128(2,4) и 131(2,4,6). Created by stiklin. algebra-ru

128(2,4) и 131(2,4,6)...

stiklin @stiklin

July 2022 1 36 Report

128(2,4) и 131(2,4,6)

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oganesbagoyan

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Task/25894187
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4cos(π/4)*ctg(π/3)*tg(π/3) -3tg(π/4) =4*(√2)/2 *1 -3*1 =2√2 - 3.
--------
(2cos(π/6) -ctg(π/4) +sin²(π/3) +ctg²(π/3) )⁻¹ =
(2*(√3)/2 -1 + ( (√3) /2 )²  - (1/√3)² )⁻¹ = (√3  -1 + 3/4 - 1/3)  )⁻¹  =(√3  -7/12)⁻¹ =
1/ (√3  -7/12) = 12/(12√3 -7) =12(12√3 +7) /383 .
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sin²x -sin⁴x + cos⁴x = sin²x (1 - sin²x ) +   cos⁴x =sin²x *cos²x + cos⁴x =
cos²x (cos²x + cos²x) =cos²x *1 = cos²x.
или
sin²x-sin⁴x +cos⁴x =sin²x -(sin⁴x - cos⁴x ) =sin²x -(sin²x -cos²x )(sin²x + cos²x) =
sin²x -(sin²x - cos²x )*1= sin²x -sin²x + cos²x=cos²x.
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(tgα+tgβ)/(ctgα+ctgβ) =tgα*tgβ(1/tgα+1/tgβ)/(ctgα+ctgβ) =
tgα*tgβ(ctgα +ctgβ) / (ctgα+ctgβ) =tgα*tgβ.
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(sin²α -cos²α + cos⁴α )/(cos²α - sin²α +sin⁴α) =
( sin²α -cos²α (1 - cos²α ) ) / (cos²α - sin²α(1 -sin²α) ) =
(sin²α -cos²α *sin²α ) / (cos²α - sin²α*cos²α) ) =
sin²α(1 -cos²α) /cos²α(1 -sin²α )   = sin⁴x /cos⁴x =tg⁴x.
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Удачи !

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