Bantuiin pakek caranya ya kak ? bentuk (-cos x - √3 sin x ) dapat diubah dalam bentuk ? a. 2 cos ( x - 4/3π) b. -2 cos ( x + 4/3π) c. 2cos(x+1/3π) d. -2 cos ( x -7/6π) e. 2cos(x-7/6π)
Takamori37
Bentuk berikut: a.sin x + b.cos x = √(a²+b²)cos(x - tan⁻¹a/b) Maka, dengna a = -√3 dan b = -1 Maka, = √(-√3)²+(-1)²). cos(x - tan⁻¹ (-√3/-1)) = √3+1. cos(x - (π+π/3)) = √4. cos(x - 7π/3) = 2.cos(x - 7π/3) = -2.cos(x - 2π - π/3) = -2.cos(x-π/3) Alias: = -2.[-cos(x - π/3 - π)] dengan [Pengurangan π] = 2.cos(x - 4π/3) [A]
a.sin x + b.cos x = √(a²+b²)cos(x - tan⁻¹a/b)
Maka, dengna a = -√3 dan b = -1
Maka,
= √(-√3)²+(-1)²). cos(x - tan⁻¹ (-√3/-1))
= √3+1. cos(x - (π+π/3))
= √4. cos(x - 7π/3)
= 2.cos(x - 7π/3)
= -2.cos(x - 2π - π/3)
= -2.cos(x-π/3)
Alias:
= -2.[-cos(x - π/3 - π)] dengan [Pengurangan π]
= 2.cos(x - 4π/3) [A]