Jika A+B = 30°, maka Cos A =

Berikut ini adalah pertanyaan dari lavankayla pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Jika A+B = 30°, maka

$\boxed{\bf{Cos\left(30^{\circ}-B\right)=\frac{1}{2}\left(\sqrt{3}\cos B+\sin B\right)}}$

$\:$

Trigonometri

Pendahuluan

A.) Definisi

.) Perbandingan Trigonometri

Pada segitiga siku-siku ABC, berlaku :

*Gambar ke-1

$\small\mathbf{\left(a.\right)\ \ \sin\alpha=\frac{y}{r}=\frac{de}{mi}}$

$\small\mathbf{\left(b.\right)\ \ \cos\alpha=\frac{x}{r}=\frac{sa}{mi}}$

$\small\mathbf{\left(c.\right)\ \ \tan\alpha=\frac{y}{x}=\frac{de}{sa}}$

$\small\mathbf{\left(d.\right)\ \ \csc\alpha=\frac{1}{\sin\alpha}=\frac{r}{y}}$

$\small\mathbf{\left(e.\right)\ \ \sec\alpha=\frac{1}{\cos\alpha}=\frac{r}{x}}$

$\small\mathbf{\left(f.\right)\ \ \cot\alpha=\frac{1}{\tan\alpha}=\frac{y}{x}}$

B.) Sudut dan Kuadran

1.) Pembagian Daerah

$\boxed{\begin{array}{c|c|c|c|c}\underline{\mathbf{Kuadran}}&\underline{\mathbf{I}}&\underline{\mathbf{II}}&\underline{\mathbf{III}}&\underline{\mathbf{IV}}\\&&&\\\mathbf{absis(x)}&\mathbf{+}&\mathbf{-}&\mathbf{-}&\mathbf{+}\\&&&\\\mathbf{Ordinat(y)}&\mathbf{+}&\mathbf{+}&\mathbf{-}&\mathbf{-}\end{array}}$

2.) Tanda-tanda Fungsi

$\boxed{\begin{array}{c|c|c|c|c}\underline{\mathbf{Kuadran}}&\underline{\mathbf{I}}&\underline{\mathbf{II}}&\underline{\mathbf{III}}&\underline{\mathbf{IV}}\\&&&\\\mathbf{sin}&\mathbf{+}&\mathbf{+}&\mathbf{-}&\mathbf{-}\\&&&\\\mathbf{cos}&\mathbf{+}&\mathbf{-}&\mathbf{-}&\mathbf{+}\\&&&\\\mathbf{tan}&\mathbf{+}&\mathbf{-}&\mathbf{+}&\mathbf{-}\end{array}}$

3.) Sudut-sudut Istimewa

$\boxed{\begin{array}{c|c|c|c|c}\underline{\mathbf{Kuadran}}&\underline{\mathbf{0^{\circ}}}&\underline{\mathbf{30^{\circ}}}&\underline{\mathbf{45^{\circ}}}&\underline{\mathbf{60^{\circ}}}\\&&&\\\mathbf{sin}&\mathbf{0}&\mathbf{\frac{1}{2}}&\mathbf{\frac{1}{2}\sqrt{2}}&\mathbf{\frac{1}{2}\sqrt{3}}\\&&&\\\mathbf{cos}&\mathbf{1}&\mathbf{\frac{1}{2}\sqrt{3}}&\mathbf{\frac{1}{2}\sqrt{2}}&\mathbf{\frac{1}{2}}\\&&&\\\mathbf{tan}&\mathbf{0}&\mathbf{\frac{1}{3}\sqrt{3}}&\mathbf{1}&\mathbf{\sqrt{3}}\end{array}}$ $\boxed{\begin{array}{c}\underline{\mathbf{90^{\circ}}}\\\\\mathbf{1}\\\\\mathbf{0}\\\\\infty\end{array}}$

4.) Sudut Berelasi

a. Kalau kita gunakan (90°± ...) atau (270°± ...)

1.) Fungsi berubah

$\boxed{\begin{array}{c|c}\underline{\mathbf{Mula-mula}}&\underline{\mathbf{Perubahan}}\\\\\mathbf{sin}&\mathbf{+/-cos}\\\\\mathbf{cos}&\mathbf{+/-sin}\\\\\mathbf{tan}&\mathbf{+/-cot}\end{array}}$

2.) Tanda +/- mengikuti kuadran

b. kalau kita gunakan (180°± ...) atau (360°− ...)

