Помогите решить!упростить: (sin10x+sin6x+sin2x)/(cos6x+cos6x+cos2x). Created by dilshodbekatanazarov. algebra-ru

(cos6x+cos6x+cos2x)...

dilshodbekatanazarov @dilshodbekatanazarov

August 2022 2 34 Report

Помогите решить!

упростить: (sin10x+sin6x+sin2x)/(cos6x+cos6x+cos2x)

$\dfrac{sin10x+sin6x+sin2x}{cos10x+cos6x+cos2x}=\dfrac{2\, sin6x\cdot cos4x+sin6x}{2\, cos6x\cdot cos4x+cos6x}=\\\\\\=\dfrac{sin6x(2cos4x+1)}{cos\, 6x(2cos4x+1)}=\dfrac{sin6x}{cos\, 6x}=tg6x$

$\star \ \ \ \dfrac{sin10x+sin6x+sin2x}{cos6x+cos6x+cos2x}=\dfrac{sin6x(2cos4x+1)}{2cos6x+cos2x}=\\\\=\dfrac{sin6x(2cos4x+1)}{2(4cos^32x-3cos2x)+cos2x}=\dfrac{sin6x(2cos4x+1)}{8cos^32x-5cos2x}=\dfrac{sin6x(2cos4x+1)}{cos2x(8cos^22x-5)}$

1 votes Thanks 0

dilshodbekatanazarov это же не окончательный ответ

NNNLLL54

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на фото.......

Объяснение:

3 votes Thanks 1

dilshodbekatanazarov спасибо

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