Rumus Trigonometri Sudut Rangkap atau Ganda merupakan lanjutan dari Rumus Jumlah dan Selisih Sudut Trigonometri
Setelah sebelumnya kita mempelajari tentang Rumus Jumlah dan Selisih Sudut Trigonometri↝
, materi kita lanjutkan ke Rumus Trigonometri Sudut Rangkap . Sudut rangkap yang dimaksud adalah 2 α 2\alpha 2 α 1 2 α \dfrac12 \alpha 2 1 α
Seperti halnya rumus jumlah dan selisih dua sudut, rumus sudut rangkap digunakan untuk menentukan nilai trigonometri untuk suatu sudut yang bukan merupakan sudut istimewa (0 ∘ , 3 0 ∘ , 4 5 ∘ , 6 0 ∘ , 9 0 ∘ 0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ 0 ∘ , 3 0 ∘ , 4 5 ∘ , 6 0 ∘ , 9 0 ∘
Bagaimana menggunakan rumus sudut rangkap ini? simak ulasan berikut.
1. Rumus Trigonometri Sudut Rangkap (Ganda) Berikut rumus-rumus trigonometri sudut rangkap sin 2 α , cos 2 α , tan 2 α \sin 2\alpha, \cos 2\alpha, \tan 2\alpha sin 2 α , cos 2 α , tan 2 α
sin 2 α = 2 sin α cos α cos 2 α = 2 cos 2 α − 1 cos 2 α = 1 − 2 sin 2 α tan 2 α = 2 tan α 1 − tan 2 α \sin 2\alpha=2\sin\alpha \cos\alpha\\ \cos 2\alpha=2\cos^2\alpha-1 \\ \cos 2\alpha = 1-2\sin^2\alpha \\ \tan 2\alpha = \frac{2\tan\alpha}{1-\tan^2\alpha} sin 2 α = 2 sin α cos α cos 2 α = 2 cos 2 α − 1 cos 2 α = 1 − 2 sin 2 α tan 2 α = 1 − tan 2 α 2 tan α
Rumus tersebut dapat kita peroleh atau kita buktikan dengan menggunakan akan rumus trigonometri jumlah dua sudut.
Pembuktian rumus trigonometri sudut rangkap : ★ \bigstar ★ Rumus Sinus sudut rangkap sin 2 α = 2 sin α cos α \sin 2\alpha = 2 \sin \alpha \cos \alpha sin 2 α = 2 sin α cos α
Ingat rumus sinus jumlah sudut: sin ( A + B ) = sin A cos B + cos A sin B \sin (A+B) = \sin A \cos B + \cos A \sin B sin ( A + B ) = sin A cos B + cos A sin B
sin 2 α = sin ( α + α ) = sin α cos α + cos α sin α = 2 sin α cos α \begin{align*}\sin 2\alpha &= \sin ( \alpha + \alpha ) \\ &= \sin \alpha \cos \alpha + \cos \alpha \sin \alpha \\ &= 2 \sin \alpha \cos \alpha \end{align*} sin 2 α = sin ( α + α ) = sin α cos α + cos α sin α = 2 sin α cos α
Sehingga terbukti : sin 2 α = 2 sin α cos α \sin 2\alpha = 2 \sin \alpha \cos \alpha sin 2 α = 2 sin α cos α
★ \bigstar ★ Rumus Cosinus sudut rangkap : cos 2 α = cos 2 α − sin 2 α \cos 2\alpha = \cos ^2 \alpha - \sin ^2 \alpha cos 2 α = cos 2 α − sin 2 α
Ingat rumus cos ( A + B ) = cos A cos B − sin 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
