{1 - \tan \alpha \ast \tan \beta }③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:\tan 105^{{\circ} }=\tan (45^{{\circ} }+ 60^{{\circ} }) = \dfrac{\tan 45^{{\circ} } + \tan 60^{{\circ} }}{1 - \tan 45^{{\circ} }\ast \tan 60...
{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 - \ta...
5andsin(α−β)=513, whereαlie between 0 andπ4, then find that value oftan2α. (α+β)=45,sin(α−β)=5130αβπ4thentan2= ……… If cos(alpha+beta)=4/5; sin(alpha-beta)=5/13 and alpha,beta lie between 0π/4 then find the value of tan2alpha Ifcos(α+β)=45li...
1.$$ . \frac { 1 } { 2 } \left[ \cos ( \alpha + \beta ) + \cos ( \alpha - \beta ) \right] $$ $$ 2 . - \frac { 1 } { 2 } \left[ \cos ( \alpha + \beta ) - \cos ( \alpha - \beta ) \right] $$ 3.$$ . \frac { 1 } { 2 } \left[ \...
关于三角函数有如下公式:$$ \sin ( \alpha + \beta ) = \sin \alpha \cos \beta + \cos \alpha \sin \beta $$,$$ \sin ( \alpha - \beta ) = \sin \alpha \cos \beta - \cos \alpha \sin \beta ; $$$ \cos ( \alpha + \beta ) = \cos \alpha \cos \beta - \sin \alpha...
{ 1 + \tan \alpha \tan \beta } $$ 2.二倍角公式: (1)$$ \sin 2 \alpha = 2 \sin \alpha \cos \alpha $$ (2)$$ \cos 2 \alpha = 2 \cos ^ { 2 } \alpha - 1 = 1 - 2 \sin ^ { 2 } \alpha = \cos ^ { 2 } \alpha - \sin ^ { 2 } a $$ (3)...
【解析】$$ \sin \alpha \sin \beta = 1 , $$ ∴$$\left\{ \begin{matrix} \sin \alpha = 1 \\ \sin \beta = 1 \end{matrix} \right.$$或$$\left\{ \begin{matrix} \sin \alpha = - 1 \\ \sin \beta = - 1 \end{matrix} \right.$$, ∴$$ \cos \alpha \cos \beta...
要点3 三角变换公式 (1)$$ \sin ( \alpha + \beta ) = \sin \alpha \cos \beta + \cos \alpha \sin \beta ; $$ (2)$$ \sin ( \alpha - \beta ) = \sin \alpha \cos \beta - \cos \alpha \sin \beta $$ (3)$$ \sin 2 \alpha = 2 \sin \alpha \cos \alpha $$...
百度试题 结果1 题目【题目】已知$$ \sin \alpha \sin \beta = 1 $$,那么$$ \cos ( \alpha + \beta ) $$的值等于[ ] A. 1 B.-1 C.0 D. ±1 相关知识点: 试题来源: 解析 【解析】B 反馈 收藏
百度试题 结果1 题目(1)$$ \sin ( \alpha \pm \beta ) = \_ $$ 相关知识点: 试题来源: 解析 $$ \sin \alpha \cos \beta = \cos \alpha \sin \beta $$ 反馈 收藏