3. Diketahui tan A tan 3 A ​ = K , tunjukkan bahwa: a. sin A sin 3 A ​ = K − 1 2 K ​

Pembahasan

Ingat! tan α = sin α cos α ​ sin ( α − β ) ​ = ​ sin α ⋅ cos β − cos α ⋅ sin β ​ sin 2 a = 2 ⋅ sin α ⋅ cos β Oleh karena tan A tan 3 A ​ = K , maka dengan mengurangi kedua ruas dengan 1, diperoleh: t a n A t a n 3 A ​ − 1 t a n A t a n 3 A ​ − t a n A t a n A ​ t a n A t a n 3 A − t a n A ​ c o s A s i n A ​ c o s 3 A s i n 3 A ​ − c o s A s i n A ​ ​ ( c o s 3 A s i n 3 A ​ − c o s A s i n A ​ ) × s i n A c o s A ​ ( c o s 3 A ⋅ c o s A s i n 3 A ⋅ c o s A ​ − c o s 3 A ⋅ c o s A c o s 3 A ⋅ s i n A ​ ) × s i n A c o s A ​ ( c o s 3 A ⋅ c o s A s i n 3 A ⋅ c o s A − c o s 3 A ⋅ s i n A ​ ) × s i n A c o s A ​ ( c o s 3 A ⋅ c o s A s i n ( 3 A − A ) ​ ) × s i n A c o s A ​ c o s 3 A ⋅ c o s A s i n 2 A ​ × s i n A c o s A ​ c o s 3 A ⋅ c o s A s i n 2 A ​ × s i n A c o s A ​ c o s 3 A ⋅ s i n A s i n 2 A ​ c o s 3 A ⋅ s i n A 2 ⋅ s i n A ⋅ c o s A ​ c o s 3 A ⋅ s i n A 2 ⋅ s i n A ⋅ c o s A ​ c o s 3 A 2 ⋅ c o s A ​ c o s 3 A c o s A ​ ​ = = = = = = = = = = = = = = = ​ K − 1 K − 1 K − 1 K − 1 K − 1 K − 1 K − 1 K − 1 K − 1 K − 1 K − 1 K − 1 K − 1 K − 1 2 K − 1 ​ ( 1 ) ​ Kemudian, perhatikan kembali tan A tan 3 A ​ = K ,dan karena tan A = cos A sin A ​ , maka t a n A t a n 3 A ​ c o s A s i n A ​ c o s 3 A s i n 3 A ​ ​ c o s 3 A s i n 3 A ​ × s i n A c o s A ​ s i n A s i n 3 A ​ × c o s 3 A c o s A ​ ​ = = = = ​ K K K K ( 2 ) ​ Dengan menyubtitusi (1) ke (2), diperoleh s i n A s i n 3 A ​ × 2 K − 1 ​ s i n A s i n 3 A ​ s i n A s i n 3 A ​ ​ = = = ​ K K × K − 1 2 ​ K − 1 2 K ​ ​ Dengan demikian, telah ditunjukkan bahwa sin A sin 3 A ​ = K − 1 2 K ​ .