化简(1 cos2α) (tanα 2-cotα 2)
来源:学生作业帮助网 编辑:作业帮 时间:2024/10/02 18:13:23
取α=45°,带入原式,左边=1,右边=0,左右不等.所以该式并非恒成立.只有在特定值下才成立.即该式为三角函数方程.设:tanα=x,根据万能公式有:sin2α=2x/(1+x^2)cos2α=(1
数学之美团为你解答cos(2α)-2cos(2θ)=(1-tan²α)/(1+tan²α)-2(1-tan²θ)/(1+tan²θ)=(1-tan²α
1/(cos2α)+tan2α=1/(cos²α-sin²α)+tan2α=(sin²α+cos²α)/(cos²α-sin²α)+tan2
(1-cos4α)/sin4α*cos2α/(1+cos2α)=2(sin2α)^2/(2sin2αcos2α)*cos2α/(1+cos2α)[(sin2α)^2表示sin2α的平方]=sin2α/
∵1+tanα1−tanα=2013,∴1cos2α+tan2α=cos2α+sin2αcos2α−sin2α+2tanα1−tan2α=1+tan2α1−tan2α+2tanα1−tan2α=(1+
sin2α=(1+cos2α)*tanα1-cos2α=sin2α*tanα两式相加:sin2α+1-cos2α=(1+cos2α+sin2α)*tanα(把公因式tanα提出)所以tanα=(1+s
证明: 左边= =1+2sinαcosα/cos²a-sin²a =[sin²a+cos²a+2sinαcosα]/[cos²a-sin
由题知,设tanα/(tanα-1)=-1所以,tanα=1/2tan2α=(2tanα)/(1-tan²α)=1/(3/4)=4/3所以,sin2α=4/5,cos2α=3/5或sin2α
(1+tanα)/(1-tanα)=2010,anα=2009/2011,(1/cos2α)+tan2α=1/(2cos²α-1)+2anα/(1-tan²α)=1/[2/(1+t
设长方体ABCD-A1B1C1D1,长,宽,高分别为a,b,c;*是指乘α,β,γ分别为角CA1B1,CA1A,CA1D,则满足Cosα^2+cosβ^2+cosγ^2=1(a^2/(a^2+b^2+
1.tan(π/4+α)=2[tan(π/4)+tanα]/[1-tan(π/4)*tanα]=(1+tanα)/(1-tanα)=2tanα=1/32.(sin2α·cosα-sinα)/(sin2
1+cos2αtanα2−cotα2=1+2cos2α−1sinα2cosα2−cosα2sinα2=2cos2αsin2α2−cos2α2sinα2cosα2=2cos2α−cosα12sinα=-
sin2α/(1-cos2α)-1/tanα=2sinacosa/2sin²a-1/tana=cosa/sina-cosa/sina=0
cos(2α)/cos(2β)=cos[(α+β)+(α-β)]/cos[(α+β)-(α-β)] &nb
原式子=[cos2(a+b)+cos2(a-b)]/2-[1+cos2(a-b)]/2=[cos2(a+b)-1]/2=-[sin(a+b)]^2sin(a+b)=2tan(a+b)/2/(1+tan
(1-tanα)/(1+tanα)=(1-sinα/cosα)/(1+sinα/cosα)=((cosα-sinα)/cosα)/((cosα+sinα)/cosα)=(cosα-sinα)/(cos
用a代替1/sin2a=(sin²a+cos²a)/2sinacosa=sin²a/2sinacosa+cos²a/2sinacosa=(1/2)(sina/c
tan^2α=2tan^2β+11+tan²α=2(1+tan²β)sec²α=2sec²β1/cos²α=2/cos²β2cos²
1cos2α+tan2α=sin2α+1cos2α=1+2sinαcosαcos2α−sin2α =(sinα+cosα)2(cosα+sinα)(cosα−sinα)=sinα+cosαc