【解析】 (1)因为$$ \tan \alpha = 2 = \frac { \sin \alpha } { \cos \alpha } > 0 $$, ∴α是第一或第三象限角. 当α是第一象限角时,结合$$ \sin ^ { 2 } \alpha + \cos ^ { 2 } \alpha = 1 $$,可得 $$\left\{ \begin{matrix} \sin \alpha = \frac { 2 \sqr...
答案见上6.AD解析:由题意知$$ \sin \frac { \alpha } { 2 } \cos \frac { \alpha } { 2 } = 1 + 2 \cos ^ { 2 } \frac { \alpha } { 2 } - 1 $$ 故有$$ 2 \sin \frac { \alpha } { 2 } \cos \frac { \alpha } { 2 } - \cos ^ { 2 } \frac { ...
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
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AD因为$$ 2 \sin \alpha = 1 + \cos \alpha $$,所以$$ 4 \sin \frac { a } { 2 } \cos \frac { a } { 2 } = $$ 幼$$ 2 \cos ^ { 2 } \frac { a } { 2 } $$0.当$$ \cos \frac { \alpha } { 2 } = 0 $$时,tan$$ \frac { \alpha } { 2 } $$...
【解析】∵$$ 2 \sin \alpha = 1 + \cos \alpha $$, ∴两边平方,整理可得:$$ 5 \cos ^ { 2 } \alpha + 2 \cos \alpha - 3 = 0 $$ ∴解得:$$ \cos \alpha = - 1 $$,或 $$ \frac { 3 } { 5 } $$, .当$$ \cos \alpha = 1 $$时,$$ \alpha = 2 k \pi...
(1)(sina+cosa)/(sina-cosa)=(tana+1)/(tana-1)=(2+1)/(2-1)=3综上,结论是:(sinα+cosα)/(sinα-cosα)=3(2)(2sin^2a-3cos^2a)/(sin^2a+4cos^2a)=(2tan^2α-3)/(tan^2α-4)=(2*4-3)/(4-4)=5/8综上,结论是:(2sin^2α-3cos^2α)/(sin^2α+4cos^2α)=5...
{ 2 } \alpha = 4 \sin \alpha , $$ ∴解得:$$ \sin \alpha = 0 $$,或 $$ \frac { 4 } { 5 } $$, ∴$$ \cos \alpha = - 1 $$,或 $$ \frac { 3 } { 5 } $$, ∴$$ \tan \alpha = \frac { \sin \alpha } { \cos \alpha } = 0 $$,或 $$ \frac {...
3.B$$ 2 \sin \alpha = 1 + \cos \alpha $$,即$$ 4 \sin \frac { \alpha } { 2 } \cos \frac { \alpha } { 2 } = 2 \cos ^ { 2 } \frac { \alpha } { 2 } $$,当cos$$ \frac { \alpha } { 2 } $$ =0时,tan $$ \frac { \alpha } { 2 } $$不...
【题目】已知$$ \tan \alpha = 2 $$,求下列各式的值:(1)角a为第三象限角时,求sinα,cosα的值;(2) $$ \frac { 4 \sin