第501题

参考答案：\(\because \left ( {\sin {x-\cos {x}}} \right )^{2}=1-2\cos {x}\sin {x}=\frac {49} {25}\)，

\(\therefore \)\(\sin x - \cos x = \pm \frac{7}{5}\)

\(\because x\)为第四象限角， \(\sin x < 0\) ，\(\cos x > 0\)

\(\therefore \sin {x}-\cos {x}<0，\therefore \sin {x}-\cos {x}=-\frac {7} {5}\)，联立\(\cos {x}+\sin {x}=\frac {1} {5}\)得：\(\left \{ \begin{array}{*{20}{l}} {\sin {x}=-\frac {3} {5}} \\ {\cos {x}=\frac {4} {5}} \end{array} \right .\)

\(\therefore \)\(\tan x = \frac{{\sin x}}{{\cos x}} = - \frac{3}{4}\)．

解析：

第502题

参考答案：1

第503题

参考答案：\( - 1\)

第504题

A.\(\sin {α}\)

B.\(-\frac {1} {\cos {α}}\)

C.\(-\frac {1} {\sin {α}}\)

D.\(-\cos {α}\)

参考答案：D

第505题

A.\(-2\)

B.2

C.\(-\frac {3} {2}\)

D.\(\frac {3} {2}\)

参考答案：D

第506题

参考答案：由题意得，\(f(\alpha ) = \frac{{\cos \alpha \cdot ( - \cos \alpha ) \cdot ( - \sin \alpha )}}{{ - \cos \alpha \cdot \cos \alpha \cdot ( - \tan \alpha )}} = \cos \alpha \)；

解析：

第507题

参考答案：若\(f(\frac{\pi }{3} - \alpha ) = \frac{1}{3}\)，则\(\cos (\frac{\pi }{3} - \alpha ) = \frac{1}{3}\)，

\(\therefore \)\({\cos^{2}}(\frac {\pi } {6}+\alpha )={\cos^{2}}\left [ {\frac {\pi } {2}-(\frac {\pi } {3}-\alpha )} \right ]={\sin^{2}}(\frac {\pi } {3}-\alpha )=1-{\cos^{2}}(\frac {\pi } {3}-\alpha )=1-\frac {1} {9}=\frac {8} {9}\)，

\(\cos {(}\frac {2\pi } {3}+\alpha )=\cos {\left [ {\pi -(\frac {\pi } {3}-\alpha )} \right ]}=-\cos {(}\frac {\pi } {3}-\alpha )=-\frac {1} {3}\)，

则 \({\cos ^2}(\frac{\pi }{6} + \alpha ) + \cos (\frac{{2\pi }}{3} + \alpha ) = \frac{8}{9} - \frac{1}{3} = \frac{5}{9}\)．

解析：

第508题

参考答案：\(\because f\left ( {α} \right )=\frac {\cos {\left ( {\frac {\pi } {2}+α} \right )\sin {\left ( {α-\frac {\pi } {2}} \right )}}} {\cos {\left ( {-\pi -α} \right )\tan {\left ( {\pi -α} \right )}}}=\frac {\left ( {-\sin {α}} \right )\left ( {-\cos {α}} \right )} {\left ( {-\cos {α}} \right )\left ( {-\tan {α}} \right )}=\cos {α}\)

\(\therefore f\left ( {\frac {2017\pi } {3}} \right )=\cos {\left ( {672\pi +\frac {\pi } {3}} \right )}=\cos {\frac {\pi } {3}}=\frac {1} {2}\)

第509题

若\(\tan {θ=2}\)，求\(\frac {\cos {\left ( {2\pi -θ} \right )}} {\cos {\left ( {\pi +θ} \right )\sin {\left ( {\frac {\pi } {2}+θ} \right )-\sin {\left ( {\frac {3\pi } {2}+θ} \right )}}}}+\frac {\cos {\left ( {\pi -θ} \right )}} {\cos {θ\left [ {\sin {\left ( {\frac {3\pi } {2}-θ} \right )-1}} \right ]}}\)的值．

参考答案：\(\frac {\cos {\left ( {2\pi -θ} \right )}} {\cos {\left ( {\pi +θ} \right )\sin {\left ( {\frac {\pi } {2}+θ} \right )-\sin {\left ( {\frac {3\pi } {2}+θ} \right )}}}}+\frac {\cos {\left ( {\pi -θ} \right )}} {\cos {θ\left [ {\sin {\left ( {\frac {3\pi } {2}-θ} \right )-1}} \right ]}}\)

\(=\frac {\cos {θ}} {\left ( {-\cos {θ}} \right )\cos {θ-\left ( {-\cos {θ}} \right )}}+\frac {-\cos {θ}} {\cos {θ\left ( {-\cos {θ-1}} \right )}}=\frac {1} {1-\cos {θ}}+\frac {1} {1+\cos {θ}}=\frac {2} {\sin^{2} {θ}}=2+\frac {2} {\tan^{2} {θ}}=2+\frac {2} {{2}^{2}}=\frac {5} {2}\)

