已知sin(π 4 a)=根号3 2
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sin(a+π/3)+sina=(-4/5)*根号3展开左边,即√3sin(a+π/6)=(-4/5)*根号3∴sin(a+π/6)=-4/5∵-π/2再问:麻烦下左边的式子怎么拆开请帮下。。怎么得到
sina=√15/4,cosa=-1/4sin(π-a)=sina=√15/4,sin(a+π/2)=cosa=-1/4,=>sin(π-a)/((sin(a+π/2)+cosa+1)=√15/4/(
sin(A+π/4)=7√2/10,A属于(π/4,π/2)cos(A+π/4)=-√2/10cosA=cos[(A+π/4)-π/4]=cos(A+π/4)cosπ/4-sin(A+π/4)sinπ
因为sin(a+π/2)=cosa,所以sin(2π/3+a)=sin[π/2+(π/6+a)]=cos(π/6+a)由sin(π/6+a)=(根号3)/3>0,得cos(π/6+a)=(±根号6)/
是四份之π还是π分之四?再问:四分之π再答:你的题给的有点特殊,所以只能按照我理解的给答案,如果有错可以追问f(x)=ab={√2sin(π/4+x)+1}x{√2sin(π/4+x)-1}-√3co
sin(a+π/3)+sina=sinacos(π/3)+sin(π/3)cosa+sina=-4√3/5化简得到√3sina+cosa=-8/5再利用sina^2+cosa^2=1带入解得cosa再
(sina-cosa)²=21-2sinacosa=2得:2sinacosa=-1则:(sina+cosa)²=1+2sinacosa=0所以,sina+cosa=0即:sina=
sin(a+π/6)+cosa=sina*(v3/2)+cosa*(1/2)+cosa=v3[sina*(1/2)+cosa*(v3/2)]=v3sin(a+π/3)=(4/5)*v3,——》sin(
[2cos^2(a/2)-sina-1]/根号2sin(π/4+a)=(cosa-sina)/[√2(√2/2sina+√2/2cosa)](2cos²a/2-1=cosa,sin(π/4+
解cos(4π/3+a)=cos(π+π/3+a)=-cos(π/3+a)=-cos[π/2-(π/6-a)]=-sin(π/6-a)=-√3/3
cos(a-π/6)+sina=4√3/5展开cosacosπ/6+sinasinπ/6+sina=4√3/5√3/2*cosa+3/2*sina=4√3/51/2*cosa+√3/2*sina=4/
根据:cos(a-π/6)+sina=(4*根号3)/5有:cosacosπ/6+sinasinπ/6+sina=(4*根号3)/5(根号3)/2cosa+3/2sina=(4*根号3)/5得到:co
sin(α+π/6)=sinα·cos(π/6)+cosα·sin(π/6)=(√3/2)sinα+(1/2)cosα;所以sin(α+π/6)+cosα=(√3/2)sinα+(3/2)cosα;即
由a范围则cos(a-π/4)>0sin²+cos²=1所以cos(a-π/4)=7√2/10cosa=cos(a-π/4+π/4)=cos(a-π/4)cosπ/4-sin(a-
方便起见,用a,b来表示α,β由题意得:sina=√2sinb√3cosa=√2cosb两式平方相加得:sin²a+3cos²a=2即:1+2cos²a=2得:cos
因为向量a和b垂直,所以,a*b=0,整理,得到4sin(a+π/6)+4cosa-根号3=0进一步整理,得到sin(a+π/6)+cosa=根号3/4,然后把sin(a+π/6)展开,得到√3/2s
(1)展开,得sinAcosπ/4+cosAsinπ/4+sinAcosπ/4-cosAsinπ/4=2sinAcosπ/4=√2×sinA=√2/3所以,sinA=1/3.(2)原式=[sinAco
sin(3π-a)=√2sin(2π+b)sina=√2sinbsin²a=2sin²b√3cos(-a)=-√2cos(π+b)-√3cosa=√2cosb3cos²a
√3sina-cosa(用辅助角公式)=2sin(a-π/6)=2sin(a-π/6+π/2-π/2)=2sin(a+π/3-π/2)=-2sin[π/2-(a+π/3)]=-2cos(a+π/3)=