Найти (sin⁴ x + cos⁴ x) / (sin⁶ x + cos⁶ x) , если tg x=2
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cos² x = 1/(1+tg²x) = 1/(1+2²) =1/5 ;
sin² x =cos²x*tq²x = 1/5* 4 =4/5 .
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(sin⁴ x + cos⁴ x)/(sin⁶ x + cos⁶ x) =( (sin²x)² +(cos²x)²) / ( (sin²x)³ +(cos²x) ³ ) =
(16/25 +1/25) /(64/125 +1/125) =(17/25)/(13/25) = 17 / 13 .
или по другому
(sin⁴ x + cos⁴ x) / (sin⁶ x + cos⁶ x) =cos⁴ x (tg⁴x +1 ) / cos⁶x(tg⁶x+1)=
=(tg²x+1)* (tg⁴x +1 ) / (tg⁶x+1)=(2²+1) (2⁴ +1) / (2⁶ +1) =5*17/65 =
ответ : 17 / 13 .