已知sin(π/4+α)sin(π/4-α)=1/6,α∈(π/2,π),求sin4α.
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sin(π/4 + α)sin(π/4 - α)= sin(π/4 + α)sin[π/2 -(π/4 + α)]
=sin(π/4 + α)cos(π/4 + α) = 1/6,
∴2sin(π/4 + α)cos(π/4 + α)= 1/3,
∴sin[2(π/4 + α)] = sin(π/2 + 2α) = sinπ/2 * cos2α + cosπ/2 * sin2α
=cos2α = 1/3,
∵又α∈(π/2,π),
∴sin2α = 二倍根号二/3,
∴sin4α = 2sin2αcos2α = 2* 1/3 * 二倍根号二/3 = 4倍根号二/9
=sin(π/4 + α)cos(π/4 + α) = 1/6,
∴2sin(π/4 + α)cos(π/4 + α)= 1/3,
∴sin[2(π/4 + α)] = sin(π/2 + 2α) = sinπ/2 * cos2α + cosπ/2 * sin2α
=cos2α = 1/3,
∵又α∈(π/2,π),
∴sin2α = 二倍根号二/3,
∴sin4α = 2sin2αcos2α = 2* 1/3 * 二倍根号二/3 = 4倍根号二/9
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