Ответы на вопрос:
Во-первых, разложим косинусы в сумму cos(pi/4 - x) = cos(pi/4)*cos(x) + sin(pi/4)*sin(x) = = 1/√2*cos x + 1/√2*sin x = 1/√2*(cos x + sin x) cos(pi/4 + x) = cos(pi/4)*cos(x) - sin(pi/4)*sin(x) = = 1/√2*cos x - 1/√2*sin x = 1/√2*(cos x - sin x) подставляем cos^6(pi/4-x) + cos^6(pi/4+x) = 1/2^3*(cos x+sin x)^6 + 1/2^3*(cos x-sin x)^6 = = 1/8*[(cos x+sin x)^6 + (cos x-sin x)^6] = 0,5 (cos x + sin x)^6 + (cos x - sin x)^6 = 4 во-вторых, разложим сумму кубов [(cos x + sin x)^2 + (cos x - sin x)^2] * [(cos x + sin x)^4 - - (cos x + sin x)^2*(cos x - sin x)^2 + (cos x - sin x)^4] = 4 первая скобка cos^2 x + 2cos x*sin x + sin^2 x + cos^2 x - 2cos x*sin x + sin^2 x = = (cos^2 x + sin^2 x) + (2cos x*sin x - 2cos x*sin x) + (cos^2 x + sin^2 x) = 2 вторая скобка (cos x+sin x)^4 - (cos x+sin x)^2*(cos x-sin x)^2 + (cos x-sin x)^4 = = ((cos x+sin x)^2)^2 + ((cos x-sin x)^2)^2 - (cos x+sin x)^2*(cos x-sin x)^2 = = (1+2cos x*sin x)^2 + (1-2cos x*sin x)^2 - (1+2cos x*sin x)(1-2cos x*sin x) = = (1+sin 2x)^2 + (1-sin 2x)^2 - (1-sin^2 2x) = 1 + 2sin 2x + sin^2 2x + + 1 - 2sin 2x + sin^2 2x - 1 + sin^2 2x = 1 + 3sin^2 2x подставляем в уравнение 2(1 + 3sin^2 2x) = 4 1 + 3sin^2 2x = 2 3sin^2 2x = 1 sin^2 2x = 1/3 sin 2x = 1/√3 2x = (-1)^n*arcsin(1/√3) + pi*k x = (-1)^n*1/2*arcsin(1/√3) + pi/2*k
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