Bài tập 6.53 trang 192 SBT Toán 10

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 HỌC247