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Jika sinAcosB=p dan sin(A−B)=q. Nilai dari cotanAtanB=...

Pertanyaan

Jika sinAcosB=p dan sin(AB)=q. Nilai dari cotanAtanB=... 

  1. 1pq 

  2. 1+pq 

  3. 1qp 

  4. 1+qp 

  5. qp1 

Pembahasan Soal:

Diketahui,

Rumus Perbandingan Trigonometri untuk Selisih Dua Sudut (Sinus)

sin(AB)=sinAcosBcosAsinB

Identitas Trigonometri

tanAcotanA==cosAsinAsinAcosA

Berdasarkan rumus tersebut, diperoleh sebagai berikut

Diketahui sinAcosB=p dan sin(AB)=q

► Menentukan cosAsinB

sin(AB)qcosAsinB===sinAcosBcosAsinBpcosAsinBpq

► Menentukan Nilai dari cotanAtanB

cotanAtanB=====sinAcosAcosBsinBsinAcosBcosAsinBppqpp(1pq)1pq 

Dengan demikian, nilai dari cotanAtanB=1pq

Oleh karena itu, jawaban yang benar adalah A. 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

R. Utami

Mahasiswa/Alumni Universitas Negeri Malang

Terakhir diupdate 12 September 2021

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Pertanyaan yang serupa

Tentukan ekspresi aljabar dari setiap ekspresi berikut. c.

Pembahasan Soal:

Ingat perbandingan sisi trigonometri berikut:

sin space theta equals fraction numerator sisi space depan over denominator sisi space miring end fraction rightwards arrow cosec space theta equals fraction numerator sisi space miring over denominator sisi space depan end fraction cos space theta equals fraction numerator sisi space samping over denominator sisi space miring end fraction rightwards arrow sec space theta equals fraction numerator sisi space miring over denominator sisi space depan end fraction tan space theta equals fraction numerator sisi space depan over denominator sisi space samping end fraction rightwards arrow cotan space theta equals fraction numerator sisi space miring over denominator sisi space depan end fraction 

Diketahui ekspresi sin space open parentheses cos to the power of negative 1 end exponent space x minus sin to the power of negative 1 end exponent space 3 x close parentheses. Misalkan:

- Untuk cos to the power of negative 1 end exponent space x 

table attributes columnalign right center left columnspacing 0px end attributes row theta equals cell cos to the power of negative 1 end exponent space x end cell row cell cos space theta end cell equals x row cell cos space theta end cell equals cell x over 1 space rightwards arrow fraction numerator sisi space samping over denominator sisi space miring end fraction end cell end table 

- Untuk sin to the power of negative 1 end exponent space 3 x 

table attributes columnalign right center left columnspacing 0px end attributes row beta equals cell sin to the power of negative 1 end exponent space 3 x end cell row cell sin space beta end cell equals cell 3 x end cell row cell sin space beta end cell equals cell fraction numerator 3 x over denominator 1 end fraction space rightwards arrow space fraction numerator sisi space depan over denominator sisi space miring end fraction end cell end table 

Perhatikan gambar berikut:

Maka:

p equals square root of 1 squared minus x squared end root p equals square root of 1 minus x squared end root    dan   q equals square root of 1 squared minus open parentheses 3 x close parentheses squared end root q equals square root of 1 minus 9 x squared end root 

Sehingga untuk sin space open parentheses x minus y close parentheses equals sin space x space cos space y minus cos space x space sin space y, digunakan sebagai berikut:

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin space open parentheses cos to the power of negative 1 end exponent space x minus sin to the power of negative 1 end exponent space 3 x close parentheses end cell equals cell sin space open parentheses theta minus beta close parentheses end cell row blank equals cell sin space theta space cos space beta minus cos space theta space sin space beta end cell row blank equals cell fraction numerator sisi space depan space theta over denominator sisi space miring space theta end fraction times fraction numerator sisi space samping space beta over denominator sisi space miring space beta end fraction minus fraction numerator sisi space samping space theta over denominator sisi space miring space theta end fraction times fraction numerator sisi space depan space beta over denominator sisi space miring space beta end fraction end cell row blank equals cell fraction numerator square root of 1 minus x squared end root over denominator 1 end fraction times fraction numerator square root of 1 minus 9 x squared end root over denominator 1 end fraction minus x over 1 times fraction numerator 3 x over denominator 1 end fraction end cell row blank equals cell open parentheses square root of 1 minus x squared end root close parentheses open parentheses square root of 1 minus 9 x squared end root close parentheses minus 3 x squared end cell end table end style  

