Đáp án:
1) B=-1
C=-1
2) C=1
D=91/2
Giải thích các bước giải:
$$\eqalign{
& 1)\,\,{0^0} < x < {90^0} \cr
& B = {\cos ^4}x - {\sin ^4}x - 2{\cos ^2}x \cr
& \,\,\,\,\, = \left( {{{\cos }^2}x - {{\sin }^2}x} \right)\left( {{{\cos }^2}x + {{\sin }^2}x} \right) - 2{\cos ^2}x \cr
& \,\,\,\,\, = \cos 2x - 2{\cos ^2}x \cr
& \,\,\,\,\, = 2{\cos ^2}x - 1 - 2{\cos ^2}x = - 1 \cr
& C = 2\left( {{{\sin }^6}x + {{\cos }^6}x} \right) - 3\left( {{{\sin }^4}x + {{\cos }^4}x} \right) \cr
& \,\,\,\, = 2{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^3} - 6{\sin ^2}x{\cos ^2}x\left( {{{\sin }^2}x + {{\cos }^2}x} \right) - 3{\left( {{{\sin }^2}x + {{\cos }^2}} \right)^2} + 6{\sin ^2}x{\cos ^2}x \cr
& \,\,\,\, = 2 - 3 = - 1 \cr
& 2)\,\,C = \cot {1^0}.\cot {2^0}.\cot {3^0}...\cot {87^0}\cot {88^0}.\cot {89^0} \cr
& \,\,\,\,\,\,C = \cot {1^0}.\cot {2^0}.\cot {3^0}...\cot {45^0}.\tan {44^0}.\tan {43^0}....\tan {1^0} \cr
& \,\,\,\,\,\,C = \left( {\cot {1^0}.\tan {1^0}} \right).\left( {\cot {2^0}.\tan {2^0}} \right)...\left( {\cot {{44}^0}\tan {{44}^0}} \right).cot{45^0} \cr
& \,\,\,\,\,\,C = 1.1.1.1 = 1 \cr
& \,\,\,\,\,\,D = {\cos ^2}{1^0} + {\cos ^2}{2^0} + ... + {\cos ^2}{89^0} \cr
& \,\,\,\,\,\,D = {\cos ^2}{1^0} + {\cos ^2}{2^0} + ... + {\cos ^2}{44^0} + {\cos ^2}{45^0} + {\sin ^2}{44^0} + ... + {\sin ^2}{1^0} \cr
& \,\,\,\,\,\,D = \left( {{{\cos }^2}{1^0} + {{\sin }^2}{1^0}} \right) + ... + \left( {{{\cos }^2}{{44}^0} + {{\sin }^2}{{44}^0}} \right) + {\cos ^2}{45^0} \cr
& \,\,\,\,\,D = 1 + 1 + ... + 1 + {\left( {{{\sqrt 2 } \over 2}} \right)^2} \cr
& \,\,\,\,\,\,D = 45 + {1 \over 2} = {{91} \over 2} \cr} $$
