已知x属于(0,π/2),tanx=1/2,求tan2x和sin(2x+π/3)的值
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已知x属于(0,π/2),tanx=1/2,求tan2x和sin(2x+π/3)的值
已知x属于(0,π/2),tanx=1/2,求tan2x和sin(2x+π/3)的值
已知x属于(0,π/2),tanx=1/2,求tan2x和sin(2x+π/3)的值
tan2x=2tanx/(1-tanx^2)=4/3
tan2x=4/3=sin2x/cos2x
sin2x^2+cos2x^2=1 x属于(0,π/2) 2x属于(0,π) tan2x>0
sin2x=4/5 cos2x=3/5
sin(2x+π/3)=sin2x cosπ/3+cos2x sinπ/3=(4+3√3)/10
tan2x=2tanx/(1-tan^2x)=4/3
sin(2x+π/3)=sin2xcosπ/3+cos2xsinπ/3=1/2sin2x+√3/2cos2x
=1/2*2tanx/(1+tan^2x)+√3/2(1-tan^2x)/(1+tan^2x)
=2/5+3√3/10
tanx=1/2
tan2x=2tanx/[1-(tanx)^2]=1/(1-1/4)=4/3
x属于(0,π/2),2x属于(0,π)
由tan2x>0知2x<π/2
则cos2x=1/√[1+(tan2x)^2]=1/√(1+16/9)=3/5
sin2x=√[1-(cos2x)^2]=4/5
sin(2x+π/3)=sin2xcos(π/3)+cos2xsin(π/3)
=(4/5)*(1/2)+(3/5)*(√3/2)
=(4+3√3)/10
tanx=1/2,所以x<π/4,tan2x>0
tan2x=2 tanx/(1+tan^2 x)=4/5
tanx=1/2,所以x<π/4,cos2x>0
sin(2x+π/3)=sin2x /2 + cos2x *√3 /2=(4+√3)/10
tan2x=2tanx/(1-tan²x)=4/3
sin(2x+π/3)=1/2sin2x+√3/2cos2x
=sinxcosx+√3/2(cos²x-sin²x)
=[sinxcosx+√3/2(cos²x-sin²x)]/(s...
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tan2x=2tanx/(1-tan²x)=4/3
sin(2x+π/3)=1/2sin2x+√3/2cos2x
=sinxcosx+√3/2(cos²x-sin²x)
=[sinxcosx+√3/2(cos²x-sin²x)]/(sin²x+cos²x)
=[tanx+√3/2(1-tan²x)]/(tan²x+1)
=8√3/15+2/5
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