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Tentukan nilai Rdanα dari setiap persamaan di bawah ini. 2sin(θ)+3cos(θ)=Rsin(θ−α)

Pertanyaan

Tentukan nilai Rdanα dari setiap persamaan di bawah ini.

2sin(θ)+3cos(θ)=Rsin(θα) 

Pembahasan Soal:

Diketahui 2sin(θ)+3cos(θ)=Rsin(θα), akan dicari nilai Rdanα

Ingat bahwa

sin(θα)=sinθcosαcosθsinαAsinθBcosθ=Rsin(θα),denganR=A2+B2

Diperhatikan

2sin(θ)+3cos(θ)=Rsin(θα)2sin(θ)+3cos(θ)=RsinθcosαRcosθsinα 

Diperoleh

2=Rcosα3=Rsinα 

Lebih lanjut, diperoleh

negative 3 equals R space sin space alpha 2 equals R space cos space alpha horizontal strike horizontal strike horizontal strike horizontal strike horizontal strike horizontal strike horizontal strike horizontal strike horizontal strike horizontal strike horizontal strike horizontal strike horizontal strike horizontal strike horizontal strike horizontal strike horizontal strike horizontal strike horizontal strike blank end strike end strike end strike end strike end strike end strike end strike end strike end strike end strike end strike end strike end strike end strike end strike end strike end strike end strike end strike space divided by minus 3 over 2 equals fraction numerator R space sin space alpha over denominator R space cos space alpha end fraction minus 3 over 2 equals tan space alpha alpha almost equal to 56 comma 30 degree 

dan

33RR====RsinαRsin56,30sin56,3033,6 

Dengan demikian, diperoleh bahwa R=3,6danα56,30

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

H. Firmansyah

Mahasiswa/Alumni Universitas Gadjah Mada

Terakhir diupdate 30 Agustus 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Pada segitiga ABC diketahui bahwa perbandingan sisi-sisi a:b:c=2:3:4. Tentukan sin(A+B).

Pembahasan Soal:

Ingat,

Sudut dalam Segitiga

A+B+C=180

Aturan Cos-Perbandingan Sisi

c2=a2+b22abcosC

Rumus Perbandingan Trigonometri untuk Selisih Dua Sudut (Sinus)

sin(AB)=sinAcosBcosAsinB

Berdasarkan rumus tersebut, diperoleh sebagai berikut

Karena segitiga ABC maka, 

A+B+CA+B==180180C

Bentuk sederhana sin(A+B)

sin(A+B)====sin(180C)sin180sinCcos180sinC0(1sinC)sinC

Menghitung sudut C dengan aturan cos
Diketahui bahwa perbandingan sisi-sisi a:b:c=2:3:4
Misal : a=2xb=3xc=4x

c2(4x)216x216x23x2cosCcosCcosCC=========a2+b22abcosC(2x)2+(3x)22(2x)(3x)cosC4x2+9x212x2cosC13x212x2cosC12x2cosC12x23x2410,25105

Menentukan nilai sin(A+B)
Diperoleh bahwa sin(A+B)=sinC dan besar C=105, sehingga

sin(A+B)=sinC=sin105=0,97

Dengan demikian, nilai dari sin(A+B) adalah 0,97. 

0

Roboguru

Jika A dan B adalah sudut lancip yang memenuhi  dan ,  nilai dari ....

Pembahasan Soal:

Karena begin mathsize 14px style A plus B equals 1 over 6 pi end style, maka begin mathsize 14px style B equals 1 over 6 pi minus A end style 

Karena begin mathsize 14px style sin invisible function application A equals 2 square root of 3 sin invisible function application B end style, maka

begin mathsize 14px style sin invisible function application A equals 2 square root of 3 sin invisible function application B sin invisible function application A equals 2 square root of 3 sin invisible function application open parentheses 1 over 6 pi minus A close parentheses sin invisible function application A equals 2 square root of 3 open parentheses sin invisible function application 1 over 6 pi cos invisible function application A minus cos invisible function application 1 over 6 pi sin invisible function application A close parentheses end style  

Perhatikan bahwa begin mathsize 14px style 1 over 6 pi equals 1 over 6 times 180 degree equals 30 degree end style. Maka

