Тригонометрия #39 В ответе должен быть 3). Created by Jklt. algebra-ru

Тригонометрия #39 В ответе должен быть 3)...

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

1 votes Thanks 1

армения20171 (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

2 votes Thanks 2

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