1.) Fungsi tetap

$\boxed{\begin{array}{c|c}\underline{\mathbf{Mula-mula}}&\underline{\mathbf{Perubahan}}\\\\\mathbf{sin}&\mathbf{+/-sin}\\\\\mathbf{cos}&\mathbf{+/-cos}\\\\\mathbf{tan}&\mathbf{+/-tan}\end{array}}$

C.) Rumus trigonometri 2 sudut

$\bf{\sin\left(A+B\right)=\sin A\cos B+\cos A\sin B}$

$\bf{\sin\left(A-B\right)=\sin A\cos B-\cos A\sin B}$

$\bf{\cos\left(A+B\right)=\cos A\cos B-\sin A\sin B}$

$\bf{\cos\left(A-B\right)=\cos A\cos B+\sin A\sin B}$

$\bf{\tan\left(A+B\right)=\frac{\tan A+\tan B}{1-\tan A\tan B}}$

$\bf{\tan\left(A+B\right)=\frac{\tan A-\tan B}{1+\tan A\tan B}}$

$\:$

Pembahasan

Diketahui :

A+B = 30°.

Ditanya :

Cos A = ...?

Jawaban :

A+B = 30°

A = 30° - B

Ingat kembali bahwa

$\bf{\cos\left(A-B\right)=\cos A\cos B+\sin A\sin B}$

$\to$ maka

Cos A

$\bf{Cos\left(30^{\circ}-B\right)=\cos\left(30^{\circ}\right)\cos B+\sin\left(30^{\circ}\right)\sin B}$

$\bf{Cos\left(30^{\circ}-B\right)=\frac{1}{2}\sqrt{3}\cos B+\frac{1}{2}\sin B}$

$\boxed{\bf{Cos\left(30^{\circ}-B\right)=\frac{1}{2}\left(\sqrt{3}\cos B+\sin B\right)}}$

$\:$

Pelajari Lebih Lanjut :

Contoh soal mencari sisi samping : yomemimo.com/tugas/48680192
Nilai dari cos²45° + sin²240° = : yomemimo.com/tugas/52094991
Jika cos (72,24°) = 1/5. maka sin (17,76°) ialah : yomemimo.com/tugas/52659460
Brpkh sin(30°) × cos(60°) : yomemimo.com/tugas/52206983

$\:$

Detail Jawaban :

Grade : SMA

Kode Kategorisasi : 10.2.6

Kelas : 10

Kode Mapel : 2

Pelajaran : Matematika

Bab : 6

Sub Bab : Bab 6 – Trigonometri Dasar

$\:$

Kata Kunci : Trigonometri 2 sudut.