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin invisible function application A end cell equals cell 2 square root of 3 open parentheses sin invisible function application 1 over 6 pi cos invisible function application A minus cos invisible function application 1 over 6 pi sin invisible function application A close parentheses end cell row cell sin invisible function application A end cell equals cell 2 square root of 3 open parentheses sin invisible function application 30 degree cos invisible function application A minus cos invisible function application 30 degree sin invisible function application A close parentheses end cell row cell sin invisible function application A end cell equals cell 2 square root of 3 open parentheses 1 half cos invisible function application A minus 1 half square root of 3 sin invisible function application A close parentheses end cell row cell sin invisible function application A end cell equals cell square root of 3 cos invisible function application A minus 3 sin invisible function application A end cell row cell sin invisible function application A plus 3 sin invisible function application A end cell equals cell square root of 3 cos invisible function application A end cell row cell 4 sin invisible function application A end cell equals cell square root of 3 cos invisible function application A end cell row cell fraction numerator sin invisible function application A over denominator cos invisible function application A end fraction end cell equals cell fraction numerator square root of 3 over denominator 4 end fraction end cell row cell tan invisible function application A end cell equals cell fraction numerator square root of 3 over denominator 4 end fraction end cell end table end style 

Selanjutnya, yang ditanyakan adalah

undefined 

Untuk mencari nilai dari sin A dan cos A perhatikan segitiga berikut

Perhatikan bahwa diketahui A  merupakan sudut lancip. Karena begin mathsize 14px style tan invisible function application A equals fraction numerator square root of 3 over denominator 4 end fraction end style, maka

Karena segitiga di atas merupakan segitiga siku-siku, maka nilai x bisa didapat menggunakan teorema Pythagoras. Sehingga

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row x equals cell square root of 4 squared plus open parentheses square root of 3 close parentheses squared end root end cell row x equals cell square root of 16 plus 3 end root end cell row x equals cell square root of 19 end cell end table end style 

Sehingga didapatkan begin mathsize 14px style sin invisible function application A equals fraction numerator square root of 3 over denominator x end fraction equals fraction numerator square root of 3 over denominator square root of 19 end fraction space d a n space cos invisible function application A equals 4 over x equals fraction numerator 4 over denominator square root of 19 end fraction end style 

 

Selanjutnya

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin invisible function application A end cell equals cell 2 square root of 3 sin invisible function application B end cell row cell sin invisible function application B end cell equals cell fraction numerator 1 over denominator 2 square root of 3 end fraction times sin invisible function application A end cell row cell sin invisible function application B end cell equals cell fraction numerator 1 over denominator 2 square root of 3 end fraction times fraction numerator square root of 3 over denominator square root of 19 end fraction end cell row cell sin invisible function application B end cell equals cell fraction numerator 1 over denominator 2 square root of 19 end fraction end cell row blank blank blank end table end style 

Untuk mencari nilai dari cos B perhatikan segitiga berikut

Perhatikan bahwa diketahui B merupakan sudut lancip. Karena begin mathsize 14px style sin invisible function application B equals fraction numerator 1 over denominator 2 square root of 19 end fraction end style, maka

Karena segitiga di atas merupakan segitiga siku-siku, maka nilai y bisa didapat menggunakan teorema Pythagoras. Sehingga

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row y equals cell square root of open parentheses 2 square root of 19 close parentheses squared minus 1 squared end root end cell row y equals cell square root of 4 times 19 minus 1 end root end cell row y equals cell square root of 76 minus 1 end root end cell row y equals cell square root of 75 end cell row y equals cell square root of 25 times 3 end root end cell row y equals cell 5 square root of 3 end cell row blank blank blank row blank blank blank row blank blank blank end table end style  

Sehingga didapatkan begin mathsize 14px style cos invisible function application B equals fraction numerator y over denominator 2 square root of 19 end fraction equals fraction numerator 5 square root of 3 over denominator 2 square root of 19 end fraction end style 

 

