Giải bài 3.1 trang 32 sách bài tập toán 10 – Kết nối tri thức với cuộc sống

Đề bài

Tính giá trị của các biểu thức:

a) \(A = \sin {45^ \circ } + 2\sin {60^ \circ } + \tan {120^ \circ } + \cos {135^ \circ }.\)

b) \(B = \tan {45^ \circ }.\cot {135^ \circ } – \sin {30^ \circ }.\cos {120^ \circ } – \sin {60^ \circ }.\cos {150^ \circ }.\)

c) \(C = {\cos ^2}{5^ \circ } + {\cos ^2}{25^ \circ } + {\cos ^2}{45^ \circ } + {\cos ^2}{65^ \circ } + {\cos ^2}{85^ \circ }.\) \(\)

d) \(D = \frac{{12}}{{1 + {{\tan }^2}{{73}^ \circ }}} – 4\tan {75^ \circ }.\cot {105^ \circ } + 12{\sin ^2}{107^ \circ } – 2\tan {40^ \circ }.\cos {60^ \circ }.\tan {50^ \circ }.\)

e) \(E = 4\tan {32^ \circ }.\cos {60^ \circ }.\cot {148^ \circ } + \frac{{5{{\cot }^2}108}}{{1 + {{\tan }^2}{{18}^ \circ }}} + 5{\sin ^2}{72^ \circ }.\)

Lời giải chi tiết

a) \(A = \sin {45^ \circ } + 2\sin {60^ \circ } + \tan {120^ \circ } + \cos {135^ \circ }\)

\(\begin{array}{l}A = \frac{{\sqrt 2 }}{2} + 2.\frac{{\sqrt 3 }}{2} + \left( { – \sqrt 3 } \right) + \left( {\frac{{ – \sqrt 2 }}{2}} \right)\\A = \frac{{\sqrt 2 }}{2} + \sqrt 3  + \left( { – \sqrt 3 } \right) + \left( {\frac{{ – \sqrt 2 }}{2}} \right)\\A = 0\end{array}\)

b) \(B = \tan {45^ \circ }.\cot {135^ \circ } – \sin {30^ \circ }.\cos {120^ \circ } – \sin {60^ \circ }.\cos {150^ \circ }\)

\(\begin{array}{l}B = 1.\left( { – 1} \right) – \frac{1}{2}.\left( {\frac{{ – 1}}{2}} \right) – \frac{{\sqrt 3 }}{2}.\left( {\frac{{ – \sqrt 3 }}{2}} \right)\\B =  – 1 + \frac{1}{4} + \frac{3}{4} = 0\end{array}\)

c) \(C = {\cos ^2}{5^ \circ } + {\cos ^2}{25^ \circ } + {\cos ^2}{45^ \circ } + {\cos ^2}{65^ \circ } + {\cos ^2}{85^ \circ }\)

\(\begin{array}{l}C = {\cos ^2}\left( {{{90}^ \circ } – {{85}^ \circ }} \right) + {\cos ^2}\left( {{{90}^ \circ } – {{65}^ \circ }} \right) + {\cos ^2}{45^ \circ } + {\cos ^2}{65^ \circ } + {\cos ^2}{85^ \circ }\\C = {\sin ^2}{85^ \circ } + {\cos ^2}{85^ \circ } + {\sin ^2}{65^ \circ } + {\cos ^2}{85^ \circ } + {\sin ^2}{45^ \circ }\\C = 1 + 1 + {\left( {\frac{{\sqrt 2 }}{2}} \right)^2} = 2 + \frac{1}{2} = \frac{5}{2}\end{array}\)

d) \(D = \frac{{12}}{{1 + {{\tan }^2}{{73}^ \circ }}} – 4\tan {75^ \circ }.\cot {105^ \circ } + 12{\sin ^2}{107^ \circ } – 2\tan {40^ \circ }.\cos {60^ \circ }.\tan {50^ \circ }\)

\(\begin{array}{l}D = 12{\cos ^2}{73^ \circ } + 12{\sin ^2}\left( {{{180}^ \circ } – {{73}^ \circ }} \right) – 4\tan {75^ \circ }.\cot \left( {{{180}^ \circ } – {{75}^ \circ }} \right) – 2\tan {40^ \circ }.\tan \left( {{{90}^ \circ } – {{40}^ \circ }} \right)\cos {60^ \circ }\\D = 12{\cos ^2}{73^ \circ } + 12{\sin ^2}{73^ \circ } – 4\tan {75^ \circ }\left( { – \cot {{75}^ \circ }} \right) – 2\tan {40^ \circ }.\cot {40^ \circ }.\frac{1}{2}\\D = 12 + 4 – 1 = 15\end{array}\)

e) \(E = 4\tan {32^ \circ }.\cos {60^ \circ }.\cot {148^ \circ } + \frac{{5{{\cot }^2}108}}{{1 + {{\tan }^2}{{18}^ \circ }}} + 5{\sin ^2}{72^ \circ }.\)

\(\begin{array}{l}E = 4\tan {32^ \circ }.\cot \left( {{{180}^ \circ } – {{32}^ \circ }} \right).\cos {60^ \circ } + \frac{{5{{\cot }^2}\left( {{{180}^ \circ } – {{72}^ \circ }} \right)}}{{1 + {{\tan }^2}{{18}^ \circ }}} + 5{\sin ^2}\left( {{{90}^ \circ } – {{18}^ \circ }} \right)\\E =  – 4\tan {32^ \circ }.\cot {32^ \circ }.\cos {60^ \circ } + 5{\cot ^2}{72^ \circ }.{\cos ^2}{18^ \circ } + 5{\cos ^2}{18^ \circ }\\E =  – 4\cos {60^ \circ } + 5{\cot ^2}\left( {{{90}^ \circ } – {{18}^ \circ }} \right).{\cos ^2}{18^ \circ } + 5{\cos ^2}{18^ \circ }\\E =  – 4\cos {60^ \circ } + 5{\tan ^2}{18^ \circ }.{\cos ^2}{18^ \circ } + 5{\cos ^2}{18^ \circ }\\E =  – 4\cos {60^ \circ } + 5{\sin ^2}{18^ \circ } + 5{\cos ^2}{18^ \circ }\\E =  – 4.\frac{1}{2} + 5 =  – 2 + 5 = 3\end{array}\)