cos2π/3为什么等于-1/2
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a=(√3sinx/2,cosx/2),b=(cosx/2,-cosx/2)f(x)=a·b=√3sin(x/2)cos(x/2)-(cosx/2)^2=(√3/2)sinx-(1/2)cosx-1/
取α=45°,带入原式,左边=1,右边=0,左右不等.所以该式并非恒成立.只有在特定值下才成立.即该式为三角函数方程.设:tanα=x,根据万能公式有:sin2α=2x/(1+x^2)cos2α=(1
原式=sinπ7(cos2π7+cos4π7+cos6π7)sinπ7=sinπ7cos2π7+sinπ7cos4π7+sinπ7cos6π7sinπ7=12(sin3π7−sinπ7)+12(sin
∵θ∈(π,3π/2),则θ/2∈(π/2,3π/4)∴cosθ/2<0∴cosθ=2cosθ/2-1=-3/5cosθ/2=1/5∴cosθ/2=-√5/5
cosπ/5*cos2π/5=(2sinπ/5*cosπ/5*cos2π/5)/(2sinπ/5)=(sin2π/5*cos2π/5)/(2sinπ/5)=(2sin2π/5*cos2π/5)/2*(
用a和b左边=cos[(a+b)+(a-b)]cos[(a+b)-(a-b)]=[cos(a+b)cos(a-b)-sin(a+b)sin(a-b)][cos(a+b)cos(a-b)+sin(a+b
先把它看成分母为1的分数,(cosπ/5)(cos2π/5)/1,然后分子分母同时乘sin(π/5),这样分子上可以用一下sin的二倍角公式式子变为:sin(2π/5)*cos(2π/5)/2sin(
这一类题可用对称法:设sinπ/5×sin2π/5=mcosπ/5×cos2π/5=n则4mn=(2sinπ/5×cosπ/5)×(2sin2π/5×cos2π/5)=sin2π/5×sin4π/5=
cos^2(π/2-a)+cos2^2(π/6+a)=sin^2a+1/2(1+cos(π/3+2a)=1/2(1-cos(π/3+2a)+1/2(1+cos(π/3+2a)=1再问:cos^2(3/
sin^2α+cos2α=1/4sin^2a+cos^2a-sin^2a=1/4cos^2a=1/4cosa=1/2因为a∈(0,π/2)所以sina=根3/2tana=sina/cosa=根3记得采
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cosθ=-3/5,θ∈(π,3π/2),则有θ/2∈(π/2,3π/4),则有cos(θ/2)<0.∵cosθ=2cos^2(θ/2)-1,∴有cos(θ/2)=-√[(1+cosθ)/2]=-√[
令:a+π/3=bcos2b=1-2(sinb)^2=1-(1/3)^2=7/9
tan[π/4+α]=(tanπ/4+tanα)/(1-tanπ/4tanα)=(1+sinα/cosα)/(1-sinα/cosα)=(cosα+sinα)/(cosα-sinα)=[2cos^2(
因为sin2α+cos2α=1(sin2a+cos2a)^2=1所以(sin2a)^2+2*sin2acon2a+(con2a)^2=1所以2*sin2acon2a=0所以sin4a=0又(sin4a
tan(π/4-a)=[1-tana]/[1+tana]=3,则tana=-1/2.而sin2a-coa2a=[2sinacosa-cos²a+sin²a]/[sin²a
你确定题目没错吗.我觉得-cos2=(1-sin2)=sin2这样的等式应该是不对的若(1-sin2)=sin2那么可解得sin2=1/2你可以用计算器算下.sin2是不等于1/2的