15.已知$$ \sin \alpha + \cos \beta = 1 , \cos \alpha + \sin \beta = 0 , $$,则sin(α+$$ \beta ) = \_ . $$ 相关知识点: 试题来源: 解析 15答案 -$$ \frac { 1 } { 2 } $$ 考什么? 命题人考查三角函数的求值及恒等变形. 这么考$$ \sin \alpha + \cos...
15.已知$$ \sin \alpha - \cos \beta = 1 , \cos \alpha - \sin \beta = 0 $$,则$$ \cos ( \
答案见上例2(1)因为$$ \sin \alpha + \cos \beta = 1 \cdots \cdots $$①$$ \cos \alpha + \sin \beta $$ =0·...② 所以$$ \textcircled { 1 } ^ { 2 } + \textcircled { 2 } ^ { 2 } $$得$$ 1 + 2 ( \sin \alpha \cos \beta + \cos \alpha \sin \beta...
【题目】15.已知$$ \sin \alpha + \cos \beta = 1 \cos \alpha + \sin \beta = 0 $$,则$$ \sin
{ 2 } \beta + 2 \cos \alpha \sin \beta = 0 $$②, ①②两式相加可得$$ \sin ^ { 2 } \alpha + \cos ^ { 2 } \alpha + \sin ^ { 2 } \beta + \cos ^ { 2 } \beta + 2 ( \sin \alpha \cos \beta + $$ $$ \cos \alpha \sin \beta ) = 1 $$,∴$$ ...
{ 2 } \beta + 2 \sin \alpha \cos \beta $$ =1①, 将$$ \cos \alpha + \sin \beta = 0 $$两边平方得$$ \cos ^ { 2 } \alpha + \sin ^ { 2 } \beta + 2 \cos \alpha \sin \beta = $$ 0②, 难点,不能灵活应用$$ \sin ^ { 2 } \alpha + \cos ^ { 2 } ...
答案见上$$ - \frac { 1 } { 2 } $$解析 因为$$ \sin \alpha + \cos \beta = 1 $$,所以$$ \sin ^ { 2 } \alpha + \cos ^ { 2 } \beta + $$ $$ 2 \sin \alpha \cos \beta = 1 $$①,因为$$ \cos \alpha + \sin \beta = 0 $$,所以$$ \cos ^ { 2 } ...
答案见上8.解析因为$$ \sin \alpha + \cos \beta = 1 $$ ①,$$ \cos \alpha + \sin \beta = 0 $$ ②,所以 $$ \textcircled { 1 } ^ { 2 } + \textcircled { 2 } ^ { 2 } $$得$$ 1 + 2 ( \sin \alpha \cos \beta + \cos \alpha \sin \beta ) + 1 = 1 ...
答案见上1.-$$ \frac { 1 } { 2 } $$解析:∵$$ \sin \alpha + \cos \beta = 1 , $$,① $$ \cos \alpha + \sin \beta = 0 $$,②∴$$ \textcircled { 1 } ^ { 2 } + \textcircled { 2 } ^ { 2 } $$,得$$ 1 + 2 ( \sin \alpha \cos \beta + \cos \...
答案见上[(1)∵$$ \because \sin \alpha + \cos \beta = 1 , \cos \alpha + \sin \beta $$ =0,∴$$ \sin ^ { 2 } \alpha + \cos ^ { 2 } \beta + 2 \sin \alpha \cos \beta = 1 $$, ① $$ \cos ^ { 2 } \alpha + \sin ^ { 2 } \beta + 2 \cos \alpha...