所以:sin\alpha =\dfrac{-1\pm \root \of {7} }{4},cos\alpha =\dfrac{1\pm \root \of {7} }{4}所以sin(\alpha +\beta )=sin\alpha cos\beta +sin\beta cos\alpha 代入上式得sin(\alpha +\beta )=1
已知\( \sin\alpha = \frac{1}{2} \),\( \cos\beta = -\frac{1}{2} \),且 \( \alpha \) 和 \( \beta \) 都在第二象限,则 \( \tan(\alpha - \beta) \) 的值为___。相关知识点: 试题来源: 解析 \( \frac{24}{25} \) ...
已知\( \sin \alpha = \frac{1}{2} \),\( \cos \beta = \frac{\sqrt{3}}{2} \),则 \( \sin (\alpha + \beta) \) 的值为: A. \( \frac{1}{2} \) B. \( \frac{\sqrt{3}}{2} \) C. \( \frac{1}{4} \) D. \( -\frac{1}{4} \) ...
已知\( \sin(\alpha) = \frac{1}{2} \),\( \cos(\beta) = -\frac{\sqrt{3}}{2} \),且\( \alpha, \beta \)均为第二象限角,则\( \cos(\alpha - \beta) \)的值为( ) A. \( \frac{1}{4} \) B. \( -\frac{1}{4} \) C. \( \frac{7}{8} \) D. \(...
已知\( \sin\alpha = \frac{1}{2} \),\( \cos\beta = -\frac{1}{2} \),且 \( \alpha \) 和 \( \beta \) 都在第二象限,则 \( \sin(\alpha + \beta) \) 的值为: A. 地持支器传向准通今热别八地持支器传向准通今热别八\( \frac{\sqrt{3}}{2} \)地持支器传...
已知\sin (\alpha - \beta )=2\cos (\alpha \beta ), \tan (\alpha - \beta )=1\div 2,则\tan \alpha - \tan \beta = () A. 4\div 7\ \ B. 7\div 4\ \ C. 7\div 6\ \ D. 4\div 5\ \ 相关知识点: 试题来源: 解析 D ...
(1)已知0\lt \alpha \lt \beta \lt \dfrac{\pi }{2},\sin \alpha = \dfrac{3}{5},\cos (\alph
已知$\cos {\alpha }=-\dfrac {1} {2}$,$\sin {\beta }=-\dfrac {\sqrt {3}} {3}$,且$\alpha ∈
例2 已知角{\alpha}的终边过点P(\frac{1}{2},{\cos}^{2}{\theta}),角{\beta}的终边过点Q({\sin}^{2}{\theta
\because \dfrac{1+\sin \alpha -\cos \alpha }{\sin \alpha }\cdot \dfrac{1+\sin \beta -\cos \beta }{\sin \beta }=2,\therefore \left( 1+\tan \dfrac{\alpha }{2} \right)\left( 1+\tan \dfrac{\beta }{2} \right)=2,\therefore \tan \dfrac{\alpha }{2}\tan \dfrac{\...