Chứng minh rằng các biểu thức sau là những số không phụ thuộc α
a) A = 2(sin6α + cos6α) - 3(sin4α + cos4α)
b) B = 4(sin4α + sin4α) - cos4α
c) C = 8(cos8α - sin8α) - cos6α - 7cos2α
a) A = 2(sin2α + cos2α)(sin4α + cos4α - sin2αcos2α) - 3(sin4α + cos4α)
= - sin4α - cos4α - 2sin2αcos2α
= - (sin2α + cos2α)2 = -1
b) B = 4[(sin2α + cos2α)2 - 2sin2αcos2α] - cos4α
= 4[(1 - sin22α)/2] - 1 + 2sin22α = 3
c) C =\(8\left( {{{\cos }^4}\alpha - {{\sin }^4}\alpha } \right)\left( {{{\cos }^4}\alpha + {{\sin }^4}\alpha } \right) - \cos 6\alpha - 7\cos 2\alpha \)
\(\begin{array}{l}
= 8\left( {{{\cos }^2}\alpha - {{\sin }^2}\alpha } \right)\left( {{{\cos }^2} + {{\sin }^2}\alpha } \right)\left[ {{{\left( {{{\cos }^2}\alpha + {{\sin }^2}\alpha } \right)}^2} - 2{{\sin }^2}\alpha {{\cos }^2}\alpha } \right] - \cos 6\alpha - 7\cos 2\alpha \\
= 8\cos 2\alpha \left( {1 - \frac{1}{2}{{\sin }^2}2\alpha } \right) - \cos 6\alpha - 7\cos 2\alpha \\
= \cos 2\alpha - 4\cos 2\alpha {\sin ^2}2\alpha - \cos \left( {4\alpha + 2\alpha } \right)\\
= \cos 2\alpha - 2\sin 4\alpha \sin 2\alpha - \cos 4\alpha \cos 2\alpha + \sin 4\alpha \sin 2\alpha \\
= \cos 2\alpha - \left( {\cos 4\alpha \cos 2\alpha + \sin 4\alpha \sin 2\alpha } \right)\\
= \cos 2\alpha - \cos 2\alpha = 0
\end{array}\)
-- Mod Toán 10