已知cos(75°+a)=三分之一
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因为(sina+cosa)^2=1+2sinacosa=1/9所以sinacosa=-4/9那么sina和cosa可以看成方程:x^2-1/9x-4/9=0的两个根解方程得:x1=-1/9,x2=4/
a为第三象限角,cos(75°+a)=1/3>0则75°+a为第四象限角∴sin(75°+a)=-√1-(1/3)²=-2√2/3cos(-255°-a)+sin(435°+a)=-cos(
问什么cosa吗:cos75=cos(45+30)=cos45cos30-sin45sin30=v6/4-v3/4然后算出sin75cos(75°+a)=cos75cosa-sin75sina=1/3
根据已知条件,sin(π-a)-cos(π+a)=sqrt(2)/3,因为sin(π-a)=sin(a),cos(π+a)=-cos(a),所以上式等价于sin(a)+cos(a)=sqrt(2)/3
cos(a+30°)=12/13,30°
f(x)=sin(2x+三分之派)+sin(2x-三分之派)+2cos方x-1=sin2x*cos(3分之π)+cos2x*sin(3分之π)+sin2x*cos(3分之π)-cos2x*sin(3分
cos(六分之π+a)=cos[二分之π-(三分之π-a)]=sin(三分之π-a)=二分之一
f(x)=1/2cos2x+根号3/2sin2x-(cos2x+1)=根号3/2sin2x-1/2cos2x-1=sin(2x-Pai/6)-1f(a)=sin(2a-Pai/6)-1=-1/5sin
(8+15*根号3)/34
∵cos(x-π/6)=-√3/3∴cosx+cos(x-π/3)=cosx+cosxcosπ/3+sinxsinπ/3=cosx+(1/2)cosx+(√3/2)sinx=(3/2)cosx+(√3
cos(75°+a)=-cos(180°-75°-a)=-cos(105°-a)=1/3cos(75°+a)=sin(90°-75°-a)=sin(15°-a)=1/3sin(a-15°)=-1/3c
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cos(x-π/6)=-√3/3cosx+cos(x-π/3)=cos(x-π/6+π/6)+cos(x-π/6-π/6)=cos(x-π/6)cosπ/6-sin(x-π/6)sinπ/6+cos(
sina+cosa=根号下2/3sina-cosa=-3/4sina+cosa-(sina-cosa)=根号下2/3-(-3/4)cosa=根号下2/6+3/8(1+sin2a+cos2)/(1+ta
因为cos(a+b)=cosacosb-sinasinbcos(a-b)=cosacosb+sinasinb相加得cos(a+b)+cos(a-b)=2cosacosb即cosacosb=[cos(a
sinA=3/4,于是cosA=√7/4cos(A+B)=cosAcosB-sinAsinB=√7/4cosB-3/4sinB=-2/3即√7/4√[1-(sinB)^2]=3/4sinB-2/3两边
f(x)=2cos(x+π/3)[sin(x+π/3)-√3cos(x+π/3)]=2cos(x+π/3)sin(x+π/3)-2√3cos²(x+π/3)=sin(2x+2π/3)-√3[
cos(105-a)=cos[180-(75+a)]=-cos(75+a)=-1/3
(1)已知abc成等比数列,那么可以设a=1,b=2,c=4,∠A∠B都是锐角且COSB=三分之四,根据正弦定理,可得出sinB=√7/4,sinA=√7/8,cosA=√57/8,答案不等于7分之4