cos 2 α = cos ( α + α ) = cos α cos α − sin α sin α = cos 2 α − sin 2 α \begin{align*} \cos 2\alpha &= \cos (\alpha + \alpha ) \\ &= \cos \alpha \cos \alpha - \sin \alpha \sin \alpha \\ &= \cos ^2 \alpha - \sin ^2 \alpha \end{align*} cos 2 α = cos ( α + α ) = cos α cos α − sin α sin α = cos 2 α − sin 2 α
Sehingga terbukti : cos 2 α = cos 2 α − sin 2 α \cos 2\alpha = \cos ^2 \alpha - \sin ^2 \alpha cos 2 α = cos 2 α − sin 2 α
Dengan menggunakan rumus identitas: sin 2 A + cos 2 A = 1 → sin 2 A = 1 − cos 2 A \sin ^2 A + \cos ^2 A = 1 \rightarrow \sin ^2 A = 1 - \cos ^2 A sin 2 A + cos 2 A = 1 → sin 2 A = 1 − cos 2 A
★ \bigstar ★ Rumus : cos 2 α = 2 cos 2 α − 1 \cos 2\alpha = 2\cos ^2 \alpha - 1 cos 2 α = 2 cos 2 α − 1
cos 2 α = cos 2 α − sin 2 α = cos 2 α − ( 1 − cos 2 α ) = 2 cos 2 α − 1 \begin{align*}\cos 2\alpha &= \cos ^2 \alpha - \sin ^2 \alpha \\ &= \cos ^2 \alpha - (1 - \cos ^2 \alpha ) \\ &= 2\cos ^2 \alpha - 1 \end{align*} cos 2 α = cos 2 α − sin 2 α = cos 2 α − ( 1 − cos 2 α ) = 2 cos 2 α − 1
Terbukti : cos 2 α = 2 cos 2 α − 1 \cos 2\alpha = 2\cos ^2 \alpha - 1 cos 2 α = 2 cos 2 α − 1
★ \bigstar ★ Rumus : cos 2 α = 1 − 2 sin 2 α \cos 2\alpha = 1 - 2\sin ^2 \alpha cos 2 α = 1 − 2 sin 2 α
cos 2 α = cos 2 α − sin 2 α = ( 1 − sin 2 α ) − sin 2 α = 1 − 2 sin 2 α \begin{align*}\cos 2\alpha &= \cos ^2 \alpha - \sin ^2 \alpha \\ &= ( 1 - \sin ^2 \alpha ) - \sin ^2 \alpha \\ &= 1 - 2\sin ^2 \alpha \end{align*} cos 2 α = cos 2 α − sin 2 α = ( 1 − sin 2 α ) − sin 2 α = 1 − 2 sin 2 α
Terbukti : cos 2 α = 1 − 2 sin 2 α \cos 2\alpha = 1 - 2\sin ^2 \alpha cos 2 α = 1 − 2 sin 2 α
★ \bigstar ★ Rumus Sudut Rangkap Tangen : tan 2 α = 2 tan α 1 − tan 2 α \tan 2\alpha = \frac{2\tan \alpha }{1 - \tan ^2 \alpha } tan 2 α = 1 − t a n 2 α 2 t a n α
Ingat rumus : tan ( A + B ) = tan A + tan B 1 − tan A tan B \tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} tan ( A + B ) = 1 − t a n A t a n B t a n A + t a n B
tan 2 α = tan ( α + α ) = tan α + tan α 1 − tan α tan α = 2 tan α 1 − tan 2 α \begin{align*} \tan 2\alpha &= \tan ( \alpha + \alpha ) \\ &= \frac{\tan \alpha + \tan \alpha }{1 - \tan \alpha \tan \alpha } \\ &= \frac{2\tan \alpha }{1 - \tan ^2 \alpha } \end{align*} tan 2 α = tan ( α + α ) = 1 − tan α tan α tan α + tan α = 1 − tan 2 α 2 tan α
