Matematika

Pertanyaan

[tex] \frac{cosx-cos3x}{sinx.cosx} [/tex]

a.-4sinx
b.-2sinx
c.2cosx
d.2sinx
e.4sinx

2 Jawaban

  • IdenTitas TrigoNometRi

    (cos x - cos 3x) / sin x cos x
    = -(cos 3x - cos x) / (sin 2x)/2
    = -2(-2 sin (3x + x)/2 sin (3x - x)/2) / sin 2x
    = 4 sin (4x/2) sin (2x/2) / sin 2x
    = 4 sin x

    Opsi E
  • Trigonometri.

    cos A - cos B = -2 sin 1/2 (A + B) sin 1/2 (A - B)
    2 sin A cos A = sin 2A

    [tex]\displaystyle \frac{\cos x-\cos 3x}{\sin x \cos x}\\ =\frac{-2\sin \frac{1}{2}(x+3x)\sin \frac{1}{2}(x-3x)}{\frac{1}{2}\sin 2x}\\ =\frac{-2\sin 2x \sin (-x)}{\frac{1}{2}\sin 2x}\\=4\sin x[/tex]

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