Prove sin24(cos18)^2+sin36(cos12)^2=cos12sin72 ,

where 24,18,36,12 are in degrees.

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We use here four formulas.

sin(2a)=2sin(a)cos(a)

sin(a)+sin(b)=2sin((a+b)/2)cos((a-b)/2)

cos(2a)=2cos^2(a)-1

and

sin(90-a)=cos(a)

`we have`

`sin(24)cos^2(18)+sin(36)cos^2(12)=cos(12)sin(72)`

`LHS=sin(24)cos^2(18)+sin(36)cos^2(12)`

`=2sin(12)cos(12)cos^2(18)+sin(36)cos^2(12)`

`=cos(12){2sin(12)cos^2(18)+sin(36)cos(12)}`

`=cos(12){2sin(12)(cos(36)+1)/2+sin(36)cos(12)}`

`=cos(12){sin(12)cos(36)+sin(12)+sin(36)cos(12)}`

`=cos(12){sin(12+36)+sin(12)}`

`=cos(12){sin(48)+sin(12)}`

`=cos(12){2sin((48+12)/2)cos((48-12)/2)}`

`=cos(12){2sin(30)cos(18)}`

`=cos(12){2(1/2)sin(90-18)}`

`=cos(12)sin(72)`

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