$Jika A+B = 30°, maka[tex]\boxed{\bf{Cos\left(30^{\circ}-B\right)=\frac{1}{2}\left(\sqrt{3}\cos B+\sin B\right)}}[/tex][tex] \: [/tex]TrigonometriPendahuluanA.) Definisi.) Perbandingan TrigonometriPada segitiga siku-siku ABC, berlaku : *Gambar ke-1[tex]\small\mathbf{\left(a.\right)\ \ \sin\alpha=\frac{y}{r}=\frac{de}{mi}} [/tex][tex]\small\mathbf{\left(b.\right)\ \ \cos\alpha=\frac{x}{r}=\frac{sa}{mi}} [/tex][tex]\small\mathbf{\left(c.\right)\ \ \tan\alpha=\frac{y}{x}=\frac{de}{sa}} [/tex][tex]\small\mathbf{\left(d.\right)\ \ \csc\alpha=\frac{1}{\sin\alpha}=\frac{r}{y}}[/tex][tex]\small\mathbf{\left(e.\right)\ \ \sec\alpha=\frac{1}{\cos\alpha}=\frac{r}{x}}[/tex][tex]\small\mathbf{\left(f.\right)\ \ \cot\alpha=\frac{1}{\tan\alpha}=\frac{y}{x}}[/tex]B.) Sudut dan Kuadran1.) Pembagian Daerah [tex]\boxed{\begin{array}{c|c|c|c|c}\underline{\mathbf{Kuadran}}&\underline{\mathbf{I}}&\underline{\mathbf{II}}&\underline{\mathbf{III}}&\underline{\mathbf{IV}}\\&&&\\\mathbf{absis(x)}&\mathbf{+}&\mathbf{-}&\mathbf{-}&\mathbf{+}\\&&&\\\mathbf{Ordinat(y)}&\mathbf{+}&\mathbf{+}&\mathbf{-}&\mathbf{-}\end{array}}[/tex]2.) Tanda-tanda Fungsi[tex]\boxed{\begin{array}{c|c|c|c|c}\underline{\mathbf{Kuadran}}&\underline{\mathbf{I}}&\underline{\mathbf{II}}&\underline{\mathbf{III}}&\underline{\mathbf{IV}}\\&&&\\\mathbf{sin}&\mathbf{+}&\mathbf{+}&\mathbf{-}&\mathbf{-}\\&&&\\\mathbf{cos}&\mathbf{+}&\mathbf{-}&\mathbf{-}&\mathbf{+}\\&&&\\\mathbf{tan}&\mathbf{+}&\mathbf{-}&\mathbf{+}&\mathbf{-}\end{array}}[/tex]3.) Sudut-sudut Istimewa[tex]\boxed{\begin{array}{c|c|c|c|c}\underline{\mathbf{Kuadran}}&\underline{\mathbf{0^{\circ}}}&\underline{\mathbf{30^{\circ}}}&\underline{\mathbf{45^{\circ}}}&\underline{\mathbf{60^{\circ}}}\\&&&\\\mathbf{sin}&\mathbf{0}&\mathbf{\frac{1}{2}}&\mathbf{\frac{1}{2}\sqrt{2}}&\mathbf{\frac{1}{2}\sqrt{3}}\\&&&\\\mathbf{cos}&\mathbf{1}&\mathbf{\frac{1}{2}\sqrt{3}}&\mathbf{\frac{1}{2}\sqrt{2}}&\mathbf{\frac{1}{2}}\\&&&\\\mathbf{tan}&\mathbf{0}&\mathbf{\frac{1}{3}\sqrt{3}}&\mathbf{1}&\mathbf{\sqrt{3}}\end{array}} [/tex] [tex] \boxed{\begin{array}{c}\underline{\mathbf{90^{\circ}}}\\\\\mathbf{1}\\\\\mathbf{0}\\\\\infty\end{array}} [/tex]4.) Sudut Berelasia. Kalau kita gunakan (90°± ...) atau (270°± ...) 1.) Fungsi berubah [tex]\boxed{\begin{array}{c|c}\underline{\mathbf{Mula-mula}}&\underline{\mathbf{Perubahan}}\\\\\mathbf{sin}&\mathbf{+/-cos}\\\\\mathbf{cos}&\mathbf{+/-sin}\\\\\mathbf{tan}&\mathbf{+/-cot}\end{array}}[/tex] 2.) Tanda +/- mengikuti kuadranb. kalau kita gunakan (180°± ...) atau (360°− ...) 