3.提示不正确.例如:当$$ \alpha = \frac { \pi } { 2 } , \beta = \frac { \pi } { 4 } $$时, $$ \cos ( \alpha - \beta ) = \cos \frac { \pi } { 4 } = \frac { \sqrt { 2 } } { 2 } , $$ 而$$ \cos \alpha - \cos \beta = \cos \frac { \p...
百度试题 结果1 题目【题目】和差化积公式$$ \cos \alpha - \cos \beta = \_ $$ 相关知识点: 试题来源: 解析 【解析】 $$ - 2 \sin \frac { \alpha + \beta } { 2 } \sin \frac { \alpha - \beta } { 2 } $$ 反馈 收藏 ...
যদি cos(alpha-beta)+1=0 হয়, তবে দেখাও যে, cosalpha+cosbeta=0 এবং sinalpha+sinbeta=0।
For all value ofα,β,γprove that ,cosα+cosβ+cosγ+cos(α+β+γ)=4cos(α+β)2⋅cos(β+γ)2⋅cos(γ+α)2 View Solution If alpha,beta and gamma are such that alpha+beta+gamma=0, then |(1,cos gamma,cosbeta),(cosgamma,1,cos alpha),(cosbeta,cos alpha,1)| View Solut...
答案 一般情况下不相等.例如,当$$ \alpha = 6 0 ^ { \circ } \beta = $$ $$ 3 0 ^ { \circ } $$时,$$ \cos ( 6 0 ^ { \circ } - 3 0 ^ { \circ } ) = \cos 3 0 ^ { \circ } = \frac { \sqrt { 3 } } { 2 } , \cos 6 0 ^ { \circ } - \co...
解析 [思考]提示:一般情况下不相等.但在特殊 情况下也有相等的时候.例如:当$$ \alpha = 0 ^ { \circ } $$, $$ \beta = 6 0 ^ { \circ } $$时,$$ \cos ( 0 ^ { \circ } - 6 0 ^ { \circ } ) = \cos 0 ^ { \circ } - \cos 6 0 ^ { \circ } $$. ...
} { 2 } \cos \frac { \alpha x - \beta x } { 2 } + \sin \frac { \alpha x + \beta x } { 2 } \sin \frac { \alpha x - \beta x } { 2 } \right] \\ = - 2 \sin \frac { \alpha x + \beta x } { 2 } \sin \frac { \alpha x - \beta x } { 2 ...
【解析】取$$ \alpha = 3 0 ^ { \circ } , \beta = 6 0 ^ { \circ } $$,则 $$ \cos ( \alpha - \beta ) = \cos ( 3 0 ^ { \circ } - 6 0 ^ { \circ } ) = \frac { \sqrt { 3 } } { 2 } , $$ 但$$ \cos \alpha - \cos \beta = \frac { \sqrt ...
【解析】 由已知得$$ \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 $$ ...
答案见上1.(1)cos acosβ+sin asinβ 2)cos acosβ-sin asinβ 3) sin acosβ-cos asinβ (4)sin acosβ+cos asinβ (5)$$ \frac { \tan \alpha - \tan \beta } { 1 + \tan \alpha \tan \beta } $$ (6) $$ \frac { \tan \alpha + \tan \beta } { 1 - \tan \...