2. Jika sin ( θ + B ) sin ( θ + A ) ​ = sin 2 B sin 2 A ​ ​ , buktikan bahwa: tan 2 θ = tan A ⋅ tan B .

Pembahasan

Ingat kembali: sin ( α + β ) ​ = ​ sin α ⋅ cos β + cos α ⋅ sin β ​ sin ( α − β ) ​ = ​ sin α ⋅ cos β − cos α ⋅ sin β ​ ​​​​ sin 2 α = 2 ⋅ sin α ⋅ cos α tan α = cos α sin α ​ Dengan menggunakan rumus-rumus di atas, dapat diperoleh s i n ( θ + B ) s i n ( θ + A ) ​ ( s i n ( θ + B ) s i n ( θ + A ) ​ ) 2 ( s i n ( θ + B ) ) 2 ( s i n ( θ + A ) ) 2 ​ sin 2 B ⋅ ( sin ( θ + A ) ) 2 ​ = = = = ​ s i n 2 B s i n 2 A ​ ​ s i n 2 B s i n 2 A ​ s i n 2 B s i n 2 A ​ sin 2 A ⋅ ( sin ( θ + B ) ) 2 ( 1 ) ​ Dengan menjabarkan ruas kiri pada (1), diperoleh: ​ = = = = = = ​ sin 2 B ⋅ ( sin ( θ + A ) ) 2 sin 2 B ⋅ ( sin θ ⋅ cos A + cos θ ⋅ sin A ) 2 sin 2 B ⋅ ( sin 2 θ ⋅ cos 2 A + cos 2 θ ⋅ sin 2 A + 2 ⋅ sin θ ⋅ cos A ⋅ cos θ ⋅ sin A ) sin 2 B ⋅ ( sin 2 θ ⋅ cos 2 A + cos 2 θ ⋅ sin 2 A + 2 ⋅ sin A ⋅ cos A ⋅ sin θ ⋅ cos θ ) sin 2 B ⋅ sin 2 θ ⋅ cos 2 A + sin 2 B ⋅ cos 2 θ ⋅ sin 2 A + sin 2 B ⋅ 2 ⋅ sin A ⋅ cos A ⋅ sin θ ⋅ cos θ 2 ⋅ sin B ⋅ cos B ⋅ sin 2 θ ⋅ cos 2 A + 2 ⋅ sin B ⋅ cos B ⋅ cos 2 θ ⋅ sin 2 A + 2 ⋅ sin B ⋅ cos B ⋅ 2 ⋅ sin A ⋅ cos A ⋅ sin θ ⋅ cos θ 2 ⋅ sin B ⋅ cos B ⋅ sin 2 θ ⋅ cos 2 A + 2 ⋅ sin B ⋅ cos B ⋅ cos 2 θ ⋅ sin 2 A + 4 ⋅ sin B ⋅ cos B ⋅ sin A ⋅ cos A ⋅ sin θ ⋅ cos θ ( 2 ) ​ Dengan menjabarkan ruas kananpada (1), diperoleh: ​ = = = = = = ​ sin 2 A ⋅ ( sin ( θ + B ) ) 2 sin 2 A ( sin θ ⋅ cos B + cos θ ⋅ sin B ) 2 sin 2 A ⋅ ( sin 2 θ ⋅ cos 2 B + cos 2 θ ⋅ sin 2 B + 2 ⋅ sin θ ⋅ cos B ⋅ cos θ ⋅ sin B ) sin 2 A ⋅ ( sin 2 θ ⋅ cos 2 B + cos 2 θ ⋅ sin 2 B + 2 ⋅ sin B ⋅ cos B ⋅ sin θ ⋅ cos θ ) sin 2 A ⋅ sin 2 θ ⋅ cos 2 B + sin 2 B ⋅ cos 2 θ ⋅ sin 2 B + sin 2 A ⋅ 2 ⋅ sin B ⋅ cos B ⋅ sin θ ⋅ cos θ 2 ⋅ sin A ⋅ cos A ⋅ sin 2 θ ⋅ cos 2 B + 2 ⋅ sin A ⋅ cos A ⋅ cos 2 θ ⋅ sin 2 B + 2 ⋅ sin A ⋅ cos A ⋅ 2 ⋅ sin B ⋅ cos B ⋅ sin θ ⋅ cos θ 2 ⋅ sin A ⋅ cos A ⋅ sin 2 θ ⋅ cos 2 B + 2 ⋅ sin A ⋅ cos A ⋅ cos 2 θ ⋅ sin 2 B + 4 ⋅ sin A ⋅ cos A ⋅ sin B ⋅ cos B ⋅ sin θ ⋅ cos θ ( 3 ) ​ Ubah (2) dan (3) pada (1) sehingga diperoleh: c o s 2 θ s i n 2 θ ​ ( c o s θ s i n θ ​ ) 2 ( c o s θ s i n θ ​ ) 2 ( c o s θ s i n θ ​ ) 2 ( c o s θ s i n θ ​ ) 2 maka,diperoleh: ( tan θ ) 2 tan 2 θ ​ = = = = = = = ​ ( 2 ⋅ s i n B ⋅ c o s B ⋅ c o s 2 A − 2 ⋅ s i n A ⋅ c o s A ⋅ c o s 2 B ) ( 2 ⋅ s i n A ⋅ c o s A ⋅ s i n 2 B − 2 ⋅ s i n B ⋅ c o s B ⋅ s i n 2 A ) ​ 2 ⋅ c o s A ⋅ c o s B ⋅ ( s i n B ⋅ c o s A − s i n A ⋅ c o s B ) 2 ⋅ s i n A ⋅ s i n B ⋅ ( c o s A ⋅ s i n B − c o s B ⋅ s i n A ) ​ 2 ​ ⋅ c o s A ⋅ c o s B ⋅ ( c o s A ⋅ s i n B − s i n A ⋅ c o s B ) ​ 2 ​ ⋅ s i n A ⋅ s i n B ⋅ ( c o s A ⋅ s i n B − s i n A ⋅ c o s B ) ​ ​ c o s A ⋅ c o s B s i n A ⋅ s i n B ​ c o s A s i n A ​ ⋅ c o s B s i n B ​ karena t a n α = c o s α s i n α ​ tan A ⋅ tan B tan A ⋅ tan B ​ Dengan demikian, telah ditunjukan bahwa tan 2 θ = tan A ⋅ tan B .