1.) Fungsi tetap[tex]\boxed{\begin{array}{c|c}\underline{\mathbf{Mula-mula}}&\underline{\mathbf{Perubahan}}\\\\\mathbf{sin}&\mathbf{+/-sin}\\\\\mathbf{cos}&\mathbf{+/-cos}\\\\\mathbf{tan}&\mathbf{+/-tan}\end{array}}[/tex]C.) Rumus trigonometri 2 sudut[tex]\bf{\sin\left(A+B\right)=\sin A\cos B+\cos A\sin B}[/tex][tex]\bf{\sin\left(A-B\right)=\sin A\cos B-\cos A\sin B}[/tex][tex]\bf{\cos\left(A+B\right)=\cos A\cos B-\sin A\sin B}[/tex][tex]\bf{\cos\left(A-B\right)=\cos A\cos B+\sin A\sin B}[/tex][tex]\bf{\tan\left(A+B\right)=\frac{\tan A+\tan B}{1-\tan A\tan B}}[/tex][tex]\bf{\tan\left(A+B\right)=\frac{\tan A-\tan B}{1+\tan A\tan B}}[/tex][tex] \: [/tex][tex] \: [/tex]PembahasanDiketahui :A+B = 30°.Ditanya :Cos A = ...?Jawaban :A+B = 30°A = 30° - BIngat kembali bahwa[tex]\bf{\cos\left(A-B\right)=\cos A\cos B+\sin A\sin B}[/tex][tex]\to[/tex] makaCos A[tex]\bf{Cos\left(30^{\circ}-B\right)=\cos\left(30^{\circ}\right)\cos B+\sin\left(30^{\circ}\right)\sin B}[/tex][tex]\bf{Cos\left(30^{\circ}-B\right)=\frac{1}{2}\sqrt{3}\cos B+\frac{1}{2}\sin B}[/tex][tex]\boxed{\bf{Cos\left(30^{\circ}-B\right)=\frac{1}{2}\left(\sqrt{3}\cos B+\sin B\right)}}[/tex][tex] \: [/tex][tex] \: [/tex]Pelajari Lebih Lanjut :Contoh soal mencari sisi samping : https://brainly.co.id/tugas/48680192Nilai dari cos²45° + sin²240° = : https://brainly.co.id/tugas/52094991Jika cos (72,24°) = 1/5. maka sin (17,76°) ialah : https://brainly.co.id/tugas/52659460Brpkh sin(30°) × cos(60°) : https://brainly.co.id/tugas/52206983[tex] \: [/tex][tex] \: [/tex]Detail Jawaban :Grade : SMAKode Kategorisasi : 10.2.6Kelas : 10Kode Mapel : 2Pelajaran : MatematikaBab : 6Sub Bab : Bab 6 – Trigonometri Dasar[tex] \: [/tex]Kata Kunci : Trigonometri 2 sudut.$ $Jika A+B = 30°, maka[tex]\boxed{\bf{Cos\left(30^{\circ}-B\right)=\frac{1}{2}\left(\sqrt{3}\cos B+\sin B\right)}}[/tex][tex] \: [/tex]TrigonometriPendahuluanA.) Definisi.) Perbandingan TrigonometriPada segitiga siku-siku ABC, berlaku : *Gambar ke-1[tex]\small\mathbf{\left(a.\right)\ \ \sin\alpha=\frac{y}{r}=\frac{de}{mi}} [/tex][tex]\small\mathbf{\left(b.\right)\ \ \cos\alpha=\frac{x}{r}=\frac{sa}{mi}} [/tex][tex]\small\mathbf{\left(c.\right)\ \ \tan\alpha=\frac{y}{x}=\frac{de}{sa}} [/tex][tex]\small\mathbf{\left(d.\right)\ \ \csc\alpha=\frac{1}{\sin\alpha}=\frac{r}{y}}[/tex][tex]\small\mathbf{\left(e.\right)\ \ \sec\alpha=\frac{1}{\cos\alpha}=\frac{r}{x}}[/tex][tex]\small\mathbf{\left(f.\right)\ \ \cot\alpha=\frac{1}{\tan\alpha}=\frac{y}{x}}[/tex]B.) Sudut dan Kuadran1.) Pembagian Daerah [tex]\boxed{\begin{array}{c|c|c|c|c}\underline{\mathbf{Kuadran}}&\underline{\mathbf{I}}&\underline{\mathbf{II}}&\underline{\mathbf{III}}&\underline{\mathbf{IV}}\\&&&\\\mathbf{absis(x)}&\mathbf{+}&\mathbf{-}&\mathbf{-}&\mathbf{+}\\&&&\\\mathbf{Ordinat(y)}&\mathbf{+}&\mathbf{+}&\mathbf{-}&\mathbf{-}\end{array}}[/tex]2.) Tanda-tanda Fungsi[tex]\boxed{\begin{array}{c|c|c|c|c}\underline{\mathbf{Kuadran}}&\underline{\mathbf{I}}&\underline{\mathbf{II}}&\underline{\mathbf{III}}&\underline{\mathbf{IV}}\\&&&\\\mathbf{sin}&\mathbf{+}&\mathbf{+}&\mathbf{-}&\mathbf{-}\\&&&\\\mathbf{cos}&\mathbf{+}&\mathbf{-}&\mathbf{-}&\mathbf{+}\\&&&\\\mathbf{tan}&\mathbf{+}&\mathbf{-}&\mathbf{+}&\mathbf{-}\end{array}}[/tex]3.) Sudut-sudut