Maka didapat bahwa

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin invisible function application A end cell equals cell fraction numerator square root of 3 over denominator square root of 19 end fraction end cell row cell cos invisible function application A end cell equals cell fraction numerator 4 over denominator square root of 19 end fraction end cell row cell sin invisible function application B end cell equals cell fraction numerator 1 over denominator 2 square root of 19 end fraction end cell row cell cos invisible function application B end cell equals cell fraction numerator 5 square root of 3 over denominator 2 square root of 19 end fraction end cell row blank blank blank row blank blank blank end table end style 

Sehingga

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell cos invisible function application open parentheses A minus B close parentheses end cell equals cell cos invisible function application A cos invisible function application B plus sin invisible function application A sin invisible function application B end cell row blank equals cell fraction numerator 4 over denominator square root of 19 end fraction times fraction numerator 5 square root of 3 over denominator 2 square root of 19 end fraction plus fraction numerator square root of 3 over denominator square root of 19 end fraction times fraction numerator 1 over denominator 2 square root of 19 end fraction end cell row blank equals cell fraction numerator 20 square root of 3 over denominator 2 open parentheses 19 close parentheses end fraction plus fraction numerator square root of 3 over denominator 2 open parentheses 19 close parentheses end fraction end cell row blank equals cell fraction numerator 20 square root of 3 over denominator 38 end fraction plus fraction numerator square root of 3 over denominator 38 end fraction end cell row blank equals cell fraction numerator 20 square root of 3 plus square root of 3 over denominator 38 end fraction end cell row blank equals cell fraction numerator 21 square root of 3 over denominator 38 end fraction end cell end table end style 

Jadi, jawaban yang tepat adalah B.

0

Roboguru

sin(x−45∘)=...

Pembahasan Soal:

Ingat kembali rumus perbandingan trigonometri untuk selisih dua sudut berikut.

  sin(AB)=sinAcosBcosAsinB 

Penyelesaian soal di atas yaitu.

sin(x45)===sinxcos45cosxsin45sinx212cosx212212(sinxcosx)      

Oleh karena itu, jawaban yang benar adalah B.

0

Roboguru

Nilai dari  ...

Pembahasan Soal:

Perbandingan trigonometri sudut-sudut berelasi yaitu:

table row cell sin space open parentheses 90 degree minus alpha close parentheses end cell equals cell cos space alpha end cell row cell sin space open parentheses 90 degree plus alpha close parentheses end cell equals cell cos space alpha end cell end table

Sehingga diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space 105 degree end cell equals cell cos space open parentheses 90 degree plus 15 degree close parentheses end cell row blank equals cell cos space 15 degree end cell row blank equals cell cos space open parentheses 45 degree minus 30 degree close parentheses end cell row blank equals cell cos space 45 degree times cos space 30 times plus sin space 45 degree times sin space 30 degree end cell row blank equals cell 1 half square root of 2 times 1 half square root of 3 plus 1 half square root of 2 times 1 half end cell row blank equals cell 1 fourth square root of 6 plus 1 fourth square root of 2 end cell row blank blank blank row cell sin space 15 degree end cell equals cell sin space open parentheses 45 degree minus 30 degree close parentheses end cell row blank equals cell sin space 45 degree times cos space 30 times negative cos space 45 degree times sin space 30 degree end cell row blank equals cell 1 half square root of 2 times 1 half square root of 3 minus 1 half square root of 2 times 1 half end cell row blank equals cell 1 fourth square root of 6 minus 1 fourth square root of 2 end cell row blank blank blank row cell sin space 75 degree end cell equals cell sin space open parentheses 90 degree minus 15 degree close parentheses end cell row blank equals cell cos space 15 degree end cell row blank equals cell cos space open parentheses 45 degree minus 30 degree close parentheses end cell row blank equals cell cos space 45 degree times cos space 30 times plus sin space 45 degree times sin space 30 degree end cell row blank equals cell 1 half square root of 2 times 1 half square root of 3 plus 1 half square root of 2 times 1 half end cell row blank equals cell 1 fourth square root of 6 plus 1 fourth square root of 2 end cell row blank blank blank row cell cos space 15 degree end cell equals cell cos space open parentheses 45 degree minus 30 degree close parentheses end cell row blank equals cell cos space 45 degree times cos space 30 times plus sin space 45 degree times sin space 30 degree end cell row blank equals cell 1 half square root of 2 times 1 half square root of 3 plus 1 half square root of 2 times 1 half end cell row blank equals cell 1 fourth square root of 6 plus 1 fourth square root of 2 end cell end table

