(2)升幂公式:1+cos\alpha =2cos^2\frac{\alpha }{2},1-cos\alpha =2sin^2\frac{\alpha }{2}。综上所述,本题答案为(1)\frac{1-cos2\alpha }{2};\frac{1+cos2\alpha }{2},(2)2cos^2\frac{\alpha }{2};2sin^2\frac{\alpha }{2}。
sin^2α +cos^2α =1的变形公式:$sin^2\alpha =1-cos^2\alpha =(1+cos\alpha )(1-cos\alpha )⟹ (si
求助三道高一数学三角函数题1、用tan(alpha)表示tan(alpha/2)2、化简:sin(50)(1+根号3tan(10))(角度)3、求证:(1-tan^2(alpha/2))/(1+tan^2(alpha/2))=cos(alpha)详细过程,谢谢
2sin(2α)=cos(2α)+1 解α (復數求解) α=22πn1+arctan(34),n1∈Z α=πn2+2π,n2∈Z 解α α=22πn1+arcsin(54),n1∈Z α=πn2+2π,n2∈Z
点2 (1) $$ \frac { 1 + \cos 2 \alpha } { 2 } $$ $$ \frac { 1 - \cos 2 \alpha } { 2 } $$ (2)2$$ 2 \cos ^ { 2 } a $$ 2.$$ \sin ^ { 2 } $$a (3)t(3)tan(α±β)(1千tan atanβ)$$ \frac { \tan \alpha - \tan \beta } { \tan (...
1.倍角公式(1)$$ S _ { 2 \alpha } : \sin 2 \alpha = \_ ; $$(2)$$ _ { 2 \alpha } : \cos 2 \alpha = \cos ^ { 2 } \alpha - \sin ^ { 2 } \alpha = 2 \cos ^ { 2 } \alpha - 1 = \_ $$___;(3)$$ T _ { 2 \alpha } : \tan 2 \alp...
4.C 因为$$ 2 \sin 2 \alpha = 1 + \cos 2 \alpha $$,所以$$ 2 \sin 2 \alpha = 2 \cos ^ { 2 } \alpha $$ 所以$$ 2 \cos \alpha ( 2 \sin \alpha - \cos \alpha ) = 0 , $$ 解得$$ \cos \alpha = 0 $$或$$ \tan \alpha = \frac { 1 } { 2 }...
【题目】二倍角的正弦、余弦和正切公式:(1)$$ \sin 2 \alpha = \_ $$(2)$$ 2 ) \cos 2 \alpha = \_ = \_ = \_ ; $$$ ( \cos ^ { 2 } \alpha = \_ , \sin ^ { 2 } \alpha = \_ ) . $$(3)$$ \tan 2 \alpha = \_ . $$ 相关...
知识点二 二倍角的正弦、余弦、正切公式(1)$$ \sin 2 \alpha = \_ ; $$(2)$$ \cos 2 \alpha = \_ = \_ - 1 = 1 - \_ $$(3)$$ ) \tan 2 \alpha = ( \alpha \neq \frac { k \pi } { 2 } + \frac { \pi } { 4 } ( \alpha \neq \frac { k \pi } { 2 }...
正切公式(1)$$ \sin 2 \alpha = \_ ; $$(2)$$ ) \cos 2 \alpha = \_ = \_ - 1 = 1 - \_ $$(3)$$ \tan 2 \alpha = ( \alpha \neq \frac { k \pi } { 2 } + \frac { \pi } { 4 } $$且$$ \alpha \neq k \pi + \frac { \pi } { 2 } , k \in Z ...