Istimewa[tex]\boxed{\begin{array}{c|c|c|c|c}\underline{\mathbf{Kuadran}}&\underline{\mathbf{0^{\circ}}}&\underline{\mathbf{30^{\circ}}}&\underline{\mathbf{45^{\circ}}}&\underline{\mathbf{60^{\circ}}}\\&&&\\\mathbf{sin}&\mathbf{0}&\mathbf{\frac{1}{2}}&\mathbf{\frac{1}{2}\sqrt{2}}&\mathbf{\frac{1}{2}\sqrt{3}}\\&&&\\\mathbf{cos}&\mathbf{1}&\mathbf{\frac{1}{2}\sqrt{3}}&\mathbf{\frac{1}{2}\sqrt{2}}&\mathbf{\frac{1}{2}}\\&&&\\\mathbf{tan}&\mathbf{0}&\mathbf{\frac{1}{3}\sqrt{3}}&\mathbf{1}&\mathbf{\sqrt{3}}\end{array}} [/tex] [tex] \boxed{\begin{array}{c}\underline{\mathbf{90^{\circ}}}\\\\\mathbf{1}\\\\\mathbf{0}\\\\\infty\end{array}} [/tex]4.) Sudut Berelasia. Kalau kita gunakan (90°± ...) atau (270°± ...) 1.) Fungsi berubah [tex]\boxed{\begin{array}{c|c}\underline{\mathbf{Mula-mula}}&\underline{\mathbf{Perubahan}}\\\\\mathbf{sin}&\mathbf{+/-cos}\\\\\mathbf{cos}&\mathbf{+/-sin}\\\\\mathbf{tan}&\mathbf{+/-cot}\end{array}}[/tex] 2.) Tanda +/- mengikuti kuadranb. kalau kita gunakan (180°± ...) atau (360°− ...) 1.) Fungsi tetap[tex]\boxed{\begin{array}{c|c}\underline{\mathbf{Mula-mula}}&\underline{\mathbf{Perubahan}}\\\\\mathbf{sin}&\mathbf{+/-sin}\\\\\mathbf{cos}&\mathbf{+/-cos}\\\\\mathbf{tan}&\mathbf{+/-tan}\end{array}}[/tex]C.) Rumus trigonometri 2 sudut[tex]\bf{\sin\left(A+B\right)=\sin A\cos B+\cos A\sin B}[/tex][tex]\bf{\sin\left(A-B\right)=\sin A\cos B-\cos A\sin B}[/tex][tex]\bf{\cos\left(A+B\right)=\cos A\cos B-\sin A\sin B}[/tex][tex]\bf{\cos\left(A-B\right)=\cos A\cos B+\sin A\sin B}[/tex][tex]\bf{\tan\left(A+B\right)=\frac{\tan A+\tan B}{1-\tan A\tan B}}[/tex][tex]\bf{\tan\left(A+B\right)=\frac{\tan A-\tan B}{1+\tan A\tan B}}[/tex][tex] \: [/tex][tex] \: [/tex]PembahasanDiketahui :A+B = 30°.Ditanya :Cos A = ...?Jawaban :A+B = 30°A = 30° - BIngat kembali bahwa[tex]\bf{\cos\left(A-B\right)=\cos A\cos B+\sin A\sin B}[/tex][tex]\to[/tex] makaCos A[tex]\bf{Cos\left(30^{\circ}-B\right)=\cos\left(30^{\circ}\right)\cos B+\sin\left(30^{\circ}\right)\sin B}[/tex][tex]\bf{Cos\left(30^{\circ}-B\right)=\frac{1}{2}\sqrt{3}\cos B+\frac{1}{2}\sin B}[/tex][tex]\boxed{\bf{Cos\left(30^{\circ}-B\right)=\frac{1}{2}\left(\sqrt{3}\cos B+\sin B\right)}}[/tex][tex] \: [/tex][tex] \: [/tex]Pelajari Lebih Lanjut :Contoh soal mencari sisi samping : https://brainly.co.id/tugas/48680192Nilai dari cos²45° + sin²240° = : https://brainly.co.id/tugas/52094991Jika cos (72,24°) = 1/5. maka sin (17,76°) ialah : https://brainly.co.id/tugas/52659460Brpkh sin(30°) × cos(60°) : https://brainly.co.id/tugas/52206983[tex] \: [/tex][tex] \: [/tex]Detail Jawaban :Grade : SMAKode Kategorisasi : 10.2.6Kelas : 10Kode Mapel : 2Pelajaran : MatematikaBab : 6Sub Bab : Bab 6 – Trigonometri Dasar[tex] \: [/tex]Kata Kunci : Trigonometri 2 sudut.$ $Jika A+B = 30°, maka[tex]\boxed{\bf{Cos\left(30^{\circ}-B\right)=\frac{1}{2}\left(\sqrt{3}\cos B+\sin B\right)}}[/tex][tex] \: [/tex]TrigonometriPendahuluanA.) Definisi.) Perbandingan TrigonometriPada segitiga siku-siku ABC, berlaku : *Gambar ke-1[tex]\small\mathbf{\left(a.