Jadi, sin space open parentheses cos to the power of negative 1 end exponent space x minus sin to the power of negative 1 end exponent space 3 x close parentheses equals open parentheses square root of 1 minus x squared end root close parentheses open parentheses square root of 1 minus 9 x squared end root close parentheses minus 3 x squared.

0

Roboguru

Dengan menggunakan formula  dan pemisalan, hitunglah .

Pembahasan Soal:

Ingat perbandingan sisi trigonometri berikut:

sin space theta equals fraction numerator sisi space depan over denominator sisi space miring end fraction rightwards arrow cosec space theta equals fraction numerator sisi space miring over denominator sisi space depan end fraction cos space theta equals fraction numerator sisi space samping over denominator sisi space miring end fraction rightwards arrow sec space theta equals fraction numerator sisi space miring over denominator sisi space depan end fraction tan space theta equals fraction numerator sisi space depan over denominator sisi space samping end fraction rightwards arrow cotan space theta equals fraction numerator sisi space miring over denominator sisi space depan end fraction 

Diketahui ekspresi sin space open square brackets cos to the power of negative 1 end exponent space 3 over 5 minus tan to the power of negative 1 end exponent space 7 over 13 close square brackets. Misalkan:

table attributes columnalign right center left columnspacing 0px end attributes row theta equals cell cos to the power of negative 1 end exponent space 3 over 5 end cell row cell cos space theta end cell equals cell 3 over 5 rightwards arrow fraction numerator sisi space samping over denominator sisi space miring end fraction end cell end table    dan     table attributes columnalign right center left columnspacing 0px end attributes row beta equals cell tan to the power of negative 1 end exponent space 7 over 13 end cell row cell tan space beta end cell equals cell 7 over 13 rightwards arrow fraction numerator sisi space depan over denominator sisi space samping end fraction end cell end table 

Perhatikan gambar berikut:

Maka:

x equals square root of 3 squared plus 4 squared end root x equals square root of 9 plus 16 end root x equals square root of 25 x equals 5    dan     y equals square root of 7 squared plus 13 squared end root y equals square root of 49 plus 169 end root y equals square root of 218 

Sehingga formula sin space open parentheses x minus y close parentheses equals sin space x space cos space y minus cos space x space sin space y, dapat digunakan sebagai berikut:

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin space open square brackets cos to the power of negative 1 end exponent space 3 over 5 minus tan to the power of negative 1 end exponent space 7 over 13 close square brackets end cell equals cell sin space open parentheses theta minus beta close parentheses end cell row blank equals cell sin space theta space cos space beta minus cos space theta space sin space beta end cell row blank equals cell fraction numerator sisi space depan space theta over denominator sisi space miring space theta end fraction times fraction numerator sisi space samping space beta over denominator sisi space miring space beta end fraction minus fraction numerator sisi space samping space theta over denominator sisi space miring space theta end fraction times fraction numerator sisi space depan space beta over denominator sisi space miring space beta end fraction end cell row blank equals cell 4 over 5 times fraction numerator 13 over denominator square root of 218 end fraction minus 3 over 5 times fraction numerator 7 over denominator square root of 218 end fraction end cell row blank equals cell fraction numerator 52 over denominator 5 square root of 218 end fraction minus fraction numerator 21 over denominator 5 square root of 218 end fraction end cell row blank equals cell fraction numerator 31 over denominator 5 square root of 218 end fraction cross times fraction numerator square root of 218 over denominator square root of 218 end fraction end cell row blank equals cell fraction numerator 31 square root of 218 over denominator 5 cross times 218 end fraction end cell row blank equals cell fraction numerator 31 square root of 218 over denominator 1.090 end fraction end cell end table end style 

Jadi, sin space open square brackets cos to the power of negative 1 end exponent space 3 over 5 minus tan to the power of negative 1 end exponent space 7 over 13 close square brackets equals fraction numerator 31 square root of 218 over denominator 1.090 end fraction.

