caranya  gimana ya?? 1.{sin(-330⁰)-tan 210} - (cos 300⁰ sin 225⁰) 2.sin 7/6π tan 4/3 π ÷ cos -1/3π

1.sin (-330) = -sin 330 = -sin (360-30) = -sin (-30) = 1/2
tan 210 = tan (180+30) = tan 30 = 1/3[tex] \sqrt{3} [/tex]
cos 300 = cos (360-60) = -cos 60 =  -1/2 (di kuadran 4) = 1/2
sin 225 = sin (180+45) = sin 45 = 1/2[tex] \sqrt{2} [/tex]
jadi 1/2-1/2[tex] \sqrt{3} [/tex] - 1/2.1/2 = 3-2[tex] \sqrt{3} [/tex]/6 - 1/4 = 3-4[tex] \sqrt{3} [/tex] /12

2.[tex] \pi [/tex] = 180
sin 7/6.180 = sin 210 = sin (180+30) = sin 30 = 1/2
tan 4/3.180 = tan 240 = tan (180+60) = tan 60 = [tex] \sqrt{3} [/tex]
cos -1/3.180 = -cos 60 = - 1/2
jadi 1/2.[tex] \sqrt{3} [/tex].-2 = -[tex] \sqrt{3} [/tex]

maaf kalau salah ya..