Подробное решение, пожалуйста! ( sin 8a - sin2a ) ( cos 2a - cos 8a ) / 1 - cos6a. Created by Morkva04. algebra-ru

1 - cos6a...

Morkva04 @Morkva04

August 2022 1 7 Report

Подробное решение, пожалуйста!
( sin 8a - sin2a ) ( cos 2a - cos 8a ) / 1 - cos6a

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nafanya2014

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$sin8\alpha-sin2\alpha =2\cdot sin\frac{8\alpha -2\alpha }{2} \cdot cos\frac{8\alpha +2\alpha }{2} =2\cdot sin3\alpha \cdot cos5\alpha$

$cos2\alpha-cos8\alpha =-2\cdot sin\frac{2\alpha +8\alpha }{2} \cdot sin \frac{2\alpha -8\alpha }{2} =-2\cdot sin5\alpha \cdot sin(-3\alpha) =\\\\=2\cdot sin 5\alpha \cdot sin3\alpha$

$1-cos6\alpha =2sin^23\alpha$

$\frac{(sin8\alpha-sin2\alpha)(cos2\alpha -cos8\alpha ) }{1-cos6\alpha } =\frac{2\cdot sin3\alpha\cdot cos5\alpha \cdot 2sin5\alpha \cdot sin 3\alpha }{2sin^23\alpha } =2 \cdot sin5 \alpha \cdot cos5\alpha =sin10\alpha$

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Morkva04 Спасибо!

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