0

Roboguru

Diketahui  sudut lancip dengan tan . Nilai sin

Pembahasan Soal:

 

0

Roboguru

Jika , buktikan bahwa .

Pembahasan Soal:

Ingat rumus selisih dua sudut pada sinus yaitu

sin open parentheses text A end text minus text B end text close parentheses equals sin space text A end text times cos space text B end text minus cos space text A end text space sin space text B end text

Sehingga fraction numerator text F end text over denominator sin open parentheses text A end text minus a close parentheses end fraction equals fraction numerator text W end text over denominator sin space text A end text end fraction dapat dijabarkan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator text F end text over denominator sin open parentheses text A end text minus a close parentheses end fraction end cell equals cell fraction numerator text W end text over denominator sin space text A end text end fraction end cell row cell fraction numerator text F end text over denominator sin space text A end text space cos space a minus cos space text A end text space sin space a end fraction end cell equals cell fraction numerator text W end text over denominator sin space text A end text end fraction end cell row cell text F end text sin space A end cell equals cell text W end text sin space text A end text space cos space a minus text W end text cos space text A end text space sin space a end cell row cell text F end text sin space text A end text minus text W end text sin space text A end text space cos space a end cell equals cell negative text W end text cos space text A end text space sin space a end cell row cell sin space text A end text open parentheses text F end text minus text W end text cos space a close parentheses end cell equals cell negative text W end text cos space text A end text space sin space a end cell row cell fraction numerator sin space text A end text over denominator cos space text A end text end fraction open parentheses text F end text minus text W end text cos space a close parentheses end cell equals cell fraction numerator negative text W end text cos space text A end text space sin space a over denominator cos space text A end text end fraction end cell row cell tan space text A end text open parentheses text F end text minus text W end text cos space a close parentheses end cell equals cell negative text W end text space sin space a end cell row cell tan space text A end text end cell equals cell fraction numerator negative text W end text space sin space a over denominator text F end text minus text W end text cos space a end fraction end cell row cell tan space text A end text end cell equals cell fraction numerator negative text W end text space sin space a over denominator negative left parenthesis text W end text cos space a minus text F end text right parenthesis end fraction end cell row cell tan space text A end text end cell equals cell fraction numerator text W end text space sin space a over denominator text W end text cos space a minus text F end text end fraction end cell end table


Jadi berdasarkan hitungan di atas terbukti bahwa tan space text A end text equals fraction numerator text W end text space sin space a over denominator text W end text space cos space a minus text F end text end fraction.

0

Roboguru

Seorang mencoba menentukan tinggi nyala api di puncak tugu Monas di Jakarta dengan cara mengukur sudut lihat dari suatu tempat sejauh a dari kaki tugu itu dan  seperti dalam gambar. jika x tinggi nyal...

Pembahasan Soal:

tan space alpha equals fraction numerator B D over denominator A B end fraction  B D equals A B space tan space alpha...............1 right parenthesis  tan space beta equals fraction numerator B C over denominator A B end fraction equals fraction numerator B D minus x over denominator A B end fraction  B D minus x equals A B space tan space beta...........2 right parenthesis  x equals alpha left parenthesis tan space alpha minus tan space beta right parenthesis  x equals alpha open parentheses fraction numerator sin space alpha over denominator cos space alpha end fraction minus fraction numerator sin space beta over denominator cos space beta end fraction close parentheses  x equals alpha open parentheses fraction numerator sin space alpha space cos space beta minus sin space alpha space cos space beta over denominator cos space alpha space cos space beta end fraction close parentheses  x equals alpha open parentheses fraction numerator sin space alpha left parenthesis alpha minus beta right parenthesis over denominator cos space alpha space cos space beta end fraction close parentheses

 

0

Roboguru

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