Caranya gimana ya?? - Universityku

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

Caranya gimana ya??

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

2. $pi$ = 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 = $sqrt{3}$
cos -1/3.180 = -cos 60 = – 1/2
jadi 1/2. $sqrt{3}$ .-2 = – $sqrt{3}$

maaf kalau salah ya..

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