第510题

参考答案：2

第511题

参考答案：6

第512题

参考答案：\(\because \cos {α=-\frac {4} {5}}\)，且\(\tan \alpha > 0\)，则\(\alpha \)为第三象限角，

\(\therefore \sin \alpha = - \sqrt {1 - co{s^2}\alpha } = - \sqrt {1 - {{( - \frac{4}{5})}^2}} = - \frac{3}{5}\)，\(\therefore \tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }} = \frac{3}{4}\)；

解析：

第513题

参考答案：\(\frac{{2\sin (\pi - \alpha ) + \sin (\frac{\pi }{2} - \alpha )}}{{\cos (2\pi - \alpha ) + \cos ( - \alpha )}} = \frac{{2\sin \alpha + \cos \alpha }}{{\cos \alpha + \cos \alpha }} = \frac{{2\sin \alpha + \cos \alpha }}{{2\cos \alpha }}\) \( = \frac{{2\tan \alpha + 1}}{2} = \frac{{2 \times \frac{3}{4} + 1}}{2} = \frac{5}{4}\)

解析：

第514题

A.3

B.\(-3\)

C.\(\frac {5} {3}\)

D.\(-1\)

参考答案：A

第515题

参考答案：\(f(\theta )=\frac {\sin {θ}} {\sin {θ}\cdot \left ( {-\sin {θ}} \right )\cdot \left ( {-\cos {θ}} \right )\cdot \left ( {-\frac {\sin {θ}} {\cos {θ}}} \right )}=-\frac {1} {\sin^{2} {θ}}\)．

解析：

第516题

参考答案：由\(f(\theta ) = - 3\)，有\(\sin^{2} {θ}=\frac {1} {3}\)，得\(3\sin^{2} {θ}=\sin^{2} {θ}+\cos^{2} {θ}\)，\(\tan^{2} {θ}=\frac {1} {2}\)

可得\(\tan \theta = \pm \frac{{\sqrt 2 }}{2}\)．

解析：

第517题

参考答案：\(\because \) 角\(\alpha \)终边上有一点，

又\(\sin \alpha = \frac{{\sqrt 2 }}{4}m = \frac{m}{{\sqrt {3 + {m^2}} }}\)，所以\(m = \pm \sqrt 5 \)．

当\(m = \sqrt 5 \)时，\(\cos \alpha = \frac{{ - \sqrt 3 }}{{\sqrt {3 + {m^2}} }} = - \frac{{\sqrt 6 }}{4}\)，\(\tan \alpha = \frac{m}{{ - \sqrt 3 }} = - \frac{{\sqrt {15} }}{3}\)；

当\(m = - \sqrt 5 \)时，\(\cos \alpha = \frac{{ - \sqrt 3 }}{{\sqrt {3 + {m^2}} }} = - \frac{{\sqrt 6 }}{4}\)，\(\tan \alpha = \frac{m}{{ - \sqrt 3 }} = \frac{{\sqrt {15} }}{3}\)．

解析：

第518题

参考答案：\(\frac{{\cos (\pi + \alpha )\cos (\frac{\pi }{2} + \alpha )\cos (\frac{{11\pi }}{2} - \alpha )}}{{\cos (\pi - \alpha )\sin ( - \pi - \alpha )\sin (\frac{{9\pi }}{2} + \alpha )}} = \frac{{ - \cos \alpha \cdot ( - \sin \alpha ) \cdot ( - \sin \alpha )}}{{ - \cos \alpha \cdot \sin \alpha \cdot \cos \alpha }} = \tan \alpha = \pm \frac{{\sqrt {15} }}{3}\)．

解析：

第519题

参考答案：\(f(\alpha ) = \frac{{\sin (\pi - \alpha )\cos (2\pi - \alpha )\sin ( - \alpha + \frac{{3\pi }}{2})}}{{\cos ( - \alpha - \pi )\cos ( - \alpha + \frac{{7\pi }}{2})}} = \frac{{\sin \alpha \cos \alpha ( - \cos \alpha )}}{{ - \cos \alpha ( - \sin \alpha )}} = - \cos \alpha \)

解析：

第520题

参考答案：由\(\cos (\alpha - \frac{{3\pi }}{2}) = \frac{1}{5}\)，得\( - \sin \alpha = \frac{1}{5}\)，\(\therefore \sin \alpha = - \frac{1}{5}\)，

又\(\alpha \)是第三象限角，\(\cos \alpha = - \sqrt {1 - si{n^2}\alpha } = - \sqrt {1 - {{( - \frac{1}{5})}^2}} = - \frac{{2\sqrt 6 }}{5}\)，

\(\therefore f(\alpha ) = \frac{{2\sqrt 6 }}{5}\)．

解析：