Question

(cos5a+5cos3a+10cosa)/(cosa)^5=

Answer 1

task/27151545
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(cos5a+5cos3a+10cosa)/(cosa)^5=
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решение :
* * * cos²φ =(1+cos2φ)/2  ; cosα*cosβ =( cos(α+β)+ cos(α-β) )/2 * * *
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cos⁵α =(cosα)*(cos²α)² =(cosα)* ( (1+cos2α) /2 )² =(cosα)* (1+2cos2α+cos²2α)/4 = (cosα)*(1+2cos2α+(1+cos4α)/2 )/4 = 
(cosα)*(3 +4cos2α +cos4α*cosα)/8=(3cosα+4cos2α*cosα +cos4α*cosα) /8=(3cosα+2cosα+2cos3α +(cos5α+cos3α)/2)/8 =(cos5α+5cos3α+10cosα)/16 .
Следовательно :
(cos5α+5cos3α+10cosα) / (cos⁵α ) =
(cos5α+5cos3α+10cosα) / ( (cos5α+5cos3α+10cosα)/16  ) =  16.

ответ: 16

Answer 2

Оганес, спасибо большое!)