Bài 3 trang 15 SGK Toán 11 tập 1 - Cánh diều

Đề bài

Tính các giá trị lượng giác (nếu có) có mỗi góc sau:

a)     \(\frac{\pi }{3} + k2\pi \,\,\left( {k \in Z} \right)\)

b)     \(k\pi \,\,\left( {k \in Z} \right)\)

c)     \(\frac{\pi }{2} + k\pi \,\,(k \in Z)\)

d)     \(\frac{\pi }{4} + k\pi \,\,\left( {k \in Z} \right)\)

Lời giải chi tiết

a)

\(\begin{array}{l}\cos \left( {\frac{\pi }{3} + k2\pi \,} \right) = \cos \left( {\frac{\pi }{3}} \right) = \frac{1}{2}\\\sin \left( {\frac{\pi }{3} + k2\pi \,} \right) = \sin \left( {\frac{\pi }{3}} \right) = \frac{{\sqrt 3 }}{2}\\\tan \left( {\frac{\pi }{3} + k2\pi \,} \right) = \frac{{\sin \left( {\frac{\pi }{3} + k2\pi \,\,} \right)}}{{\cos \left( {\frac{\pi }{3} + k2\pi \,\,} \right)}} = \sqrt 3 \\\cot \left( {\frac{\pi }{3} + k2\pi \,\,} \right) = \frac{1}{{\tan \left( {\frac{\pi }{3} + k2\pi \,\,} \right)}} = \frac{{\sqrt 3 }}{3}\end{array}\)

b)

 \(\begin{array}{l}\cos \left( {k\pi \,} \right) = \left[ \begin{array}{l} - 1\,\,\,\,\,\,\,\,\,\,\,\,\,;k = 2n + 1\\1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,;k = 2n\,\,\,\end{array} \right.\\\sin \left( {k\pi \,} \right) = 0\\\tan \left( {k\pi \,} \right) = \frac{{\sin \left( {k\pi \,\,} \right)}}{{\cos \left( {k\pi \,\,} \right)}} = 0\\\cot \left( {k\pi \,\,} \right)\end{array}\)

c)

\(\begin{array}{l}\cos \left( {\frac{\pi }{2} + k\pi \,} \right) = 0\\\sin \left( {\frac{\pi }{2} + k\pi \,} \right) = \left[ \begin{array}{l}\sin \left( { - \frac{\pi }{2}} \right)\, =  - 1\,\,\,\,\,\,\,;k = 2n + 1\\\sin \left( {\frac{\pi }{2}\,} \right)\, = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,;k = 2n\,\,\,\end{array} \right.\\\tan \left( {\frac{\pi }{2} + k\pi \,} \right)\\\cot \left( {\frac{\pi }{2} + k\pi \,\,} \right) = 0\end{array}\)

d)

Với k=2n+1 thì

\(\begin{array}{l}\cos \left( {\frac{\pi }{4} + k\pi \,} \right) = \cos \left( {\frac{\pi }{4} + (2n + 1)\pi \,} \right) = \cos \left( {\frac{\pi }{4} + 2n\pi  + \pi \,} \right) = \cos \left( {\frac{\pi }{4} + \pi \,} \right) =  - \cos \left( {\frac{\pi }{4}} \right) =  - \frac{{\sqrt 2 }}{2}\\\sin \left( {\frac{\pi }{4} + k\pi \,} \right) = \sin \left( {\frac{\pi }{4} + (2n + 1)\pi \,} \right) = \sin \left( {\frac{\pi }{4} + 2n\pi  + \pi \,} \right) = sin\left( {\frac{\pi }{4} + \pi \,} \right) =  - \sin \left( {\frac{\pi }{4}} \right) =  - \frac{{\sqrt 2 }}{2}\\\tan \left( {\frac{\pi }{4} + k\pi \,} \right) = 1\\\cot \left( {\frac{\pi }{4} + k\pi \,\,} \right) = 1\end{array}\)

Với k=2n thì

\(\begin{array}{l}\cos \left( {\frac{\pi }{4} + k\pi \,} \right) = co{\mathop{\rm s}\nolimits} \left( {\frac{\pi }{4} + 2n\pi \,} \right) = \cos \left( {\frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2}\\\sin \left( {\frac{\pi }{4} + k\pi \,} \right) = \sin \left( {\frac{\pi }{4} + 2n\pi \,} \right) = \sin \left( {\frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2}\\\tan \left( {\frac{\pi }{4} + k\pi \,} \right) = 1\\\cot \left( {\frac{\pi }{4} + k\pi \,\,} \right) = 1\end{array}\)