Tunjukkan bahwa: tan 6 ∘ tan 4 2 ∘ tan 6 6 ∘ tan 7 8 ∘ = 1

Pembahasan

Ingat rumus perkalian trigonometri 2 sin α sin β 2 cos α cos β ​ = = ​ cos ( α − β ) − cos ( α + β ) cos ( α + β ) + cos ( α − β ) ​ ingat tan α = cos α sin α ​ , sehingga di dapat tan 6 ∘ tan 4 2 ∘ tan 6 6 ∘ tan 7 8 ∘ = cos 6 ∘ cos 4 2 ∘ cos 6 6 ∘ cos 7 8 ∘ sin 6 ∘ sin 4 2 ∘ sin 6 6 ∘ sin 7 8 ∘ ​ Mencari sin 6 ∘ sin 4 2 ∘ sin 6 6 ∘ sin 7 8 ∘ sin 4 2 ∘ sin 7 8 ∘ sin 6 ∘ sin 6 6 ∘ ​ = = = = = = = = = = = = = = = = = ​ 2 1 ​ ( 2 sin 4 2 ∘ sin 7 8 ∘ ) 2 1 ​ ( cos ( 4 2 ∘ − 7 8 ∘ ) − cos ( 4 2 ∘ + 7 8 ∘ ) ) 2 1 ​ ( cos ( − 3 6 ∘ ) − cos ( 12 0 ∘ ) ) ingat cos ( − α ) = cos α 2 1 ​ ( cos 3 6 ∘ − ( − 2 1 ​ ) ) 2 1 ​ ( cos 3 6 ∘ + 2 1 ​ ) ingat cos 3 6 ∘ = 4 5 ​ + 1 ​ 2 1 ​ ( 4 5 ​ + 1 ​ + 2 1 ​ ) 2 1 ​ ( 4 5 ​ + 1 ​ + 4 2 ​ ) 2 × 4 5 ​ + 3 ​ 8 5 ​ + 3 ​ 2 1 ​ ( 2 sin 6 ∘ sin 6 6 ∘ ) 2 1 ​ ( cos ( 6 ∘ − 6 6 ∘ ) − cos ( 6 ∘ + 6 6 ∘ ) ) 2 1 ​ ( cos ( − 6 0 ∘ ) − cos 7 2 ∘ ) ingat cos ( − α ) = cos α 2 1 ​ ( 2 1 ​ − cos 7 2 ∘ ) ingat cos 7 2 ∘ = 4 5 ​ − 1 ​ 2 1 ​ ( 2 1 ​ − 4 5 ​ − 1 ​ ) 2 1 ​ ( 4 2 ​ − 4 5 ​ − 1 ​ ) 2 × 4 3 − 5 ​ ​ 8 3 − 5 ​ ​ ​ sehingga diperoleh sin 6 ∘ sin 4 2 ∘ sin 6 6 ∘ sin 7 8 ∘ ​ = = = = = = ​ 8 3 + 5 ​ ​ × 8 3 − 5 ​ ​ 8 ( 3 + 5 ​ ) ​ × 8 ( 3 − 5 ​ ) ​ 64 3 2 − ( 5 ​ ) 2 ​ 64 9 − 5 ​ 64 4 ​ 16 1 ​ ​ Mencari cos 6 ∘ cos 4 2 ∘ cos 6 6 ∘ cos 7 8 ∘ cos 6 ∘ cos 4 2 ∘ cos 6 6 ∘ cos 7 8 ∘ = 2 1 ​ ( 2 cos 6 ∘ cos 6 6 ∘ ) 2 1 ​ ( 2 cos 4 2 ∘ cos 7 8 ∘ ) = 4 1 ​ ( cos ( 6 ∘ + 6 6 ∘ ) + cos ( 6 ∘ − 6 6 ∘ ) ) ( cos ( 4 2 ∘ + 7 8 ∘ ) + cos ( 4 2 ∘ − 7 8 ∘ ) ) = 4 1 ​ ( cos 7 2 ∘ + cos ( − 6 0 ∘ ) ) ( cos 12 0 ∘ + cos ( − 3 6 ∘ ) ) ingat cos ( − α ) = cos α = 4 1 ​ ( cos 7 2 ∘ + cos 6 0 ∘ ) ( cos 12 0 ∘ + cos 3 6 ∘ ) = 4 1 ​ ( cos 7 2 ∘ + 2 1 ​ ) ( − 2 1 ​ + cos 3 6 ∘ ) = 4 1 ​ ( − 2 1 ​ cos 7 2 ∘ + cos 7 2 ∘ cos 3 6 ∘ − 4 1 ​ + 2 1 ​ cos 3 6 ∘ ) = 8 1 ​ ( − cos 7 2 ∘ + 2 cos 7 2 ∘ cos 3 6 ∘ − 2 1 ​ + cos 3 6 ∘ ) = 8 1 ​ ( − cos 7 2 ∘ + ( cos ( 7 2 ∘ + 3 6 ∘ ) + cos ( 7 2 ∘ − 3 6 ∘ ) ) − 2 1 ​ + cos 3 6 ∘ ) = 8 1 ​ ( − cos 7 2 ∘ + cos 10 8 ∘ + cos 3 6 ∘ − 2 1 ​ + cos 3 6 ∘ ) ingat cos ( 18 0 ∘ − α ) = − cos α = 8 1 ​ ( − cos 7 2 ∘ + cos ( 18 0 ∘ − 7 2 ∘ ) + 2 cos 3 6 ∘ − 2 1 ​ ) = 8 1 ​ ( − cos 7 2 ∘ − cos 7 2 ∘ + 2 cos 3 6 ∘ − 2 1 ​ ) = 8 1 ​ ( − 2 cos 7 2 ∘ + 2 cos 3 6 ∘ − 2 1 ​ ) = 8 1 ​ ( − 2 cos 7 2 ∘ + 2 cos 3 6 ∘ ) × cos 1 8 ∘ cos 1 8 ∘ ​ − 16 1 ​ = 8 1 ​ cos 1 8 ∘ ( − 2 cos 7 2 ∘ cos 1 8 ∘ + 2 cos 3 6 ∘ cos 1 8 ∘ ) ​ − 16 1 ​ = − 16 1 ​ + 8 1 ​ cos 1 8 ∘ ( − ( cos ( 7 2 ∘ + 1 8 ∘ ) + cos ( 7 2 ∘ − 1 8 ∘ ) ) ) ​ + cos 1 8 ∘ ( cos ( 3 6 ∘ + 1 8 ∘ ) + cos ( 3 6 ∘ − 1 8 ∘ ) ) ​ = − 16 1 ​ + 8 1 ​ cos 1 8 ∘ ( − ( cos 9 0 ∘ + cos 5 4 ∘ ) + ( cos 5 4 ∘ + cos 1 8 ∘ ) ) ​ = − 16 1 ​ + 8 1 ​ cos 1 8 ∘ ( − 0 − cos 5 4 ∘ + cos 5 4 ∘ + cos 1 8 ∘ ) ​ = − 16 1 ​ + 8 1 ​ cos 1 8 ∘ cos 1 8 ∘ ​ = − 16 1 ​ + 8 1 ​ = 16 − 1 + 2 ​ = 16 1 ​ Sehingga didapat tan 6 ∘ tan 4 2 ∘ tan 6 6 ∘ tan 7 8 ∘ ​ = = = ​ c o s 6 ∘ c o s 4 2 ∘ c o s 6 6 ∘ c o s 7 8 ∘ s i n 6 ∘ s i n 4 2 ∘ s i n 6 6 ∘ s i n 7 8 ∘ ​ 16 1 ​ 16 1 ​ ​ 1 ​ Dengan demikian, terbukti bahwa tan 6 ∘ tan 4 2 ∘ tan 6 6 ∘ tan 7 8 ∘ = 1 .

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