4. Buktikan bahwa: sin 3 θ ⋅ sin 3 θ + cos 3 θ ⋅ cos 3 θ = cos 3 2 θ

Pembahasan

Ingat kembali: rumus sinus untuk penjumlahan dua sudut: sin ( α + β ) = sin α ⋅ cos β + cos α ⋅ sin β rumus cosinus untuk penjumlahan dua sudut: cos ( α + β ) = cos α ⋅ cos β + sin α ⋅ sin β rumus sinus untuk sudut ganda: sin 2 α = 2 ⋅ sin α ⋅ cos α rumus cosinus untuk sudut ganda: cos 2 α = cos 2 α − sin 2 α identitas trigonometri: sin 2 α + cos 2 α = 1 , cos 2 α = 1 − sin 2 α , dan sin 2 α = 1 − cos 2 α rumus pemfaktoran pangkat tiga: a 2 − b 2 = ( a − b ) ( a + b ) dan a 3 − b 3 = ( a − b ) ( a 2 + b 2 + ab ) Oleh karena itu,dapat diperoleh rumus berikut: sin 3 α sin 3 α ​ = = = = = = = = = ​ sin ( 2 α + α ) sin 2 α ⋅ cos α + cos 2 α ⋅ sin α ( 2 ⋅ sin α ⋅ cos α ) ⋅ cos α + ( cos 2 α − sin 2 α ) ⋅ sin α 2 ⋅ sin α ⋅ cos 2 α + sin α ⋅ cos 2 α − sin 3 α 3 ⋅ sin α ⋅ cos 2 α − sin 3 α 3 ⋅ sin α ⋅ ( 1 − sin 2 α ) − sin 3 α 3 ⋅ sin α ⋅ 1 − 3 ⋅ sin α ⋅ sin 2 α − sin 3 α 3 ⋅ sin α − 3 ⋅ sin 3 α − sin 3 α 3 ⋅ sin α − 4 ⋅ sin 3 α ​ dan cos 3 α cos 3 α ​ = = = = = = = = = = ​ cos ( 2 α + α ) cos 2 α ⋅ cos α − sin 2 α ⋅ sin α ( 2 ⋅ cos 2 α − 1 ) ⋅ co s α − ( 2 ⋅ sin α ⋅ cos α ) ⋅ sin α 2 ⋅ cos 3 α − co s α − 2 ⋅ sin 2 α ⋅ cos α 2 ⋅ cos 3 α − co s α − 2 ⋅ ( 1 − cos 2 α ) ⋅ cos α 2 ⋅ cos 3 α − co s α − 2 ⋅ ( 1 − cos 2 α ) ⋅ cos α 2 ⋅ cos 3 α − co s α − 2 ⋅ 1 ⋅ cos α + 2 ⋅ cos 2 α ⋅ cos α 2 ⋅ cos 3 α − co s α − 2 ⋅ cos α + 2 ⋅ cos 3 α 2 ⋅ cos 3 α + 2 ⋅ cos 3 α − co s α − 2 ⋅ cos α 4 ⋅ cos 3 α − 3 ⋅ cos α ​ sehingga ​ = = = = = = = = = = = = = = = ​ sin 3 θ ⋅ sin 3 θ + cos 3 θ ⋅ cos 3 θ ( 3 ⋅ sin θ − 4 ⋅ sin 3 θ ) ⋅ sin 3 θ + ( 4 ⋅ cos 3 θ − 3 ⋅ cos θ ) ⋅ cos 3 θ 3 ⋅ sin 4 θ − 4 ⋅ sin 6 θ + 4 ⋅ cos 6 θ − 3 ⋅ cos 4 θ 4 ⋅ cos 6 θ − 4 ⋅ sin 6 θ − 3 ⋅ cos 4 θ + 3 ⋅ sin 4 θ 4 ⋅ ( cos 6 θ − sin 6 θ ) − 3 ⋅ ( cos 4 θ − sin 4 θ ) 4 ⋅ ( ( cos 2 θ ) 3 − ( sin 2 θ ) 3 ) − 3 ⋅ ( ( cos 2 θ ) 2 − ( sin 2 θ ) 2 ) 4 ⋅ ( cos 2 θ − sin 2 θ ) ⋅ ( cos 4 θ + sin 4 θ + cos 2 θ ⋅ sin 2 θ ) − 3 ⋅ ( cos 2 θ − sin 2 θ ) ⋅ ( cos 2 θ + sin 2 θ ) 4 ⋅ ( cos 2 θ − sin 2 θ ) ⋅ ( cos 4 θ + sin 4 θ + cos 2 θ ⋅ sin 2 θ ) − 3 ⋅ ( cos 2 θ − sin 2 θ ) ⋅ 1 4 ⋅ ( cos 2 θ − sin 2 θ ) ⋅ ( cos 4 θ + sin 4 θ + cos 2 θ ⋅ sin 2 θ ) − 3 ⋅ ( cos 2 θ − sin 2 θ ) ⋅ ( cos 2 θ + sin 2 θ ) 2 4 ⋅ ( cos 2 θ − sin 2 θ ) ⋅ ( cos 4 θ + sin 4 θ + cos 2 θ ⋅ sin 2 θ ) − 3 ⋅ ( cos 2 θ − sin 2 θ ) ⋅ ( cos 4 θ + sin 4 θ + 2 ⋅ cos 2 θ ⋅ sin 2 θ ) ( cos 2 θ − sin 2 θ ) ⋅ ( 4 ⋅ cos 4 θ + 4 ⋅ sin 4 θ + 4 ⋅ cos 2 θ ⋅ sin 2 θ ) + ( cos 2 θ − sin 2 θ ) ⋅ ( − 3 ⋅ cos 4 θ − 3 ⋅ sin 4 θ − 6 ⋅ cos 2 θ ⋅ sin 2 θ ) ( cos 2 θ − sin 2 θ ) ⋅ ( cos 4 θ + sin 4 θ − 2 ⋅ cos 2 θ ⋅ sin 2 θ ) ( cos 2 θ − sin 2 θ ) ⋅ ( cos 2 θ − sin 2 θ ) 2 ( cos 2 θ − sin 2 θ ) 3 ( cos 2 θ ) 3 cos 3 2 θ ​ Dengan demikian, terbukti bahwa sin 3 θ ⋅ sin 3 θ + cos 3 θ ⋅ cos 3 θ = cos 3 2 θ .

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