已知sinq-cosq=√2,求sin∧4q+cos∧4q=
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已知sinq-cosq=√2,求sin∧4q+cos∧4q=
已知sinq-cosq=√2,求sin∧4q+cos∧4q=
已知sinq-cosq=√2,求sin∧4q+cos∧4q=
∵sinq-cosq=√2,∴(sinq-cosq)^2=2,∴(sinq)^2+(cosq)^2-2sinqcosq=2,
∴1-2sinqcosq=2,∴2sinqcosq=-1,∴4(sinqcosq)^2=1,∴2(sinqcosq)^2=1/2.
∴(sinq)^4+(cosq)^4
=[(sinq)^4+2(sinqcosq)^2+(cosq)^4]-2(sinqcosq)
=[(sinq)^2+(cosq)^2]^2-1/2=1-1/2=1/2.