Помогите, пожалуйста!!. Created by vikushamur. matematika-ru

Помогите, пожалуйста!!...

vikushamur @vikushamur

October 2021 1 10 Report

Помогите, пожалуйста!!

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oganesbagoyan

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1+(tq²(-α) -1) /sin(0,5π +2α) =1+(tq²α -1) /sin(π/2 +2α) =
1+(tq²α - 1 ) /( cos² α - sin²α) = 1+ (tq²α- 1 ) / cos²α ( 1- tq²α) = 1 -1/cos²α = (cos²α - 1)/ cos²α = -sin²α /cos²α = - tq²α.
------------------------
( sin²2,5α - sin²1,5α) /(sin4α*sinα +cos3α*cos2α) =
( (1 -cos2*2,5α)/2 -(1-cos2*1,5α) / 2 ) / ( (cos3α -cos5α)/2 + (cosα +cos5α)/2 ) =(cos3α - cos5α ) / 2 / (cos3α + cosα)/2 =(cos3α -cos5α) / (cos3α + cosα) = 2sin4α* sinα / 2 cos2α*cosα = 2sin2α*cos2α *sinα / cos2α*cosα =
2sin2α *sinα / cosα = 2*2sinα *cosα*sinα / cosα = 4sin²α .

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