Nilai sudut trigonometrinya yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator sin space 105 degree plus sin space 15 degree over denominator sin space 75 degree minus cos space 15 degree end fraction end cell equals cell fraction numerator open parentheses begin display style 1 fourth end style square root of 6 plus begin display style 1 fourth end style square root of 2 close parentheses plus open parentheses begin display style 1 fourth end style square root of 6 minus begin display style 1 fourth end style square root of 2 close parentheses over denominator open parentheses begin display style 1 fourth end style square root of 6 plus begin display style 1 fourth end style square root of 2 close parentheses minus open parentheses begin display style 1 fourth end style square root of 6 plus begin display style 1 fourth end style square root of 2 close parentheses end fraction end cell row blank equals cell fraction numerator begin display style 1 fourth end style square root of 6 plus begin display style 1 fourth end style square root of 6 plus begin display style 1 fourth end style square root of 2 minus begin display style 1 fourth end style square root of 2 over denominator begin display style 1 fourth end style square root of 6 minus begin display style 1 fourth end style square root of 6 plus begin display style 1 fourth end style square root of 2 minus begin display style 1 fourth end style square root of 2 end fraction end cell row blank equals cell fraction numerator begin display style 2 over 4 end style square root of 6 over denominator 0 end fraction end cell row blank equals infinity end table

Maka, nilai dari fraction numerator sin space 105 degree plus sin space 15 degree over denominator sin space 75 degree minus cos space 15 degree end fraction equals infinity.

0

Roboguru

Diketahui:  Ditanya: Nilai jumlah dan selisih sudut berikut:

Pembahasan Soal:

Diketahui:

sin space A equals 3 over 5 comma space angle A space lancip cos space B equals fraction numerator negative 12 over denominator 13 end fraction comma space angle B space tumpul

Untuk sudut lancip, nilai trigonometri sudut A seluruhnya bernilai positif.

table attributes columnalign right center left columnspacing 0px end attributes row sa equals cell square root of 5 squared minus 3 squared end root end cell row blank equals cell square root of 25 minus 9 end root end cell row blank equals cell square root of 16 end cell row blank equals 4 end table 

maka:

cos space A equals 4 over 5 tan space A equals 3 over 4

Untuk sudut tumpul dengan nilai cos negatif maka sudut terletak di kuadran II.

table attributes columnalign right center left columnspacing 0px end attributes row de equals cell square root of 13 squared minus 12 squared end root end cell row blank equals cell square root of 169 minus 144 end root end cell row blank equals cell square root of 25 end cell row blank equals 5 end table

maka:

sin space B equals 5 over 13 tan space B equals negative 5 over 12

  • Nilai sin space left parenthesis A minus B right parenthesis:

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space left parenthesis A minus B right parenthesis end cell equals cell sin space A space cos space B minus cos space A space sin space B end cell row blank equals cell 3 over 5 times fraction numerator negative 12 over denominator 13 end fraction minus 4 over 5 times 5 over 13 end cell row blank equals cell negative 36 over 65 minus 20 over 65 end cell row blank equals cell negative fraction numerator 36 plus 20 over denominator 65 end fraction end cell row blank equals cell negative 56 over 65 end cell end table

Nilai cos space open parentheses A plus B close parentheses:

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space open parentheses A plus B close parentheses end cell equals cell cos space A space cos space B minus sin space A space sin space B end cell row blank equals cell 4 over 5 times fraction numerator negative 12 over denominator 13 end fraction minus 3 over 5 times 5 over 13 end cell row blank equals cell negative 48 over 65 minus 15 over 65 end cell row blank equals cell negative 63 over 65 end cell end table

Jadi, sin space left parenthesis A minus B right parenthesis dan cos space open parentheses A plus B close parentheses secara berturut-turut adalah negative 56 over 65 dan negative 63 over 65.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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