Question

Тригонометрия

#39
В ответе должен быть 3)

Answer 1

$displaystyle frac{sin^22x-cos^23x}{cos5x}=frac{(2sinx*cosx)^2-(4cos^3x-3cosx)^2}{cos(2x+3x)}=\\=frac{4sin^2x*cos^2x-(16cos^6x-24cos^4x+9cos^2x)}{cos3x*cos2x-sin3x*sin2x}=\\=frac{cos^2x(4(1-cos^2x)-16cos^4x+24cos^2x-9)}{(4cos^2x-3cosx)(2cos^2x-1)-(3sinx-4sin^3x)(2sinx*cosx)}=\\=frac{cos^2x(20cos^2x-16cos^4x-5)}{16cos^5x-20cos^3x+5cosx}=frac{cos^2x(20cos^2x-16cos^4x-5)}{-cosx(20cos^2x-16cos^4x-5)}=\\=frac{cos^2x}{-cosx}=-cosx$

Answer 2

(sin²2x-cos²3x)/cos5x=

1/2(1-cos4x-1-cos6x)/cos5x=

-1/2(cos4x+cos6x)/cos5x=

(-cos(4x+6x)/2*cos(6x-4x)/2)/cos5x=

(-cos5x*cosx)/cos5x=-cosx