Task/27151545
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<span>(cos5a+5cos3a+10cosa)/(cosa)^5=
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</span>решение :<span>
* * * cos</span>²φ =(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
(5*✓3-✓800)*(5*✓3+✓80)=(5*✓3)²-(✓80)²=25*3-80=75-80=-5
1)43+3=46 2)46-5=41 (43=3)-5=41
=n^2-9a^2 согласно формуле сокращённого умножения