{cos}y}-\tan y$

$=\tan x+\tan y-\tan y$

$=\tan x.$

-- Mod Toán 11 HỌC247