Matematika Sekolah Menengah Pertama Nilai dari Cos 165°(Cos 135° + Tan 15°) = .....​

Nilai dari Cos 165°(Cos 135° + Tan 15°) = .....​

Penjelasan dengan langkah-langkah:

cos 165°(cos 135° + tan 15°) = cos (180°-15°) (cos (180°-45°) + tan 15°)

= -cos 15° (-cos 45° + tan 15°)

= cos 15° cos 45° - cos 15° tan 15°

= cos (45°-30°) cos 45° - sin 15°

= (cos 45° cos 30° + sin 45° sin 30°) cos 45° - sin (45°-30°)

= ((1/2)√2 · (1/2)√3 + (1/2)√2 · (1/2)) · (1/2)√(2) - (sin 45° cos 30° - cos 45° sin 30°)

= ((1/2)·(1/2)·√(2) ·√(3) + (1/2)·(1/2)·√(2)) ·(1/2)√(2) - ((1/2)√(2) · (1/2)√(3) - (1/2)√(2) · (1/2))

= ((1/4)√(2·3) + (1/4)√(2)) ·(1/2)√(2) - ((1/2)·(1/2)·√(2) ·√(3) - (1/2)·(1/2)·√(2))

= ((1/4)√(6) + (1/4)√(2)) ·(1/2)√(2) - ((1/4)√(2·3) - (1/4)·√(2))

= ((1/4)√(6) + (1/4)√(2)) ·(1/2)√(2) - ((1/4)√(6) - (1/4)·√(2))

= (1/4)√(6)·(1/2)√(2) + (1/4√(2) ·(1/2)√(2) - (1/4)√(6) + (1/4)√(2)

= (1/4)·(1/2)·√(6) ·√(2) + (1/4)·(1/2)·√(2)·√(2) - (1/4)√(6) + (1/4)√(2)

= (1/8)√(6·2) + (1/8)√(2·2) - (1/4)√(6) + (1/4)√(2)

= (1/8)√(12) + (1/8)√(4) - (1/4)√(6) + (1/4)√(2)

= (1/8)√(4·3) + (1/8)·2 - (1/4)√(6) + (1/4)√(2)

= (1/8)·√(4) ·√(3) + 2/8 - (1/4)√(6) + (1/4)√(2)

= (1/8)·2·√(3) + (1/4) - (1/4)√(6) + (1/4)√(2)

= (2/8)√(3) + (1/4) - (1/4)√(6) + (1/4)√(2)

= (1/4)√(3) + (1/4) - (1/4)√(6) + (1/4)√(2)

Kesimpulan:

Jadi nilai dari cos 165°(cos 135° + tan 15°) = (1/4)√(3) + (1/4) - (1/4)√(6) + (1/4)√(2)

Semoga membantu

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