\right)\ \ \sin\alpha=\frac{y}{r}=\frac{de}{mi}} [/tex][tex]\small\mathbf{\left(b.\right)\ \ \cos\alpha=\frac{x}{r}=\frac{sa}{mi}} [/tex][tex]\small\mathbf{\left(c.\right)\ \ \tan\alpha=\frac{y}{x}=\frac{de}{sa}} [/tex][tex]\small\mathbf{\left(d.\right)\ \ \csc\alpha=\frac{1}{\sin\alpha}=\frac{r}{y}}[/tex][tex]\small\mathbf{\left(e.\right)\ \ \sec\alpha=\frac{1}{\cos\alpha}=\frac{r}{x}}[/tex][tex]\small\mathbf{\left(f.\right)\ \ \cot\alpha=\frac{1}{\tan\alpha}=\frac{y}{x}}[/tex]B.) Sudut dan Kuadran1.) Pembagian Daerah [tex]\boxed{\begin{array}{c|c|c|c|c}\underline{\mathbf{Kuadran}}&\underline{\mathbf{I}}&\underline{\mathbf{II}}&\underline{\mathbf{III}}&\underline{\mathbf{IV}}\\&&&\\\mathbf{absis(x)}&\mathbf{+}&\mathbf{-}&\mathbf{-}&\mathbf{+}\\&&&\\\mathbf{Ordinat(y)}&\mathbf{+}&\mathbf{+}&\mathbf{-}&\mathbf{-}\end{array}}[/tex]2.) Tanda-tanda Fungsi[tex]\boxed{\begin{array}{c|c|c|c|c}\underline{\mathbf{Kuadran}}&\underline{\mathbf{I}}&\underline{\mathbf{II}}&\underline{\mathbf{III}}&\underline{\mathbf{IV}}\\&&&\\\mathbf{sin}&\mathbf{+}&\mathbf{+}&\mathbf{-}&\mathbf{-}\\&&&\\\mathbf{cos}&\mathbf{+}&\mathbf{-}&\mathbf{-}&\mathbf{+}\\&&&\\\mathbf{tan}&\mathbf{+}&\mathbf{-}&\mathbf{+}&\mathbf{-}\end{array}}[/tex]3.) Sudut-sudut Istimewa[tex]\boxed{\begin{array}{c|c|c|c|c}\underline{\mathbf{Kuadran}}&\underline{\mathbf{0^{\circ}}}&\underline{\mathbf{30^{\circ}}}&\underline{\mathbf{45^{\circ}}}&\underline{\mathbf{60^{\circ}}}\\&&&\\\mathbf{sin}&\mathbf{0}&\mathbf{\frac{1}{2}}&\mathbf{\frac{1}{2}\sqrt{2}}&\mathbf{\frac{1}{2}\sqrt{3}}\\&&&\\\mathbf{cos}&\mathbf{1}&\mathbf{\frac{1}{2}\sqrt{3}}&\mathbf{\frac{1}{2}\sqrt{2}}&\mathbf{\frac{1}{2}}\\&&&\\\mathbf{tan}&\mathbf{0}&\mathbf{\frac{1}{3}\sqrt{3}}&\mathbf{1}&\mathbf{\sqrt{3}}\end{array}} [/tex] [tex] \boxed{\begin{array}{c}\underline{\mathbf{90^{\circ}}}\\\\\mathbf{1}\\\\\mathbf{0}\\\\\infty\end{array}} [/tex]4.) Sudut Berelasia. Kalau kita gunakan (90°± ...) atau (270°± ...) 1.) Fungsi berubah [tex]\boxed{\begin{array}{c|c}\underline{\mathbf{Mula-mula}}&\underline{\mathbf{Perubahan}}\\\\\mathbf{sin}&\mathbf{+/-cos}\\\\\mathbf{cos}&\mathbf{+/-sin}\\\\\mathbf{tan}&\mathbf{+/-cot}\end{array}}[/tex] 2.) Tanda +/- mengikuti kuadranb. kalau kita gunakan (180°± ...) atau (360°− ...) 1.) Fungsi tetap[tex]\boxed{\begin{array}{c|c}\underline{\mathbf{Mula-mula}}&\underline{\mathbf{Perubahan}}\\\\\mathbf{sin}&\mathbf{+/-sin}\\\\\mathbf{cos}&\mathbf{+/-cos}\\\\\mathbf{tan}&\mathbf{+/-tan}\end{array}}[/tex]C.) Rumus trigonometri 2 sudut[tex]\bf{\sin\left(A+B\right)=\sin A\cos B+\cos A\sin B}[/tex][tex]\bf{\sin\left(A-B\right)=\sin A\cos B-\cos A\sin B}[/tex][tex]\bf{\cos\left(A+B\right)=\cos A\cos B-\sin A\sin B}[/tex][tex]\bf{\cos\left(A-B\right)=\cos A\cos B+\sin A\sin B}[/tex][tex]\bf{\tan\left(A+B\right)=\frac{\tan A+\tan B}{1-\tan A\tan