ответ:Sin^6α-cos^6α-(3/4)*cos²2α+(3/4)*sin²α sin^6α-cos^6α-(3/4)*cos²α+(3/4)*sin²α= -(cos^6α-sin^6α)-(3/4)(cos²α-sin²α)= -(cos²α-sin²α)(cos^4α+sin²α*cos²α+sin^4α) -3/4(cos²α-sin²α)= -(cos²α-sin²α)(cos^4α+sin²α*cos²α+sin^4α -3/4)= -cos2α((cos²α+sin²α)²-cos²α*sin²α -3/4)= - cos2α(1-(1/4)*sin²2α -3/4)= - cos2α(1/4-(1/4)*sin²2α)= - (1/4)cos2α*(1-sin²2α) = - (1/4)cos2α*cos²2α= - (1/4)cos³2α .
Подробнее – на Otvet.Ws – https://otvet.ws/questions/4864874-pomogite-pozhaluista-sin6a-cos6a-34cos2a34sin2a.h
Пошаговое объяснение:
(3 1/4 +2 1/6 ) : 2 3/5 - 2/3 * 2 1/4 + 5 1/6= 5 3/4
(3 1/4 +2 1/6 ) : 2 3/5 - 2/3 * 2 1/4 + 5 1/6= 5 3/41) 3 1/4 + 2 1/6 = 3 3/12 + 2 2/12 =5 5/12
(3 1/4 +2 1/6 ) : 2 3/5 - 2/3 * 2 1/4 + 5 1/6= 5 3/41) 3 1/4 + 2 1/6 = 3 3/12 + 2 2/12 =5 5/122) 5 5/12 : 2 3/5 = 65/12 : 13/5 = (65*5)/(12*13)=
(3 1/4 +2 1/6 ) : 2 3/5 - 2/3 * 2 1/4 + 5 1/6= 5 3/41) 3 1/4 + 2 1/6 = 3 3/12 + 2 2/12 =5 5/122) 5 5/12 : 2 3/5 = 65/12 : 13/5 = (65*5)/(12*13)==(5*5)/ (12*1) =25/12= 2 1/12
(3 1/4 +2 1/6 ) : 2 3/5 - 2/3 * 2 1/4 + 5 1/6= 5 3/41) 3 1/4 + 2 1/6 = 3 3/12 + 2 2/12 =5 5/122) 5 5/12 : 2 3/5 = 65/12 : 13/5 = (65*5)/(12*13)==(5*5)/ (12*1) =25/12= 2 1/123) 2/3 * 2 1/4 = 2*9/ 3*4= 1*3/ 1*2= 3/2 = 1 1/2
(3 1/4 +2 1/6 ) : 2 3/5 - 2/3 * 2 1/4 + 5 1/6= 5 3/41) 3 1/4 + 2 1/6 = 3 3/12 + 2 2/12 =5 5/122) 5 5/12 : 2 3/5 = 65/12 : 13/5 = (65*5)/(12*13)==(5*5)/ (12*1) =25/12= 2 1/123) 2/3 * 2 1/4 = 2*9/ 3*4= 1*3/ 1*2= 3/2 = 1 1/24) 2 1/12 - 1 1/2 = 25/12 - 18/12 = 7/12
(3 1/4 +2 1/6 ) : 2 3/5 - 2/3 * 2 1/4 + 5 1/6= 5 3/41) 3 1/4 + 2 1/6 = 3 3/12 + 2 2/12 =5 5/122) 5 5/12 : 2 3/5 = 65/12 : 13/5 = (65*5)/(12*13)==(5*5)/ (12*1) =25/12= 2 1/123) 2/3 * 2 1/4 = 2*9/ 3*4= 1*3/ 1*2= 3/2 = 1 1/24) 2 1/12 - 1 1/2 = 25/12 - 18/12 = 7/125) 7/12 + 5 1/6 = 7/12 + 5 2/12= 5 9/12= 5 3/4
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