求证:cos^8(x) - sin^8(x) + (1/4)sin2xsin4x = cos2x
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求证:cos^8(x) - sin^8(x) + (1/4)sin2xsin4x = cos2x
注:开头的cos、sin上的8是8次方,x不是在8上的,是在cos、sin上的
注:开头的cos、sin上的8是8次方,x不是在8上的,是在cos、sin上的
cos^8(x) - sin^8(x) + (1/4)sin2xsin4x
= [cos^4(x) + sin^4(x)]*[cos^4(x) - sin^4(x)] + (1/4)*sin2x * 2*sin2x*cos2x
=[cos^4(x) + 2*cos^2(x)*sin^2(x) + sin^4(x) - 2*cos^2(x)*sin^2(x)] * [cos^2(x) + sin^2(x)]*[cos^2(x) - sin^2(x)] + (1/2)*(sin2x)^2 * cos2x
= [1 - 2*cos^2(x) * sin^2(x)]*cos2x + (1/2)*(sin2x)^2 * cos2x
= [1 - (1/2)* (sin2x)^2] * cos2x + (1/2)* (sin2x)^2 * cos2x
= cos2x
= [cos^4(x) + sin^4(x)]*[cos^4(x) - sin^4(x)] + (1/4)*sin2x * 2*sin2x*cos2x
=[cos^4(x) + 2*cos^2(x)*sin^2(x) + sin^4(x) - 2*cos^2(x)*sin^2(x)] * [cos^2(x) + sin^2(x)]*[cos^2(x) - sin^2(x)] + (1/2)*(sin2x)^2 * cos2x
= [1 - 2*cos^2(x) * sin^2(x)]*cos2x + (1/2)*(sin2x)^2 * cos2x
= [1 - (1/2)* (sin2x)^2] * cos2x + (1/2)* (sin2x)^2 * cos2x
= cos2x
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