B}}[/tex][tex]\bf{\tan\left(A+B\right)=\frac{\tan A-\tan B}{1+\tan A\tan B}}[/tex][tex] \: [/tex][tex] \: [/tex]PembahasanDiketahui :A+B = 30°.Ditanya :Cos A = ...?Jawaban :A+B = 30°A = 30° - BIngat kembali bahwa[tex]\bf{\cos\left(A-B\right)=\cos A\cos B+\sin A\sin B}[/tex][tex]\to[/tex] makaCos A[tex]\bf{Cos\left(30^{\circ}-B\right)=\cos\left(30^{\circ}\right)\cos B+\sin\left(30^{\circ}\right)\sin B}[/tex][tex]\bf{Cos\left(30^{\circ}-B\right)=\frac{1}{2}\sqrt{3}\cos B+\frac{1}{2}\sin B}[/tex][tex]\boxed{\bf{Cos\left(30^{\circ}-B\right)=\frac{1}{2}\left(\sqrt{3}\cos B+\sin B\right)}}[/tex][tex] \: [/tex][tex] \: [/tex]Pelajari Lebih Lanjut :Contoh soal mencari sisi samping : https://brainly.co.id/tugas/48680192Nilai dari cos²45° + sin²240° = : https://brainly.co.id/tugas/52094991Jika cos (72,24°) = 1/5. maka sin (17,76°) ialah : https://brainly.co.id/tugas/52659460Brpkh sin(30°) × cos(60°) : https://brainly.co.id/tugas/52206983[tex] \: [/tex][tex] \: [/tex]Detail Jawaban :Grade : SMAKode Kategorisasi : 10.2.6Kelas : 10Kode Mapel : 2Pelajaran : MatematikaBab : 6Sub Bab : Bab 6 – Trigonometri Dasar[tex] \: [/tex]Kata Kunci : Trigonometri 2 sudut.$ $Jika A+B = 30°, maka[tex]\boxed{\bf{Cos\left(30^{\circ}-B\right)=\frac{1}{2}\left(\sqrt{3}\cos B+\sin B\right)}}[/tex][tex] \: [/tex]TrigonometriPendahuluanA.) Definisi.) Perbandingan TrigonometriPada segitiga siku-siku ABC, berlaku : *Gambar ke-1[tex]\small\mathbf{\left(a.\right)\ \ \sin\alpha=\frac{y}{r}=\frac{de}{mi}} [/tex][tex]\small\mathbf{\left(b.\right)\ \ \cos\alpha=\frac{x}{r}=\frac{sa}{mi}} [/tex][tex]\small\mathbf{\left(c.\right)\ \ \tan\alpha=\frac{y}{x}=\frac{de}{sa}} [/tex][tex]\small\mathbf{\left(d.\right)\ \ \csc\alpha=\frac{1}{\sin\alpha}=\frac{r}{y}}[/tex][tex]\small\mathbf{\left(e.\right)\ \ \sec\alpha=\frac{1}{\cos\alpha}=\frac{r}{x}}[/tex][tex]\small\mathbf{\left(f.\right)\ \ \cot\alpha=\frac{1}{\tan\alpha}=\frac{y}{x}}[/tex]B.) Sudut dan Kuadran1.) Pembagian Daerah [tex]\boxed{\begin{array}{c|c|c|c|c}\underline{\mathbf{Kuadran}}&\underline{\mathbf{I}}&\underline{\mathbf{II}}&\underline{\mathbf{III}}&\underline{\mathbf{IV}}\\&&&\\\mathbf{absis(x)}&\mathbf{+}&\mathbf{-}&\mathbf{-}&\mathbf{+}\\&&&\\\mathbf{Ordinat(y)}&\mathbf{+}&\mathbf{+}&\mathbf{-}&\mathbf{-}\end{array}}[/tex]2.) Tanda-tanda Fungsi[tex]\boxed{\begin{array}{c|c|c|c|c}\underline{\mathbf{Kuadran}}&\underline{\mathbf{I}}&\underline{\mathbf{II}}&\underline{\mathbf{III}}&\underline{\mathbf{IV}}\\&&&\\\mathbf{sin}&\mathbf{+}&\mathbf{+}&\mathbf{-}&\mathbf{-}\\&&&\\\mathbf{cos}&\mathbf{+}&\mathbf{-}&\mathbf{-}&\mathbf{+}\\&&&\\\mathbf{tan}&\mathbf{+}&\mathbf{-}&\mathbf{+}&\mathbf{-}\end{array}}[/tex]3.) Sudut-sudut