Apakah tempoh f (theta) = tan ((13 theta) / 12) - cos ((6 theta) / 5)?

Apakah tempoh f (theta) = tan ((13 theta) / 12) - cos ((6 theta) / 5)?
Anonim

Jawapan:

# 60pi #

Penjelasan:

Tempoh #tan (13t) / 12) # --> # (12 (pi)) / 13 #

Tempoh #cos ((6t) / 5) # --> # (5 (2pi)) / 6 = (10pi) / 6 = (5pi) / 3 #

Tempoh f (t) -> kurang umum # (12pi) / 13 dan (5pi) / 3 #

# (12pi) / 13 #..x (13) = # 12pi #..x (5) -> # 60pi #

# (5pi) / 3 #..x (3) ……. = # 5pi #.x (12) -> # 60pi #

Tempoh #f (t) = 60pi #