(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}。
3.倍角公式的重要变形--升幂公式和降幂公式 (1)升幂公式$$ 1 + \cos 2 \alpha = 2 \cos ^ { 2 } \alpha , 1 - \cos 2 \alpha = $$ 2$$ 2 \sin ^ { 2 } $$,α, $$ 1 + \cos \alpha = 2 \cos ^ { 2 } \frac { \alpha } { 2 } , 1 - \cos \alpha = ...
2sec^(2)alpha-sec^(4)alpha-2cos ec^(2)alpha+cos ec^(4)alpha=cot^(4)alpha-tan^(4)alpha View Solution ((1)/(sec^(2)alpha-cos^(2)alpha)+(1)/(cos ec^(2)alpha-sin^(2)alpha))cos^(2)alpha*sin^(2)alpha=(1-cos^(2)alpha*sin^(2)alpha)/(2+cos^(2)alpha*sin^(2)alpha...
1、用tan(alpha)表示tan(alpha/2)2、化简:sin(50)(1+根号3tan(10))(角度)3、求证:(1-tan^2(alpha/2))/(1+tan^2(alpha/2))=cos(alpha)详细过程,谢谢 扫码下载作业帮搜索答疑一搜即得 答案解析 查看更多优质解析 解答一 举报 1tan(alpha)=2tan(alpha)/(1+tan^2(alpha/2))22sin(50)(1/2+...
Given: $$(1 + \sin\alpha)(1 - \sin\alpha) = \cos^2 \alpha $$ We have to verify the identity. First, we identify the known trigonometry identity...Become a member and unlock all Study Answers Start today. Try it now Create an account Ask a question Our ex...
$$$ 1 - \cos 2 \alpha = \_ . $$ 答案 4.$$ 2 \cos ^ { 2 } $$α $$ 2 \sin ^ { 2 } a $$相关推荐 1升幂公式1+cos2α= 24.升幂公式1+cos2α= 1-cos2α= 34.升幂公式$$ 1 + \cos 2 \alpha = \_ . $$$ 1 - \cos 2 \alpha = \_ . $$ 反馈...
$$ ∵α为锐角,∴$$ \sin \alpha \neq 0 $$, ∴$$ 2 \sin \alpha = \cos \alpha $$,即$$ \tan \alpha = \frac { 1 } { 2 } . $$ (法一) 由$$ \tan ( \beta - \alpha ) = \frac { \tan \beta - \tan \alpha } { 1 + \tan \beta \tan \alpha } $$ $$ ...
sin 3 alpha = 4 sin alpha sin(x + alpha) sin(x-alpha) 05:22 The general solution of 4 sin^4 x + cos^4x= 1 is 02:39 For n in Z , the general solution of (sqrt(3)-1)sintheta+(sqrt(3)+1)c... 02:34 The value of cosycos(pi/2-x)-cos(pi/2-y)cosx+sinycos(pi/2...
Let alpha=(pi)/(5) and A,=[[cos alpha,sin alpha-sin alpha,cos alpha]] then B=A^(4)-A^(3)+A^(2)-A is View Solution (cos2 alpha)/(cos^(4)alpha-sin^(4)alpha)-(cos^(4)alpha+sin^(4)alpha)/(2-sin^(2)2 alpha)= ...
$$ ∵α为锐角,∴$$ \sin \alpha \neq 0 $$, ∴$$ 2 \sin \alpha = \cos \alpha $$,即$$ \tan \alpha = \frac { 1 } { 2 } . $$ (法一) 由$$ \tan ( \beta - \alpha ) = \frac { \tan \beta - \tan \alpha } { 1 + \tan \beta \tan \alpha } $$ $$ ...