Matematika

Pertanyaan

Tolongin bagian no 28 dan 29
Tolongin bagian no 28 dan 29

2 Jawaban

  • Nomer 29.
    Ingat bahwa sin^2(x)+cos^2(x)=1
    Shg
    2.1 - 3.1 + 1 = 0
    Smoga membantu. Beri tanggapn atau like
  • 28) (sec x)/(cosec^2 x) - (cosec x)/(sec^2 x)
    = (1/cos x)/(1/sin^2 x) - (1/sin x)/(1/cos^2 x)
    = (sin^2 x)/(cos x) - (cos^2 x)/(sin x)
    = (sin^3 x - cos^3 x)/(cos x sin x)
    = (sin x - cos x)(sin^2 x + sin x cos x + cos^2 x) / (cos x sin x)
    = (sin x - cos x) (sin x cos x + cos^2 x + sin^2 x) / (sin x cos x)
    = (sin x - cos x) [(sin x cos x)/(sin x cos x) + (cos^2 x)/(sin x cos x) + (sin^2 x)/(sin x cos x)]
    = (sin x - cos x) (1 + (cos x)/(sin x) + (sin x)/(cos x))
    = (sin x - cos x) (1 + cot x + Tan x)

    29) 2(sin^6 x + cos^6 x) - 3(sin^4 x + cos^4 x) + 1
    = 2 [(sin^2 + cos^2 x)^3 - 3 sin^2 x cos^2 x (sin^2 x + cos^2 x)] - 3 [(sin^2 x + cos^2 x)^2 - 2 sin^2 x cos^2 x] + 1
    = 2 [(1)^3 - 3 sin^2 x cos^2 x (1)] - 3 [(1)^2 - 2 sin^2 x cos^2 x] + 1
    = 2(1 - 3 sin^2 x cos^2 x) - 3(1 - 2 sin^2 x cos^2 x) + 1
    = 2 - 6 sin^2 x cos^2 x - 3 + 6 sin^2 x cos^2 x + 1
    = 2 - 3 + 1 - 6 sin^2 x cos^2 x + 6 sin^2 x cos^2 x
    = 0