The problem is to prove that \(\displaystyle{{\sin{{20}}}^{\circ}{{\sin{{40}}}^{\circ}{{\sin{{60}}}^{\circ}{{\sin{{80}}}^{\circ}=}}}}{\frac{{{3}}}{{{16}}}}\)

Terrence Riddle

Terrence Riddle

Answered question

2022-04-10

The problem is to prove that
sin20sin40sin60sin80=316

Answer & Explanation

ze2m1ingkdvu

ze2m1ingkdvu

Beginner2022-04-11Added 16 answers

Denote Q=sin20sin40sin60sin80. Observe first that
sin60=sin(20+40)
=sin20cos40+cos20sin40
=sin20(2cos2201)+cos20(2sin20cos20)
We can now write
Q=sin20sin(6020)sin60sin(60+20)
=sin20(sin(60cos20cos60sin20)sin60(sin60cos20+cos60sin20)
=sin20sin60(sin260cos220cos260sin220)
=sin20sin603cos220sin2204
=sin20sin604cos22014
=sin60sin604
=sin2604=316

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