Find the area enclosed by the figure
1 answer:
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The half angle formula is:
sin x/2 = +/- √ (( 1-cos x ) / 2 )
5π/12 = 5π/6 : 2
sin 5π/12 = +/- √(( 1 - cos ( 5π/6)) / 2 )
cos 5π/6 = cos 150° = -√3/2
≈ 0.966 ( this is sin 5π/12 )
sin x/2 = +/- √ (( 1-cos x ) / 2 )
5π/12 = 5π/6 : 2
sin 5π/12 = +/- √(( 1 - cos ( 5π/6)) / 2 )
cos 5π/6 = cos 150° = -√3/2
≈ 0.966 ( this is sin 5π/12 )