1. Jika tan β = 1 + tan α ⋅ tan γ tan α + tan γ ​ , tunjukkan bahwa: sin 2 β = 1 + sin 2 α ⋅ sin 2 γ sin 2 α + sin 2 γ ​ .

Pembahasan

Ingat kembali: tan A = cos A sin A ​ sin ( A + B ) ​ = ​ sin A ⋅ cos B + cos A ⋅ sin B ​ sin ( A − B ) ​ = ​ sin A ⋅ cos B − cos A ⋅ sin B ​ cos ( A + B ) ​ = ​ cos A ⋅ cos B − sin A ⋅ sin B ​ cos ( A − B ) ​ = ​ cos A ⋅ cos B + sin A ⋅ sin B ​ sin 2 A = 1 + tan 2 A 2 ⋅ tan A ​ 2 ⋅ sin A ⋅ cos B = sin ( A + B ) + sin ( A − B ) 2 ⋅ cos 2 A = 1 + cos 2 A 2 ⋅ sin 2 A = 1 − cos 2 A Dengan menggunakan rumus-rumus di atas, dapat diperoleh tan β ​ = = = = = = ​ 1 + t a n α ⋅ t a n γ t a n α + t a n γ ​ 1 + c o s α s i n α ​ ⋅ c o s γ s i n γ ​ c o s α s i n α ​ + c o s γ s i n γ ​ ​ c o s α ⋅ c o s γ c o s α ⋅ c o s γ ​ + c o s α ⋅ c o s γ s i n α ⋅ s i n γ ​ c o s α ⋅ c o s γ s i n α ⋅ c o s γ ​ + c o s α ⋅ c o s γ c o s α ⋅ s i n γ ​ ​ c o s α ⋅ c o s γ c o s α ⋅ c o s γ + s i n α ⋅ s i n γ ​ c o s α ⋅ c o s γ s i n α ⋅ c o s γ + c o s α ⋅ s i n γ ​ ​ c o s α ⋅ c o s γ + s i n α ⋅ s i n γ s i n α ⋅ c o s γ + c o s α ⋅ s i n γ ​ c o s ( α − γ ) s i n ( α + γ ) ​ ​ sehingga diperoleh sin 2 β sin 2 β ​ = = = = = = = = = = = = = = ​ 1 + t a n 2 β 2 ⋅ t a n β ​ 1 + ( c o s ( α − γ ) s i n ( α + γ ) ​ ) 2 2 ⋅ c o s ( α − γ ) s i n ( α + γ ) ​ ​ c o s 2 ( α − γ ) c o s 2 ( α − γ ) ​ + c o s 2 ( α − γ ) s i n 2 ( α + γ ) ​ 2 ⋅ c o s ( α − γ ) s i n ( α + γ ) ​ ​ c o s 2 ( α − γ ) c o s 2 ( α − γ ) + s i n 2 ( α + γ ) ​ 2 ⋅ c o s ( α − γ ) s i n ( α + γ ) ​ ​ c o s ( α − γ ) ​ 1 2 ⋅ s i n ( α + γ ) ​ × c o s 2 ( α − γ ) + s i n 2 ( α + γ ) c o s 2 ( α − γ ) ​ c o s ( α − γ ) ​ c o s 2 ( α − γ ) + s i n 2 ( α + γ ) 2 ⋅ s i n ( α + γ ) ⋅ c o s ( α − γ ) ​ c o s 2 ( α − γ ) + s i n 2 ( α + γ ) s i n ( α + γ + α − γ ) + s i n ( α + γ − ( α − γ ) ) ​ c o s 2 ( α − γ ) + s i n 2 ( α + γ ) s i n 2 α + s i n 2 γ ​ 2 ⋅ c o s 2 ( α − γ ) + 2 ⋅ s i n 2 ( α + γ ) 2 ⋅ ( s i n 2 α + s i n 2 γ ) ​ 1 + c o s 2 ( α − γ ) + 1 − c o s 2 ( α + γ ) 2 ⋅ ( s i n 2 α + s i n 2 γ ) ​ 2 + c o s ( 2 α − 2 γ ) − c o s ( 2 α + 2 γ ) 2 ⋅ ( s i n 2 α + s i n 2 γ ) ​ Olehkarena c o s ( 2 α − 2 γ ) − c o s ( 2 α + 2 γ ) = c o s 2 α ⋅ c o s 2 γ + s i n 2 α ⋅ s i n 2 γ − ( c o s 2 α ⋅ c o s 2 γ − s i n 2 α ⋅ s i n 2 γ ) = c o s 2 α ⋅ c o s 2 γ + s i n 2 α ⋅ s i n 2 γ − c o s 2 α ⋅ c o s 2 γ + s i n 2 α ⋅ s i n 2 γ ) = c o s 2 α ⋅ c o s 2 γ − c o s 2 α ⋅ c o s 2 γ + s i n 2 α ⋅ s i n 2 γ + s i n 2 α ⋅ s i n 2 γ = 2 ⋅ s i n 2 α ⋅ s i n 2 γ makadiperoleh ​ 2 + 2 ⋅ s i n 2 α ⋅ s i n 2 γ 2 ⋅ ( s i n 2 α + s i n 2 γ ) ​ 2 ⋅ ( 1 + s i n 2 α ⋅ s i n 2 γ ) 2 ⋅ ( s i n 2 α + s i n 2 γ ) ​ 1 + s i n 2 α ⋅ s i n 2 γ s i n 2 α + s i n 2 γ ​ ​ Dengan demikian, telah ditunjukan bahwa sin 2 β = 1 + sin 2 α ⋅ sin 2 γ sin 2 α + sin 2 γ ​ .