1. Tentukan rumus cos 3 A , cos 4 A , tan 3 A ,dan tan 4 A .

Pembahasan

Ingat kembali rumuspenjumlahan dua sudut: cos ( α + b ) = cos α ⋅ cos β − sin α ⋅ sin β tan ( α + b ) = 1 − tan α ⋅ tan β tan α + tan β ​ cos 2 α = cos 2 α − sin 2 α = 2 ⋅ cos 2 α − 1 sin 2 α = 2 ⋅ sin α ⋅ cos α tan 2 α = 1 − tan 2 α 2 ⋅ tan α ​ dan identitas trigonometri sin 2 α + cos 2 α = 1 . Oleh karena itu, dapatdiperoleh rumus berikut: untuk cos 3 A cos 3 A cos 3 α ​ = = = = = = = = = ​ cos ( 2 A + A ) cos 2 A ⋅ cos A − sin 2 A ⋅ sin A ( 2 ⋅ cos 2 A − 1 ) ⋅ co s A − ( 2 ⋅ sin A ⋅ cos A ) ⋅ sin A 2 ⋅ cos 3 A − co s A − 2 ⋅ sin 2 A ⋅ cos A 2 ⋅ cos 3 A − co s A − 2 ⋅ ( 1 − cos 2 A ) ⋅ cos A 2 ⋅ cos 3 A − co s A − 2 ⋅ 1 ⋅ cos A + 2 ⋅ cos 2 A ⋅ cos A 2 ⋅ cos 3 A − co s A − 2 ⋅ cos A + 2 ⋅ cos 3 A 2 ⋅ cos 3 A + 2 ⋅ cos 3 A − co s A − 2 ⋅ cos A 4 ⋅ cos 3 A − 3 ⋅ cos A ​ untuk cos 4 A : cos 4 A cos 4 A ​ = = = = = = = = ​ cos ( 2 A + 2 A ) cos 2 A ⋅ cos 2 A − sin 2 A ⋅ sin 2 A cos 2 2 A − sin 2 2 A ( cos 2 A ) 2 − ( sin 2 A ) 2 ( cos 2 A − sin 2 A ) 2 − ( 2 ⋅ cos A ⋅ sin A ) 2 ( cos 2 A ) 2 + ( sin 2 A ) 2 − 2 ⋅ cos 2 A ⋅ sin 2 A − 2 2 ⋅ ( sin A ) 2 ⋅ ( cos A ) 2 cos 4 A + sin 4 A − 2 ⋅ sin 2 A ⋅ cos 2 A − 4 ⋅ sin 2 A ⋅ cos 2 A cos 4 A + sin 4 A − 6 ⋅ sin 2 A ⋅ cos 2 A ​ untuk tan 3 A : tan 3 A tan 3 A ​ = = = = = = = = = = = = = ​ tan ( A + 2 A ) 1 − t a n A ⋅ t a n 2 A t a n A + t a n 2 A ​ 1 − t a n A ⋅ 1 − tan 2 A 2 ⋅ tan A ​ t a n A + 1 − tan 2 A 2 ⋅ tan A ​ ​ 1 1 ​ − 1 − tan 2 A tan A ⋅ 2 ⋅ tan A ​ 1 tan A ​ + 1 − tan 2 A 2 ⋅ tan A ​ ​ 1 ⋅ ( 1 − tan 2 A ) 1 ⋅ ( 1 − tan 2 A ) ​ − 1 − tan 2 A 2 ⋅ tan 2 A ​ 1 ⋅ ( 1 − tan 2 A ) tan A ⋅ ( 1 − tan 2 A ) ​ + 1 − tan 2 A 2 ⋅ tan A ​ ​ 1 ⋅ 1 − 1 ⋅ tan 2 A 1 ⋅ 1 − 1 ⋅ tan 2 A ​ − 1 − tan 2 A 2 ⋅ tan 2 A ​ 1 ⋅ 1 − 1 ⋅ tan 2 A tan A ⋅ 1 − tan A ⋅ tan 2 A ​ + 1 − tan 2 A 2 ⋅ tan A ​ ​ 1 − tan 2 A 1 − tan 2 A ​ − 1 − tan 2 A 2 ⋅ tan 2 A ​ 1 − tan 2 A tan A − tan 3 A ​ + 1 − tan 2 A 2 ⋅ tan A ​ ​ 1 − tan 2 A 1 − tan 2 A − 2 ⋅ tan 2 A ​ 1 − tan 2 A tan A − tan 3 A + 2 ⋅ tan A ​ ​ 1 − tan 2 A 1 − 3 ⋅ tan 2 A ​ 1 − tan 2 A 3 ⋅ tan A − tan 3 A ​ ​ 1 − t a n 2 A 3 ⋅ t a n A − t a n 3 A ​ × 1 − 3 ⋅ t a n 2 A 1 − t a n 2 A ​ 1 − t a n 2 A ​ 1 3 ⋅ t a n A − t a n 3 A ​ × 1 − 3 ⋅ t a n 2 A 1 − t a n 2 A ​ 1 ​ 1 × ( 1 − 3 ⋅ t a n 2 A ) ( 3 ⋅ t a n A − t a n 3 A ) × 1 ​ 1 − 3 ⋅ t a n 2 A 3 ⋅ t a n A − t a n 3 A ​ ​ untuk tan 4 A : tan 4 A tan 4 A ​ = = = = = = = = = = = = = ​ tan ( 2 A + 2 A ) 1 − t a n 2 A ⋅ t a n 2 A t a n 2 A + t a n 2 A ​ 1 − 1 − tan 2 A 2 ⋅ tan A ​ ⋅ 1 − tan 2 A 2 ⋅ tan A ​ 1 − tan 2 A 2 ⋅ tan A ​ + 1 − tan 2 A 2 ⋅ tan A ​ ​ 1 1 ​ − ( 1 − tan 2 A ) 2 ( 2 ⋅ tan A ) 2 ​ 1 − tan 2 A 2 ⋅ tan A + 2 ⋅ tan A ​ ​ 1 × ( 1 − tan 2 A ) 2 1 × ( 1 − tan 2 A ) 2 ​ − ( 1 − tan 2 A ) 2 2 2 ⋅ tan 2 A ​ 1 − tan 2 A 4 ⋅ tan A ​ ​ ( 1 − tan 2 A ) 2 ( 1 − tan 2 A ) 2 ​ − ( 1 − tan 2 A ) 2 4 ⋅ tan 2 A ​ 1 − tan 2 A 4 ⋅ tan A ​ ​ ( 1 − tan 2 A ) 2 ( 1 − tan 2 A ) 2 − 4 ⋅ tan 2 A ​ 1 − tan 2 A 4 ⋅ tan A ​ ​ 1 − t a n 2 A ​ 1 4 ⋅ t a n A ​ × ( 1 − t a n 2 A ) 2 − 4 ⋅ t a n 2 A ( 1 − t a n 2 A ) 2 ​ ( 1 − t a n 2 A ) ​ 1 × ( ( 1 − t a n 2 A ) 2 − 4 ⋅ t a n 2 A ) 4 ⋅ t a n A × ( 1 − t a n 2 A ) ​ ( 1 − t a n 2 A ) 2 − 2 ⋅ t a n 2 A 4 ⋅ t a n A − 4 ⋅ t a n A ⋅ t a n 2 A ​ 1 2 + ( t a n 2 A ) 2 − 2 ⋅ 1 ⋅ t a n 2 A − 4 ⋅ t a n 2 A 4 ⋅ t a n A − 4 ⋅ t a n 3 A ​ 1 + t a n 4 A − 2 ⋅ t a n 2 A − 4 ⋅ t a n 2 A 4 ⋅ ( t a n A − t a n 3 A ) ​ 1 + t a n 4 A − 6 ⋅ t a n 2 A 4 ⋅ ( t a n A − t a n 3 A ) ​ ​ Dengan demikian, diperoleh rumus: cos 3 α ​ = ​ 4 ⋅ cos 3 A − 3 ⋅ cos A ​ cos 4 A ​ = ​ cos 4 A + sin 4 A − 6 ⋅ sin 2 A ⋅ cos 2 A ​ tan 3 A ​ = ​ 1 − 3 ⋅ t a n 2 A 3 ⋅ t a n A − t a n 3 A ​ ​ tan 4 A ​ = ​ 1 + t a n 4 A − 6 ⋅ t a n 2 A 4 ⋅ ( t a n A − t a n 3 A ) ​ ​

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