Buktikan bahwa Sin^2 195° Sin75° Cos 75° = 1/8 (1 - 1/2akar3)
DB45
Sin² 195 sin 75 cos 75 = = (sin 195. sin 75)(sin 195 cos 75) = - ¹/₂ ( -2 sin 195 sin 75) . {¹/₂ (2 sin 195. cos 75)} = -¹/₂ [cos(195+75)- cos(195-75)]. [¹/₂ (sin(195+75) +sin (195-75)]
= (sin 195. sin 75)(sin 195 cos 75)
= - ¹/₂ ( -2 sin 195 sin 75) . {¹/₂ (2 sin 195. cos 75)}
= -¹/₂ [cos(195+75)- cos(195-75)]. [¹/₂ (sin(195+75) +sin (195-75)]
= [-¹/₂ (cos 270 - cos 120)] . [ ¹/₂ (sin 270 + sin 120)]
= [- ¹/₂ (0 - (-¹/₂)] . [¹/₂ (-1 + ¹/₂ √3)]
= [ -¹/₂ (¹/₂)].[(-¹/₂ + ¹/₄ √3)]
= -¹/₄ (- ¹/₂ + ¹/₄√3)
= ¹/₈ - ¹/₁₆ √3
= ¹/₈ ( 1 - ¹/₂ √3)