Tentukan nilainya.a. tan315-cos240+sin150 b. (sin120x

a. 0, b. 3/4

Pembahasan

Untuk soal ini, kita perlu menghitung nilai dari ekspresi trigonometri:
a. tan315 - cos240 + sin150
b. (sin120 x cos300) / tan210

Kita gunakan nilai-nilai sudut istimewa:
Untuk bagian a:
tan315° = tan(360° - 45°) = -tan45° = -1
cos240° = cos(180° + 60°) = -cos60° = -1/2
sin150° = sin(180° - 30°) = sin30° = 1/2

Maka, tan315 - cos240 + sin150 = (-1) - (-1/2) + (1/2) = -1 + 1/2 + 1/2 = -1 + 1 = 0.

Untuk bagian b:
sin120° = sin(180° - 60°) = sin60° = √3/2
cos300° = cos(360° - 60°) = cos60° = 1/2
tan210° = tan(180° + 30°) = tan30° = 1/√3

Maka, (sin120 x cos300) / tan210 = (√3/2 x 1/2) / (1/√3) = (√3/4) / (1/√3) = (√3/4) * (√3/1) = 3/4.