Matematika

Pertanyaan

matematika sin75°+sin15°/cos105°+cos15°=

1 Jawaban

  • Trigonometri.

    sin A + sin B = 2 sin [(A + B) / 2] cos (A - B) / 2]
    cos C + cos B = 2 cos [(C + B) / 2] cos [(C - B) / 2]


    (sin 75° + sin 15°) / (cos 105°+ cos 15°)
    = {2 sin [(75
    ° + 15°) / 2] cos [(75° - 15°) / 2]} / {2 cos [(105° + 15°) / 2] cos [(105° - 15°) / 2]}
    = (2 sin 45° cos 30°) / (2 cos 60° cos 45°)
    = cos 30° / cos 60°
    = 1/2 √3 / (1/2)
    = √3

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