证明:Sin42°-cos12°+sin54°=½

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证明:Sin42°-cos12°+sin54°=½

证明:Sin42°-cos12°+sin54°=½
证明:Sin42°-cos12°+sin54°=½

证明:Sin42°-cos12°+sin54°=½
sin42°-cos12°+sin54°
=sin42°-sin78°+cos36°
=2sin(-18°)cos60°+1-2(sin18°)^2
=1-sin18°-2(sin18°)^2
=1-(√5-1)/4-2[(√5-1)/4]^2
=1-(√5-1)/4-2[(5-2√5+1)/16]
=1-(√5-1)/4-(3-√5)/4
=1/2.  证明完毕.
下面证明:sin18°=(√5-1)/4.
∵sin36°=cos54°,∴sin(2×18°)=cos(3×18°),
∴2sin18°cos18°=4(cos18°)^3-3cos18°.
显然,cos18°>0, ∴2sin18°=4(cos18°)^2-3=4-4(sin18°)^2-3,
∴4(sin18°)^2+2sin18°-1=0,∴(2sin18°)^2+2sin18°-1=0.
显然,sin18°>0,∴2sin18°=[-1+√(1+4)]/2. [只取正号,否则sin18°<0]
∴sin18°=(√5-1)/4.

sin42°-cos12°+sin54°
=sin42°-sin78°+sin54°
=-2cos60°sin18°+sin54°
=sin54°-sin18°
=2cos36°sin18°
=2cos36°sin18°*2cos18°/(2cos18°)
=sin72°/(2sin72°)
=1/2