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Câu hỏi: 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) \(\frac{\pi }{4} + k\pi \left( {k \in Z} \right)\)
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) \(\frac{\pi }{4} + k\pi \left( {k \in Z} \right)\)
Phương pháp giải
Giá trị lượng giác của các góc đặc biệt
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
\(\begin{array}{l}\cos \left( {\frac{\pi }{4} + k\pi } \right) = co{\mathop{\rm s}\nolimits} \left( {\frac{\pi }{4} + 2n\pi } \right) = \cos \left( {\frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2}\\\sin \left( {\frac{\pi }{4} + k\pi } \right) = \sin \left( {\frac{\pi }{4} + 2n\pi } \right) = \sin \left( {\frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2}\\\tan \left( {\frac{\pi }{4} + k\pi } \right) = 1\\\cot \left( {\frac{\pi }{4} + k\pi } \right) = 1\end{array}\)
Giá trị lượng giác của các góc đặc biệt
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
\(\begin{array}{l}\cos \left( {\frac{\pi }{4} + k\pi } \right) = co{\mathop{\rm s}\nolimits} \left( {\frac{\pi }{4} + 2n\pi } \right) = \cos \left( {\frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2}\\\sin \left( {\frac{\pi }{4} + k\pi } \right) = \sin \left( {\frac{\pi }{4} + 2n\pi } \right) = \sin \left( {\frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2}\\\tan \left( {\frac{\pi }{4} + k\pi } \right) = 1\\\cot \left( {\frac{\pi }{4} + k\pi } \right) = 1\end{array}\)