作业帮已知sin^2(45)° sin^2(105)°
来源：学生作业帮助网 编辑：作业帮 时间：2024/06/30 21:14:58
可设a=sinα,b=sinβ,-1
sina2+sinb2=sina-sina2/2只需求该区间就可以了令x=sina可得f(x)=x-x2/2-1=
3(sinA)^2+2(sinB)^2=5sinA(sinA)^2+(sinB)^2=5sinA/2-(sinA)^2/25sinA/2-(sinA)^2/2=-(1/2)(sinA-5/2)^2+2
sin(α-45°)=(√2/2)(sinα-cosα)=-√2/10则sinα-cosα=1/5sinα=cosα+1/5(sinα)^2+(cosα)^2=(cosα+1/5)^2+(cosα)^
0.5+(0.707-0.5)/(45-30)*(40-30)=0.638
sina-cosa=2√2/3sina=cosa+2√2/3sin^2a+cos^2a=1所以(cosa+2√2/3)^2+(cosa)^2=1即2cos^2a+(4√2/3)cosa-1/9=0所以
sin(45°-α)=-2/3√2/2cosα-√2/2sinα=-2/3sinα-cosα=2√2/3（1）式(sinα-cosα)^2=8/9sinα^2+cosα^2-2sinα*cosα=8/
sin(45°-a)=-cosa所以cosa=2/3而45°
2(sina)^2-sinacosa-3(cosa)^2=0(2sina-3cosa)(sina+cosa)=02sina-3cosa=0或sina+cosa=0sina/cosa=3/2或sina/
2sin2α+2sinαcosα1+tanα=2sinα(sinα+cosα)1+sinαcosα=2sinαcosα(sinα+cosα)sinα+cosα=2sinαcosα=k．当0＜α＜π4时
解应为（sinα+cosα）/(sinα-cosα)=2两边平方得（sin²α+cos²α+2sinαcosα）/(sin²α+cos²α-2sinαcosα)
是sin²A+sin²B-√2sinAsinB吧.由余弦定理得cosC=(a²+b²-c²)/(2ab)由正弦定理得(sin²A+sin&#
1）cos2a=7/25=1-2(sina)^23∏/4
本题题目应是要证:2tan（α+ β）=3tanα,答案见图片:
sinx=2cosx,sin^2x=4cos^2xsin^2x=4-4sin^2x,sin^2x=4/5(cosx+sinx)/(cosx-sinx)+sin^2x=(1+tanx)/(1-tanx)
依题有2sin2x=sinθ+cosθsinx的平方=sinθ*cosθ又2sin2x=4sinx*cosxsinθ*cosθ=[(sinθ+cosθ)的平方-1]/2所以有sinx的平方=[(4si
sin²1°+sin²2°+sin²3°...+sin²45°+sin²46°...+sin²89°＝sin^2(90-89)+sin^2(
答：sin^2a+sin^2(a+60)+sin^2（a+120）=3/2.证明：左边=sin^2a+sin^2(a+60)+sin^2（a+120）=sin^2a+(sinacos60+cosasi
sin(π/2-x)=cosx原式=sin^21°+……+sin^244°+1/2+cos^244°+……+cos^21°=44+1/2=89/2
sin²β=(2sinα-3sin²α)/2∵0≤sin²β≤1∴0≤(2sinα-3sin²α)/2≤10≤sinα≤2/3sin²α+sin