Sehingga terbukti : tan 2 α = 2 tan α 1 − tan 2 α \tan 2\alpha = \dfrac{2\tan \alpha }{1 - \tan ^2 \alpha } tan 2 α = 1 − tan 2 α 2 tan α
Contoh Soal Sudut Rangkap Sederhanakan bentuk-bentuk berikut
2 sin 22 , 5 ∘ cos 22 , 5 ∘ 2 \sin 22,5^\circ \cos 22,5^\circ 2 sin 22 , 5 ∘ cos 22 , 5 ∘ 2 cos 2 67 , 5 ∘ – 1 2 \cos^2 67,5^\circ – 1 2 cos 2 67 , 5 ∘ –1 2 tan 3 α 1 − tan 2 α \dfrac{2\tan 3\alpha}{1-\tan^2\alpha} 1 − tan 2 α 2 tan 3 α Penyelesaian
2 sin 22 , 5 ∘ cos 22 , 5 ∘ 2 \sin 22,5^\circ \cos 22,5^\circ 2 sin 22 , 5 ∘ cos 22 , 5 ∘ sin 2 A = 2 sin A cos A \sin 2A=2\sin A \cos A sin 2 A = 2 sin A cos A 2 sin 22 , 5 ∘ cos 22 , 5 ∘ = sin 2 ( 22 , 5 ∘ ) = sin 4 5 ∘ = 1 2 2 \begin{align*}2 \sin 22,5^\circ \cos 22,5^\circ&=\sin 2(22,5^\circ)\\&=\sin 45^\circ \\ &=\frac12 \sqrt{2}\end{align*} 2 sin 22 , 5 ∘ cos 22 , 5 ∘ = sin 2 ( 22 , 5 ∘ ) = sin 4 5 ∘ = 2 1 2 2 cos 2 67 , 5 ∘ – 1 2 \cos^2 67,5^\circ – 1 2 cos 2 67 , 5 ∘ –1 cos 2 A = 2 cos 2 A − 1 \cos 2A=2\cos^2 A -1 cos 2 A = 2 cos 2 A − 1 2 cos 2 67 , 5 ∘ – 1 = cos 2 ( 67 , 5 ∘ ) = cos 13 5 ∘ = cos ( 18 0 ∘ − 4 5 ∘ ) = − cos 4 5 ∘ = − 1 2 2 \begin{align*}2 \cos^2 67,5^\circ – 1&=\cos 2(67,5^\circ)\\&=\cos 135^\circ = \cos (180^\circ - 45^\circ) \\ &=-\cos 45^\circ =-\frac12 \sqrt{2}\end{align*} 2 cos 2 67 , 5 ∘ –1 = cos 2 ( 67 , 5 ∘ ) = cos 13 5 ∘ = cos ( 18 0 ∘ − 4 5 ∘ ) = − cos 4 5 ∘ = − 2 1 2 2 tan 3 α 1 − tan 2 α \dfrac{2\tan 3\alpha}{1-\tan^2\alpha} 1 − tan 2 α 2 tan 3 α tan 2 A = 2 tan A 1 − tan 2 A \tan 2A=\dfrac{2\tan A}{1-\tan^2A} tan 2 A = 1 − tan 2 A 2 tan A 2 tan 3 α 1 − tan 2 α = tan 2 ( 3 α ) = tan 6 α \begin{align*}\frac{2\tan 3\alpha}{1- \tan^2 \alpha }&=\tan 2(3\alpha) \\ &=\tan 6\alpha \end{align*} 1 − tan 2 α 2 tan 3 α = tan 2 ( 3 α ) = tan 6 α Diketahui sin A = 5 13 \sin A =\dfrac{5}{13} sin A = 13 5 sin 2 A \sin 2A sin 2 A cos 2 A \cos 2A cos 2 A tan 2 A \tan 2A tan 2 A
Penyelesaian
Karena sin A \sin A sin A cos A \cos A cos A
Kita gunakan rumus identitas saja yaitu cos A = 1 − sin 2 A \cos A = \sqrt{1-\sin^2A} cos A = 1 − sin 2 A
sin A = 5 13 \sin A =\dfrac{5}{13} sin A = 13 5
cos A = 1 − sin 2 A = 1 − ( 5 13 ) 2 = 1 − 25 169 = 144 169 cos A = 12 13 \begin{align*} \cos A &= \sqrt{1-\sin^2A}\\ &= \sqrt{1-\left( \frac{5}{13} \right)^2} \\ &=\sqrt{1- \frac{25}{169} } \\ &=\sqrt{\frac{144}{169}} \\ \cos A &= \frac{12}{13}\end{align*} cos A cos A = 1 − sin 2 A = 1 − ( 13 5 ) 2 = 1 − 169 25 = 169 144 = 13 12
