已知sinα是方程5x^2-7x-6=0的根
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/21 08:17:04
分子=(cosa)*(-cosa)*(-tana)²=-cos²a*sin²a/cos²a=-sin²a分母=sina*(-sina)*(-cosa)
(x-2)(5x+3)=00
由5x²-7x-6=0得x=2或x=-3/5所以sinα=-3/5,sin²α=9/25,cos²α=16/25所以sin²[(2k+1/2)π-α]+cos&
方程5x^2-7x-6=0可因式分解为(x-2)(5x+3)=0所以两个根分别为X1=2X2=-3/5因为-1=再问:不好意思,我把题目打错了,是这样的:[cos(2π-α)cos(π+α)tan^2
sinθ+cosθ=7/5sinθcosθ=m/51=(sinθ+cosθ)^2-2sinθcosθ∴1=49/25-2m/52m/5=24/25m=12/5再问:请问第二小题怎么做?谢谢!再答:(2
由根与系数的关系sina+cosa=2/3sinacosa=a/3sina^2+cosa^2=(sina+cosa)^2-2sinacosa=4/9-2a/3=1a=-5/6
毕成来也3x+y+2z=28①5x-3y+z=7②①-②*2,得-7x+7y=147y=14+7xy=2+x将y=2+x带入②,得5x-6-3x+2=72x+z=13z=13-2x将y=2+x,z=1
sinα²+cosα²=(sinα+cosα)²-2sinαcosα=(7/5)²-2×(m/5)=1解得m=12/5
由5x平方-7x-6=0得(5x+3)(x-2)=0所以x=-0.6或2因为sinα大于等于-1小于等于1所以X=-0.6所以sinα=-0.6[sin(-α-3π/2)sin(3π/2-α)tg平方
5x^2-7x-6=0(5x+3)(x-2)=0x1=-3/5,x2=2sinα是方程5x^2-7x-6=0的根sinα≠2∴sinα=-3/5sin(α+3/2*π)*sin(3/2*π-α)*ta
5x^2-7x-6=0(5x+3)(x-2)=0x=-3/5或x=2(舍去)sinα=-3/5α为第三象限角cosα=-4/5sin(-α-3/2*π)*cos(3/2*π-α)*tan^2(π-α)
sin(α+3π/2)=-cosαsin(3π/2-α)=-cosαtan(2π-α)=-tanαtan(π-α)=-tanαcos(π/2-α)=sinαcos(π/2+α)=-sinα原式=[(c
5x^2-7x-6=0的根x1=2,x2=-3/5.sinα=-3/5,所以cosa=-4/5,tana=3/4sin(α-3π/2)cos(π-α)tan(π+α)=cosa*(-cosa)*tan
5x²-7x-6=0(5x+3)(x-2)=0x=-3/5x=2>1取sinα=-3/5cos(2π-α)cos(π+α)tan²(2π-α)/sin(π-α)sin(2π-α)c
一.因为tanα=sinα/√(1-sin^2α)=3/416sin^2α=9(1-sin^2α)所以sinα=√(9/25)=3/5于是cosα=√(1-sin^2α)=4/51.sinα·cosα
〔sin(-α-3/2π)sin(3/2π-α)*tan(2π-α)〕/〔cos(π/2-α)cos(π/2+α)*cos(π-α)〕=cosαcosαtanα/sinαsinαcosα=sinα/s
x^2-5x+6=(x-2)(x-3)=0tanα=2tanβ=3tan(α+β)=(tanα+tanβ)/[1-tanαtanβ]=5/(1-6)=-1α+β=3π/4sin(α+β)=√2/2co
∵tana,tanb是方程x²-5x+6=0的两个实根∴tana+tanb=5.tana*tanb=6(根与系数关系)则tan(a+b)=(tana+tanb)/(1-tana*tanb)=
{[sin(α-3/2π)cos(3/2π-α)]/[cos(π/2-α)sin(π/2+α)]}×tan^2(π-α)={-sin(3/2π-a)cos(3/2π-α)]/[cos(π/2-α)si