【解析】【答案】 $$ \frac { 3 \sqrt { 5 } - 2 \sqrt { 7 } } { 1 2 } $$ 【解答过程】 由$$ \sin \alpha = \frac { 2 } { 3 } $$,$$ \cos \beta = - \frac { 3 } { 4 } $$,$$ \alpha \in ( \frac { \pi } { 2 } , \pi ) $$,$$ \beta \...
【解析】 由题意可得:$$ \cos \alpha = - \sqrt { 1 - \frac { 4 } { 9 } } = - \frac { \sqrt { 5 } } { 3 } $$, $$ \sin \beta = - \sqrt { 1 - \frac { 9 } { 1 6 } } = - \frac { \sqrt { 7 } } { 4 } $$ ∴$$ \sin ( \alpha + \...
tan alpha and tan beta are roots of the equation x^(2)+ax+b=0, then the value of sin^(2)(alpha+beta)+a sin(alpha+beta)cos(alpha+beta)+b cos^(2)(alpha+beta) is equal to View Solution tan alpha and tan beta are roots of the equation x^(2)+ax+b=0, then the value of s...
tan alpha and tan beta are roots of the equation x^(2)+ax+b=0, then the value of sin^(2)(alpha+beta)+a sin(alpha+beta)cos(alpha+beta)+b cos^(2)(alpha+beta) is equal to View Solution tan alpha and tan beta are roots of the equation x^(2)+ax+b=0, then the value of s...
【解析】 由已知得$$ \cos \alpha - \cos \beta = \frac { 1 } { 2 } $$,① $$ \sin \alpha - \sin \beta = - \frac { 1 } { 3 } $$.②由①$$ 2 + \textcircled { 2 } 2 $$得$$ ( \cos \alpha - \cos \beta ) ^ { 2 } + ( \sin \alpha - \sin $$...
A8DB关于三角函数有如下的公式: sin(a B)sin a cos B cos a sin Bi cos(a B)cos a cos B-sin a sin B2 tan a tan Btan(a B)31-tana·tan3 利用这些公式可将某些不是特殊角的三角函数转化为特殊角的三角函数来求值,如: tan45°+tan60°1+√3(1+√③)(1+√3)tan105°=tan(45°+60...
知识点 三个公式(1)两角和的余弦公式$$ \cos ( \alpha + \beta ) = \cos \alpha \cos \beta - \sin \alpha
{1 + \tan \alpha \tan \beta }(1+ \tan \alpha \tan \beta \neq 0)利用这些公式可以将一些不是特殊角的三角函数转化为特殊角的三角函数来求值.如:\tan 105^{{\circ} }= \tan (45^{{\circ} }+ 60^{{\circ} })= \dfrac{\tan 45^{{\circ} } + \tan 60^{{\circ} }}{1 - \...
由题意得:$$\left\{ \begin{matrix} \cos \beta + \sin \alpha = 1 . 1 1 7 6 \\ \cos \beta - \sin \alpha = 0 . 0 5 8 0 \end{matrix} \right.$$, 解得,$$ \cos \beta = 0 . 5 8 7 8 , \sin \alpha = 0 . 5 2 9 8 $$, ∵α、β都是锐角, ∴$$ \alpha...
3.已知$$ \sin ( \beta - \alpha ) \cos \beta - \cos ( \alpha - \beta ) \sin \beta = \frac { 3 } { 5 } $$,α为第三象限角,则$$ \cos ( \alpha + \frac { \pi } { 4 } ) = $$ (A) A.-$$ \frac { \sqrt { 2 } } { 1 0 } $$ B.-$$ \frac { 7...