Nilai sin 2 A \sin 2A sin 2 A
sin 2 A = 2 sin A cos A = 2 ⋅ 5 13 ⋅ 12 13 sin 2 A = 120 169 \begin{align*}\sin 2 A &=2\sin A \cos A \\ &= 2 \cdot \frac{5}{13} \cdot \frac{12}{13} \\ \sin 2A&= \frac{120}{169}\end{align*} sin 2 A sin 2 A = 2 sin A cos A = 2 ⋅ 13 5 ⋅ 13 12 = 169 120
Nilai cos 2 A \cos 2A cos 2 A
cos 2 A = cos 2 A − sin 2 A = ( 12 13 ) 2 − ( 5 13 ) 2 = 144 169 − 25 169 cos 2 A = 119 169 \begin{align*}\cos 2 A &=\cos^2 A - \sin^2 A \\ &= \left(\frac{12}{13} \right)^2- \left(\frac{5}{13} \right)^2\\ &= \frac{144}{169}-\frac{25}{169}\\ \cos 2A &= \frac{119}{169} \end{align*} cos 2 A cos 2 A = cos 2 A − sin 2 A = ( 13 12 ) 2 − ( 13 5 ) 2 = 169 144 − 169 25 = 169 119
Nilai tan 2 A \tan 2A tan 2 A
tan 2 A = sin 2 A cos 2 A = 120 169 119 169 tan 2 A = 120 119 \begin{align*}\tan 2A &=\frac{\sin 2A}{\cos 2A} \\ &=\frac{\frac{120}{169}}{\frac{119}{169}} \\ \tan 2A &=\frac{120}{119} \end{align*} tan 2 A tan 2 A = cos 2 A sin 2 A = 169 119 169 120 = 119 120
2. Rumus Trigonometri untuk Setengah Sudut Dari rumus trigonometri sudut ganda, dapat diturunkan rumus trigonometri untuk
setengah sudut, yaitu dengan menetapkan 1 2 α \dfrac12 \alpha 2 1 α α \alpha α
Rumus Trigonometri untuk sin 1 2 A , cos 1 2 A , \sin \frac{1}{2} A , \cos \frac{1}{2} A, sin 2 1 A , cos 2 1 A , tan 1 2 A \tan \frac{1}{2} A tan 2 1 A
sin 1 2 A = 1 − cos A 2 cos 1 2 A = 1 + cos A 2 tan 1 2 A = 1 − cos A 1 + cos A = sin A 1 + cos A = 1 − cos A sin A \begin{align*} \sin \frac{1}{2} A & = \sqrt{\frac{1- \cos A}{2}} \\ \cos \frac{1}{2} A & = \sqrt{\frac{1 + \cos A}{2}} \\ \tan \frac{1}{2} A & = \sqrt{\frac{1 - \cos A}{1 + \cos A}} = \frac{\sin A}{1+ \cos A} = \frac{1- \cos A}{\sin A } \end{align*} sin 2 1 A cos 2 1 A tan 2 1 A = 2 1 − cos A = 2 1 + cos A = 1 + cos A 1 − cos A = 1 + cos A sin A = sin A 1 − cos A
Pembuktian Rumus sudut 1 2 A \frac{1}{2} A 2 1 A Misalkan 2 α = A → α = 1 2 A 2\alpha = A \rightarrow \alpha = \frac{1}{2} A 2 α = A → α = 2 1 A
Substitusi bentuk permisalan di atas ke persamaan trigonometri sudut rangkap yang akan dibuktikan.
★ \bigstar ★ Rumus : sin 1 2 A = 1 − cos A 2 \sin \frac{1}{2} A = \sqrt{\frac{1- \cos A}{2}} sin 2 1 A = 2 1 − c o s A
gunakan rumus : cos 2 α = 1 − 2 sin 2 α \cos 2 \alpha = 1 - 2\sin ^2 \alpha cos 2 α = 1 − 2 sin 2 α
