(1)sin^4α-cos^4α=2sin^2α-1 (2)tan^2α-sin^2α=tan^2αsin^2α (3)cosα/1-sinα=1+sinα/cos(1)sin^4α-cos^4α=2sin^2α-1 (2)tan^2α-sin^2α=tan^2αsin^2α (3)cosα/1-sinα=1+sinα/cosα求证

来源:学生作业学帮网 编辑:学帮网 时间:2024/05/08 07:58:50

(1)sin^4α-cos^4α=2sin^2α-1 (2)tan^2α-sin^2α=tan^2αsin^2α (3)cosα/1-sinα=1+sinα/cos
(1)sin^4α-cos^4α=2sin^2α-1
(2)tan^2α-sin^2α=tan^2αsin^2α
(3)cosα/1-sinα=1+sinα/cosα
求证

(1) 设 x= (sin α)^2,
则 (cos α)^2 = 1-x.
所以 (sin α)^4 - (cos α)^4
= x^2 - (1-x)^2
= 2x -1
= 2(sin α)^2 -1.
(2) 设 x= (sin α)^2,
则 (cos α)^2 = 1-x,
(tan α)^2 = x / (1-x).
所以 (tan α)^2 -(sin α)^2 = x / (1-x) -x
= x^2 / (1-x).
=(tan α)^2 * (sin α)^2.
= = = = = = = = = =
换元法+暴力破解, sin 神马的都是浮云.
当sin, cos, tan等都带平方时可用, 因为不用考虑正负问题.
(3) 因为 (cos α)^2 = 1 -(sin α)^2
= (1 +sin α) (1 -sin α).
所以 cos α / (1 -sin α) = (1 +sin α) / cos α.
= = = = = = = = =
如果要证明 a /b = c /d,
可先证明 ad =bc.
不等式也类似, 不过要注意正负问题.

sin(α+π/3)+sinα=负5分之4根号3 α∈(-π/2,0)求cosα怎样从9/4*sin²α=9/4(1-cos²α)=3/4*cos²a+12/5*cosx+48/25化为cos²α+4/5*cosα-11/100=0呀sin(α+π/3)+sinα=-4√3/5sinαcosπ/3+cosαsinπ/3+sinα=-4√3/53/2*si 已知tanα=3,计算:(1)4sinα-2cosα/5cosα+3sinα (2)sinαcosα (3)(sinα+cosα)^2(急~)(1)4sinα-2cosα/5cosα+3sinα(2)sinαcosα(3)(sinα+cosα)^2 已知tan(π-α)=2,求sin²α-2sinαcosα-cos²α/4cos²α-3sin²α+1 tan²α-sin²α=tan²α×sin²α(cosα-1)²+sin²α=2-2cosαsin四次方x+cos四次方x=1-2sin³xcos²x已知tanα-3求:①4sinα-2cosα/5cosα+3tanα②sinαcosα③(sinα+cosα)²已知cosα=1/4求sinα 若sin^2α+sinα=1 则cos^4α+cos^2α= 求证:sinα^4+cosα=1-2sinα^2cosα^2 [1-(sinα)^4-(cosα)^4]/[1-(sinα)^6-(cos)^6)]=? 已知tan=2,那么1/2sinαcosα+cosα²α和4sin²α+2sinαcosα-cos²α? 2sinα-3cosα/4sinα-9cosα=-1,则9sin^2α-3sinαcosα-5= 已知2cos^2α+3cosαsinα-3sin^2α=1求(1)tanα (2)(2sinα-3cosα)/(4sinα-9cosα) 已知tanα=-1/2,求下列各式的值(1)4sinα-3cosα/2sinα+5cosα (2)2sin^2α-3sinαcosα-5cos^2α 已知2cos平方α+3cosαsinα-3sin平方α=1,求(1)tanα;(2)(2sinα-3cosα)/(4sinα-9cosα) 已知:tanα=-4/3,求下列各式的值(1)3sinα+2cosα/3cosα+sinα(2)2sin²α+sinα*cosα-3cos²α 求证:(1-sinα+cosα)/(1+sinα+cosα)=tan(π/4-α/2) 若sinα+sin^2α=1,则cos^2α+cos^4α+cos^6α=? 求证:1-2sinαcosα/cos²α-sin²α=tan(π/4-α) 已知tan=1/4,则(sinα+2cos)/(2sinα+cosα)= 求证 1+sinα+cosα+2sinαcosα/求证 (1+sinα+cosα+2sinαcosα)/(1+sinα+cosα)=sinα+cosα