Istimewa[tex]\boxed{\begin{array}{c|c|c|c|c}\underline{\mathbf{Kuadran}}&\underline{\mathbf{0^{\circ}}}&\underline{\mathbf{30^{\circ}}}&\underline{\mathbf{45^{\circ}}}&\underline{\mathbf{60^{\circ}}}\\&&&\\\mathbf{sin}&\mathbf{0}&\mathbf{\frac{1}{2}}&\mathbf{\frac{1}{2}\sqrt{2}}&\mathbf{\frac{1}{2}\sqrt{3}}\\&&&\\\mathbf{cos}&\mathbf{1}&\mathbf{\frac{1}{2}\sqrt{3}}&\mathbf{\frac{1}{2}\sqrt{2}}&\mathbf{\frac{1}{2}}\\&&&\\\mathbf{tan}&\mathbf{0}&\mathbf{\frac{1}{3}\sqrt{3}}&\mathbf{1}&\mathbf{\sqrt{3}}\end{array}} [/tex] [tex] \boxed{\begin{array}{c}\underline{\mathbf{90^{\circ}}}\\\\\mathbf{1}\\\\\mathbf{0}\\\\\infty\end{array}} [/tex]4.) Sudut Berelasia. Kalau kita gunakan (90°± ...) atau (270°± ...) 1.) Fungsi berubah [tex]\boxed{\begin{array}{c|c}\underline{\mathbf{Mula-mula}}&\underline{\mathbf{Perubahan}}\\\\\mathbf{sin}&\mathbf{+/-cos}\\\\\mathbf{cos}&\mathbf{+/-sin}\\\\\mathbf{tan}&\mathbf{+/-cot}\end{array}}[/tex] 2.) Tanda +/- mengikuti kuadranb. kalau kita gunakan (180°± ...) atau (360°− ...) 1.) Fungsi tetap[tex]\boxed{\begin{array}{c|c}\underline{\mathbf{Mula-mula}}&\underline{\mathbf{Perubahan}}\\\\\mathbf{sin}&\mathbf{+/-sin}\\\\\mathbf{cos}&\mathbf{+/-cos}\\\\\mathbf{tan}&\mathbf{+/-tan}\end{array}}[/tex]C.) Rumus trigonometri 2 sudut[tex]\bf{\sin\left(A+B\right)=\sin A\cos B+\cos A\sin B}[/tex][tex]\bf{\sin\left(A-B\right)=\sin A\cos B-\cos A\sin B}[/tex][tex]\bf{\cos\left(A+B\right)=\cos A\cos B-\sin A\sin B}[/tex][tex]\bf{\cos\left(A-B\right)=\cos A\cos B+\sin A\sin B}[/tex][tex]\bf{\tan\left(A+B\right)=\frac{\tan A+\tan B}{1-\tan A\tan B}}[/tex][tex]\bf{\tan\left(A+B\right)=\frac{\tan A-\tan B}{1+\tan A\tan B}}[/tex][tex] \: [/tex][tex] \: [/tex]PembahasanDiketahui :A+B = 30°.Ditanya :Cos A = ...?Jawaban :A+B = 30°A = 30° - BIngat kembali bahwa[tex]\bf{\cos\left(A-B\right)=\cos A\cos B+\sin A\sin B}[/tex][tex]\to[/tex] makaCos A[tex]\bf{Cos\left(30^{\circ}-B\right)=\cos\left(30^{\circ}\right)\cos B+\sin\left(30^{\circ}\right)\sin B}[/tex][tex]\bf{Cos\left(30^{\circ}-B\right)=\frac{1}{2}\sqrt{3}\cos B+\frac{1}{2}\sin B}[/tex][tex]\boxed{\bf{Cos\left(30^{\circ}-B\right)=\frac{1}{2}\left(\sqrt{3}\cos B+\sin B\right)}}[/tex][tex] \: [/tex][tex] \: [/tex]Pelajari Lebih Lanjut :Contoh soal mencari sisi samping : https://brainly.co.id/tugas/48680192Nilai dari cos²45° + sin²240° = : https://brainly.co.id/tugas/52094991Jika cos (72,24°) = 1/5. maka sin (17,76°) ialah : https://brainly.co.id/tugas/52659460Brpkh sin(30°) × cos(60°) : https://brainly.co.id/tugas/52206983[tex] \: [/tex][tex] \: [/tex]Detail Jawaban :Grade : SMAKode Kategorisasi : 10.2.6Kelas : 10Kode Mapel : 2Pelajaran : MatematikaBab : 6Sub Bab : Bab 6 – Trigonometri Dasar[tex] \: [/tex]Kata Kunci : Trigonometri 2 sudut.$

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Last Update: Sun, 12 Feb 23

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Jika A+B = 30°, maka Cos A = ​

Jawaban dan Penjelasan

Trigonometri

Pendahuluan

A.) Definisi

B.) Sudut dan Kuadran

C.) Rumus trigonometri 2 sudut

Pembahasan

Pelajari Lebih Lanjut :

Detail Jawaban :

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Jika A+B = 30°, maka Cos A =