cos 2 α = 1 − 2 sin 2 α cos A = 1 − 2 sin 2 1 2 A 2 sin 2 1 2 A = 1 − cos A sin 2 1 2 A = 1 − cos A 2 sin 1 2 A = 1 − cos A 2 \begin{align*}\cos 2 \alpha &= 1 - 2\sin ^2 \alpha \\ \cos A &= 1 - 2\sin ^2 \frac{1}{2} A\\2\sin ^2 \frac{1}{2} A &= 1 - \cos A \\\sin ^2 \frac{1}{2} A &= \frac{1 - \cos A}{2} \\\sin \frac{1}{2} A &= \sqrt{\frac{1 - \cos A}{2} } \end{align*} cos 2 α cos A 2 sin 2 2 1 A sin 2 2 1 A sin 2 1 A = 1 − 2 sin 2 α = 1 − 2 sin 2 2 1 A = 1 − cos A = 2 1 − cos A = 2 1 − cos A
Sehingga terbukti : sin 1 2 A = 1 − cos A 2 \sin \frac{1}{2} A = \sqrt{\frac{1- \cos A}{2}} sin 2 1 A = 2 1 − c o s A
★ \bigstar ★ Rumus : cos 1 2 A = 1 + cos A 2 \cos \frac{1}{2} A = \sqrt{\frac{1 + \cos A}{2}} cos 2 1 A = 2 1 + c o s A
gunakan rumus : cos 2 α = 2 cos 2 α − 1 \cos 2 \alpha = 2\cos ^2 \alpha - 1 cos 2 α = 2 cos 2 α − 1
cos 2 α = 2 cos 2 α − 1 cos A = 2 cos 2 1 2 A − 1 2 cos 2 1 2 A = 1 + cos A cos 2 1 2 A = 1 + cos A 2 cos 1 2 A = 1 + cos A 2 \begin{align*} \cos 2 \alpha &= 2\cos ^2 \alpha - 1 \\\cos A &= 2\cos ^2 \frac{1}{2}A - 1 \\2\cos ^2 \frac{1}{2}A &= 1 + \cos A \\\cos ^2 \frac{1}{2}A &= \frac{1 + \cos A}{2} \\\cos \frac{1}{2}A &= \sqrt{\frac{1 + \cos A}{2} } \end{align*} cos 2 α cos A 2 cos 2 2 1 A cos 2 2 1 A cos 2 1 A = 2 cos 2 α − 1 = 2 cos 2 2 1 A − 1 = 1 + cos A = 2 1 + cos A = 2 1 + cos A
Sehingga terbukti : cos 1 2 A = 1 + cos A 2 \cos \frac{1}{2} A = \sqrt{\frac{1 + \cos A}{2}} cos 2 1 A = 2 1 + c o s A
★ \bigstar ★ Rumus : tan 1 2 A = 1 − cos A 1 + cos A = sin A 1 + cos A = 1 − cos A sin A \tan \frac{1}{2} A = \sqrt{\frac{1 - \cos A}{1 + \cos A}} = \frac{\sin A}{1+ \cos A} = \frac{1- \cos A}{\sin A } tan 2 1 A = 1 + c o s A 1 − c o s A = 1 + c o s A s i n A = s i n A 1 − c o s A
gunakan rumus : tan 1 2 A = sin 1 2 A cos 1 2 A , sin 1 2 A = 1 − cos A 2 \tan \frac{1}{2}A = \frac{\sin \frac{1}{2}A }{\cos \frac{1}{2}A } , \sin \frac{1}{2} A = \sqrt{\frac{1- \cos A}{2}} tan 2 1 A = c o s 2 1 A s i n 2 1 A , sin 2 1 A = 2 1 − c o s A cos 1 2 A = 1 + cos A 2 \cos \frac{1}{2} A = \sqrt{\frac{1 + \cos A}{2}} cos 2 1 A = 2 1 + c o s A
a). Rumus Pertama : tan 1 2 A = sin 1 2 A cos 1 2 A \tan \frac{1}{2}A = \frac{\sin \frac{1}{2}A }{\cos \frac{1}{2}A } tan 2 1 A = c o s 2 1 A s i n 2 1 A
