arccsc[√(4+2√2)],用弧度制表示
求[√(4加2√2)]的反余割
arccsc√(4+2√2)
=arccot√[(4+2√2)-1]
=arccot√(3+2√2)
=arccot(√2+1)
=π/8
令θ=arccsc[√(4+2√2)]
1/sinθ=cscθ=√(4+2√2)
sin²θ=1/(4+2√2)=(4-2√2)/8
2sin²θ=1-√2/2=1-cos(π/4)
cos(π/4)=1-2sin²θ=cos2θ.
2θ=π/4
θ=π/8
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求[√(4加2√2)]的反余割
arccsc√(4+2√2)
=arccot√[(4+2√2)-1]
=arccot√(3+2√2)
=arccot(√2+1)
=π/8
令θ=arccsc[√(4+2√2)]
1/sinθ=cscθ=√(4+2√2)
sin²θ=1/(4+2√2)=(4-2√2)/8
2sin²θ=1-√2/2=1-cos(π/4)
cos(π/4)=1-2sin²θ=cos2θ.
2θ=π/4
θ=π/8