Bài 1.6 trang 15 SGK Toán 11 tập 1 - Cùng khám phá

Đề bài

Không dùng máy tính cầm tay, tính:

a) \({\sin ^2}\frac{\pi }{4} + \cos \left( { - \frac{\pi }{2}} \right);\)

b) \({\tan ^2}\left( {{{30}^0}} \right) - {\cot ^2}\left( {{{240}^0}} \right);\)

c) \({\sin ^3}\frac{\pi }{2} - \cos 5\pi ;\)

d) \(\tan \frac{{11\pi }}{3} - \cot \left( { - \frac{{21\pi }}{4}} \right)\).

Lời giải chi tiết

a)

\(\begin{array}{l}{\sin ^2}\frac{\pi }{4} + \cos \left( { - \frac{\pi }{2}} \right)\\ = {\sin ^2}\frac{\pi }{4} + \cos \frac{\pi }{2}\\ = {\left( {\frac{{\sqrt 2 }}{2}} \right)^2} + 0 = \frac{1}{2}\end{array}\)

b)

\(\begin{array}{l}{\tan ^2}\left( {{{30}^0}} \right) - {\cot ^2}\left( {{{240}^0}} \right)\\ = {\left( {\frac{1}{{\sqrt 3 }}} \right)^2} - \cot \left( {{{60}^0}} \right)\\ = \frac{1}{3} - \frac{{\sqrt 3 }}{3} = \frac{{1 - \sqrt 3 }}{3}\end{array}\)

c)

\(\begin{array}{l}{\sin ^3}\frac{\pi }{2} - \cos 5\pi \\ = {1^3} - \cos \left( {\pi  + 4\pi } \right)\\ = 1 - \cos \pi \\ = 1 - \left( { - 1} \right) = 2\end{array}\)

d)

\(\begin{array}{l}\tan \frac{{11\pi }}{3} - \cot \left( { - \frac{{21\pi }}{4}} \right)\\ = \tan \left( {\frac{2}{3}\pi  + 3\pi } \right) - \cot \left( { - \frac{\pi }{4} - 5\pi } \right)\\ = \tan \left( {\frac{{2\pi }}{3}} \right) - \cot \left( { - \frac{\pi }{4}} \right)\\ =  - \sqrt 3  + \cot \left( {\frac{\pi }{4}} \right)\\ =  - \sqrt 3  + 1\end{array}\)