tan 1 2 A = sin 1 2 A cos 1 2 A = 1 − cos A 2 1 + cos A 2 tan 1 2 A = 1 − cos A 1 + cos A \begin{align*}\tan \frac{1}{2} A &= \frac{\sin \frac{1}{2}A }{\cos \frac{1}{2}A } \ &= \frac{ \sqrt{\frac{1- \cos A}{2}} }{ \sqrt{\frac{1 + \cos A}{2}} } \ \tan \frac{1}{2}A &= \sqrt{ \frac{1- \cos A}{1 + \cos A} } \end{align*} tan 2 1 A = cos 2 1 A sin 2 1 A = 2 1 + c o s A 2 1 − c o s A tan 2 1 A = 1 + cos A 1 − cos A
b). Rumus kedua :
tan 1 2 A = 1 − cos A 1 + cos A tan 1 2 A = 1 − cos A 1 + cos A × 1 + cos A 1 + cos A = 1 − cos 2 A ( 1 + cos A ) 2 = sin 2 A ( 1 + cos A ) 2 = sin A 1 + cos A \begin{align*} \tan \frac{1}{2}A &= \sqrt{ \frac{1- \cos A}{1 + \cos A} } \\ \tan \frac{1}{2}A &= \sqrt{ \frac{1- \cos A}{1 + \cos A} \times \frac{1 + \cos A}{1 + \cos A} } \\ &= \sqrt{ \frac{1- \cos ^2 A}{(1 + \cos A)^2} } \\ &= \sqrt{ \frac{\sin ^2 A }{(1 + \cos A)^2} } \\ &= \frac{\sin A}{1+ \cos A} \end{align*} tan 2 1 A tan 2 1 A = 1 + cos A 1 − cos A = 1 + cos A 1 − cos A × 1 + cos A 1 + cos A = ( 1 + cos A ) 2 1 − cos 2 A = ( 1 + cos A ) 2 sin 2 A = 1 + cos A sin A
c). Rumus ketiga :
tan 1 2 A = 1 − cos A 1 + cos A tan 1 2 A = 1 − cos A 1 + cos A × 1 − cos A 1 − cos A = ( 1 − cos A ) 2 1 − cos 2 A = ( 1 − cos A ) 2 sin 2 A = 1 − cos A sin A \begin{align*} \tan \frac{1}{2}A &= \sqrt{\frac{1- \cos A}{1 + \cos A} } \\ \tan \frac{1}{2}A &= \sqrt{\frac{1- \cos A}{1 + \cos A} \times \frac{1 - \cos A}{1 - \cos A} } \\ &= \sqrt{\frac{(1- \cos A)^2}{1 - \cos ^2 A} } \\ &= \sqrt{\frac{(1- \cos A)^2}{\sin ^2 A } } \\ &= \frac{1- \cos A}{\sin A } \end{align*} tan 2 1 A tan 2 1 A = 1 + cos A 1 − cos A = 1 + cos A 1 − cos A × 1 − cos A 1 − cos A = 1 − cos 2 A ( 1 − cos A ) 2 = sin 2 A ( 1 − cos A ) 2 = sin A 1 − cos A
Sehingga terbukti : tan 1 2 A = 1 − cos A 1 + cos A = sin A 1 + cos A = 1 − cos A sin A \tan \frac{1}{2} A = \sqrt{\frac{1 - \cos A}{1 + \cos A}} = \frac{\sin A}{1+ \cos A} = \frac{1- \cos A}{\sin A } tan 2 1 A = 1 + c o s A 1 − c o s A = 1 + c o s A s i n A = s i n A 1 − c o s A
Contoh Soal Setengah Sudut : Hitunglah nilai dari :
sin 1 5 ∘ \sin 15^\circ sin 1 5 ∘ cos 67 , 5 ∘ \cos 67,5^\circ cos 67 , 5 ∘ tan 22 , 5 ∘ \tan 22,5^\circ tan 22 , 5 ∘ Alternatif Penyelesaian :
Nilai dari sin 1 5 ∘ \sin 15^\circ sin 1 5 ∘
Misalkan 1 2 A = 1 5 ∘ → A = 3 0 ∘ \frac{1}{2}A = 15^\circ \rightarrow A = 30^\circ 2 1 A = 1 5 ∘ → A = 3 0 ∘
