1.提示:①$$ \sin ( \alpha + \beta ) = \cos \left[ \frac { \pi } { 2 } - ( \alpha + \beta ) \right] = $$ $$ \cos \left[ ( \frac { \pi } { 2 } - \alpha ) - \beta \right] = \cos ( \frac { \pi } { 2 } - \alpha ) \cos
beta ) \right] = \cos \left[ ( \frac { \pi } { 2 } - \alpha ) - \beta \right] \\ = \cos ( \frac { \pi } { 2 } - \alpha ) \cos \beta + \sin ( \frac { \pi } { 2 } - \alpha ) \sin \beta \\ = \sin \alpha \cos \beta + \cos \alpha \sin \be...
【解析】 从$$ C _ { ( \alpha \pm \beta ) } $$,出发,结合诱导公式进行推导: $$ \sin ( \alpha + \beta ) = \cos \left[ \frac { \pi } { 2 } - ( \alpha + \beta ) \right] = \cos \left[ ( \frac { \pi } { 2 } - \alpha ) - \beta \right] $$ $$ = \cos ...
能,推导见分析提示:可以.$$ \sin ( \alpha + \beta ) $$ $$ = \cos \left[ \frac { \pi } { 2 } - ( \alpha + \beta ) \right] \\ = \cos \left[ ( \frac { \pi } { 2 } - \alpha ) - \beta \right] \\ = \cos ( \frac { \pi } { 2 } - \alpha ) \cos...
根据两角和与差的余弦公式和诱导公式,能否推导出两角和与差的正弦公式?如何推导?[即如何用 a、β的三角函数来表示$$ \sin ( \alpha + \beta ) ,
$$(2)两角差的正弦公式的推导在公式 $$ S _ { \alpha + \beta } $$中用-β代替β可以得到$$ \sin ( \alpha - \beta ) $$$ = \sin \alpha \cos ( - \beta ) + \cos \alpha \sin ( - \beta ) = \sin \alpha \cos \beta - \cos \alpha \sin \beta $$即$$ \sin ( \...
$$ \sin ( \alpha - \beta ) = \sin \left[ \alpha + ( - \beta ) \right] \\ = \sin a \cos ( - \beta ) + \cos a \sin ( - \beta ) \\ = \sin \alpha \cos \beta - \cos \alpha \sin \beta . $$ 反馈 收藏
确定了两角差的余弦公式后推导的,推导过程如下:$$ \sin ( \alpha + \beta ) = \underline { } \\ = \cos ( \frac { \pi } { 2 } - \alpha ) \cos \beta + \sin ( \frac { \pi } { 2 } - \alpha ) \sin \beta \\ = \sin \alpha \cos \beta + \cos \alpha \sin \beta ...
【题目】公式推导:(1)$$ \sin \alpha \cos \beta = \frac { 1 } { 2 } \left[ \sin ( \alpha + \beta ) + \sin ( \alpha - \beta ) \right] $$(2)$$ \sin \theta + \sin \varphi = 2 \sin \frac { \theta + \varphi } { 2 } \cos \frac { \theta - \varphi ...
在上述公式中以-β代替β,结合诱导公式,得$$ \sin \left[ \alpha + ( - \beta ) \right] = $$ $$ \sin \alpha \cos ( - \beta ) + \cos \alpha \sin ( - \beta ) = \sin \alpha \cos \beta - \cos \alpha \sin \beta $$ 即$$ \sin ( \alpha - \beta ) = \sin \alph...