优优班--学霸训练营 > 知识点挑题
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            • 1.

              给出下列等式:\(①\arcsin \dfrac{\pi }{2}=1\);\(②\arcsin \left( -\dfrac{1}{2} \right)=-\dfrac{\pi }{6}\);\(③\arcsin \left( \sin \dfrac{\pi }{3} \right)=\dfrac{\pi }{3}\);\(④\sin \left( arc\sin \dfrac{1}{2} \right)=\dfrac{1}{2}\),其中正确等式的个数是\((\)  \()\)

              A.\(1\)         
              B.\(2\)        
              C.\(3\)       
              D.\(4\)
            • 2.

              若角\(α\)是三角形的一个内角,且\(\sin α= \dfrac{1}{3}\),则\(α\)等于\((\)  \()\)

              A.\(π-\arccos \dfrac{2 \sqrt{2}}{3}\)
              B.\(\arcsin \dfrac{1}{3}\)

              C.\(\arcsin \dfrac{1}{3}\)或\(π-\arcsin \dfrac{1}{3}\)
              D.\(\arccos \dfrac{2 \sqrt{2}}{3}\)或\(π-\arccos \dfrac{2 \sqrt{2}}{3}\)
            • 3.

              已知\(0 < α < \)\(\dfrac{\pi }{{2}}\) \( < β < π\),\(\tan \dfrac{α}{2}= \dfrac{1}{2},\cos (β-α)= \dfrac{ \sqrt{2}}{10} \)

              \((1)\)求\(\dfrac{\cos 2\alpha }{\sqrt{2}\cos (\dfrac{\pi }{4}+\alpha )\sin \alpha }\) 的值

              \((2)\)求\(β\)的值

            • 4.

              已知方程\(6{{x}^{2}}+5x+1=0\)的两实根为\(mn\),则\(\arctan m+\arctan n=\) ________

            • 5.
              \(\arcsin (- \dfrac {1}{2})+\arccos (- \dfrac { \sqrt {3}}{2})+\arctan (- \sqrt {3})=\) ______ .
            • 6.
              设\(x_{1}\),\(x_{2}\)是方程\(x^{2}-x\sin \dfrac {3π}{5}+\cos \dfrac {3π}{5}=0\)的两个根,则\(\arctan x_{1}+\arctan x_{2}\)的值为 ______
            • 7.
              若\(\sin x=- \dfrac {3}{5}(π < x < \dfrac {3}{2}π)\),则\(x=(\)  \()\)
              A.\(\arcsin (- \dfrac {3}{5})\)
              B.\(π+\arcsin \dfrac {3}{5}\)
              C.\(2π-\arcsin \dfrac {3}{5}\)
              D.\(π-\arcsin \dfrac {3}{5}\)
            • 8.
              已知 \(\sin θ\)\(+\) \(\cos θ\)\(=\)\(θ\)\(∈\),则 \(θ\)的值等于\((\)    \()\)
              A.      
              B.   
              C.     
              D. 
            • 9.
              在\(\triangle ABC\)中,若\(4π\sin A-3\arccos (- \dfrac {1}{2})=0\),则\(A=\) ______ .
            • 10.
              求值:\( \begin{vmatrix} \arcsin \dfrac { \sqrt {3}}{2} & 2 \\ \arctan \dfrac { \sqrt {3}}{3} & 3\end{vmatrix} =\) ______ 弧度.
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