sin 1 2 A = 1 − cos A 2 sin 1 5 ∘ = 1 − cos 3 0 ∘ 2 = 1 − 1 2 3 2 = 2 − 3 4 = 1 2 2 − 3 \begin{align*}\sin \frac{1}{2} A &= \sqrt{\frac{1 - \cos A}{2} } \\ \sin 15^\circ &= \sqrt{\frac{1 - \cos 30^\circ}{2} } \\ &= \sqrt{\frac{1 - \frac{1}{2} \sqrt{3} }{2} } \\ &= \sqrt{\frac{2 - \sqrt{3} }{4} } \\ &= \frac{1}{2} \sqrt{2 - \sqrt{3} } \end{align*} sin 2 1 A sin 1 5 ∘ = 2 1 − cos A = 2 1 − cos 3 0 ∘ = 2 1 − 2 1 3 = 4 2 − 3 = 2 1 2 − 3
Jadi, nilai sin 1 5 ∘ = 1 2 2 − 3 \sin 15^\circ = \frac{1}{2} \sqrt{2 - \sqrt{3} } sin 1 5 ∘ = 2 1 2 − 3
Nilai dari cos 67 , 5 ∘ \cos 67,5^\circ cos 67 , 5 ∘
Misalkan 1 2 A = 67 , 5 ∘ → A = 13 5 ∘ \frac{1}{2}A = 67,5^\circ \rightarrow A = 135^\circ 2 1 A = 67 , 5 ∘ → A = 13 5 ∘
nilai cos 13 5 ∘ = cos ( 18 0 ∘ − 4 5 ∘ ) = − cos 4 5 ∘ = − 1 2 2 \cos 135^\circ = \cos ( 180^\circ - 45^\circ ) = -\cos 45^\circ = -\frac{1}{2} \sqrt{2} cos 13 5 ∘ = cos ( 18 0 ∘ − 4 5 ∘ ) = − cos 4 5 ∘ = − 2 1 2
cos 1 2 A = 1 + cos A 2 cos 67 , 5 ∘ = 1 + cos 13 5 ∘ 2 = 1 + ( − 1 2 2 ) 2 = 2 − 2 4 = 1 2 2 − 2 \begin{align*}\cos \frac{1}{2} A &= \sqrt{\frac{1 + \cos A}{2} } \\ \cos 67,5^\circ &= \sqrt{\frac{1 + \cos 135^\circ}{2} } \\ &= \sqrt{\frac{1 + (-\frac{1}{2} \sqrt{2} )}{2} } \\ &= \sqrt{\frac{2 - \sqrt{2} }{4} } \\ &= \frac{1}{2} \sqrt{2 - \sqrt{2} } \end{align*} cos 2 1 A cos 67 , 5 ∘ = 2 1 + cos A = 2 1 + cos 13 5 ∘ = 2 1 + ( − 2 1 2 ) = 4 2 − 2 = 2 1 2 − 2
Jadi, nilai cos 67 , 5 ∘ = 1 2 2 − 2 \cos 67,5^\circ = \frac{1}{2} \sqrt{2 - \sqrt{2} } cos 67 , 5 ∘ = 2 1 2 − 2
Nilai dari tan 22 , 5 ∘ \tan 22,5^\circ tan 22 , 5 ∘
Misalkan 1 2 A = 22 , 5 ∘ → A = 4 5 ∘ \frac{1}{2}A = 22,5^\circ \rightarrow A = 45^\circ 2 1 A = 22 , 5 ∘ → A = 4 5 ∘
tan 1 2 A = sin A 1 + cos A tan 22 , 5 ∘ = sin 4 5 ∘ 1 + cos 4 5 ∘ = 1 2 2 1 + 1 2 2 = 2 2 + 2 = 2 2 + 2 × 2 − 2 2 − 2 = 2 2 − 2 4 − 2 = 2 2 − 2 2 = 2 − 1 \begin{align*} \tan \frac{1}{2} A &= \frac{\sin A}{1+ \cos A} \\ \tan 22,5^\circ &= \frac{\sin 45^\circ}{1+ \cos 45^\circ} \\ &= \frac{\frac{1}{2} \sqrt{2} }{1+ \frac{1}{2} \sqrt{2} } \\ &= \frac{ \sqrt{2} }{2+ \sqrt{2} } \\ &= \frac{ \sqrt{2} }{2+ \sqrt{2} } \times \frac{2 - \sqrt{2} }{2 - \sqrt{2} } \\ &= \frac{ 2\sqrt{2} - 2 }{4 - 2} \\ &= \frac{ 2\sqrt{2} - 2 }{2} \\ &= \sqrt{2} - 1 \end{align*} tan 2 1 A tan 22 , 5 ∘ = 1 + cos A sin A = 1 + cos 4 5 ∘ sin 4 5 ∘ = 1 + 2 1 2 2 1 2 = 2 + 2 2 = 2 + 2 2 × 2 − 2 2 − 2 = 4 − 2 2 2 − 2 = 2 2 2 − 2 = 2 − 1
Jadi, nilai tan 22 , 5 ∘ = 2 − 1 \tan 22,5^\circ = \sqrt{2} - 1 tan 22 , 5 ∘ = 2 − 1
