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Tentukan nilai trigonometri berikut! d.

Pertanyaan

Tentukan nilai trigonometri berikut!

d. begin mathsize 14px style fraction numerator 2 space tan space 37 comma 5 degree over denominator 1 minus tan squared space 37 comma 5 degree end fraction end style 

Pembahasan Soal:

Dengan menerapkan rumus trigonometri sudut rangkap untuk tangen, diperoleh perhitungan sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 2 space tan space a over denominator 1 minus tan squared space a end fraction end cell equals cell tan space 2 a end cell row cell fraction numerator 2 space tan space 37 comma 5 degree over denominator 1 minus tan squared space 37 comma 5 degree end fraction end cell equals cell tan space 2 open parentheses 37 comma 5 degree close parentheses end cell row blank equals cell tan space 75 degree end cell end table end style 

Selanjutnya, dengan menerapkan rumus jumlah dua sudut untuk tangen, diperoleh:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell tan space open parentheses a plus b close parentheses end cell equals cell fraction numerator tan space a plus tan space b over denominator 1 minus tan space a space tan space b end fraction end cell row cell tan space 75 degree end cell equals cell tan space open parentheses 45 degree plus 30 degree close parentheses end cell row blank equals cell fraction numerator tan space 45 plus tan space 30 degree over denominator 1 minus tan space 45 space tan space 30 degree end fraction end cell row blank equals cell fraction numerator 1 plus begin display style 1 third end style square root of 3 over denominator 1 minus open parentheses 1 close parentheses open parentheses begin display style 1 third end style square root of 3 close parentheses end fraction end cell row blank equals cell fraction numerator 1 plus begin display style 1 third end style square root of 3 over denominator 1 minus begin display style 1 third end style begin display style square root of 3 end style end fraction cross times 3 over 3 end cell row blank equals cell fraction numerator 3 plus square root of 3 over denominator 3 minus square root of 3 end fraction end cell row blank equals cell fraction numerator 3 plus square root of 3 over denominator 3 minus square root of 3 end fraction cross times fraction numerator 3 plus square root of 3 over denominator 3 plus square root of 3 end fraction end cell row blank equals cell fraction numerator 9 plus 6 square root of 3 plus 3 over denominator 9 minus 3 end fraction end cell row blank equals cell fraction numerator 12 plus 6 square root of 3 over denominator 6 end fraction end cell row blank equals cell 2 plus square root of 3 end cell end table end style 

Dengan demikian, nilai begin mathsize 14px style fraction numerator 2 space tan space 37 comma 5 degree over denominator 1 minus tan squared space 37 comma 5 degree end fraction end style adalah begin mathsize 14px style 2 plus square root of 3 end style.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

L. Marlina

Mahasiswa/Alumni Institut Teknologi Bandung

Terakhir diupdate 31 Agustus 2021

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Pertanyaan yang serupa

Jika , dimana  sudut lancip. Hitunglah tanpa menggunakan kalkulator: a.

Pembahasan Soal:

Ingat kembali:

table attributes columnalign right center left columnspacing 0px end attributes row cell cot space straight A end cell equals cell fraction numerator 1 over denominator tan space straight A end fraction end cell row cell tan space 2 straight A end cell equals cell fraction numerator 2 space tan space straight A over denominator 1 minus tan squared space straight A end fraction end cell end table 

Diketahui tan space straight theta equals 2 over 3, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell cot space 2 straight theta end cell equals cell fraction numerator 1 over denominator tan space 2 straight theta end fraction end cell row blank equals cell fraction numerator 1 over denominator begin display style fraction numerator 2 space tan space straight theta over denominator 1 minus tan squared space straight theta end fraction end style end fraction end cell row blank equals cell fraction numerator 1 over denominator begin display style fraction numerator 2 cross times begin display style 2 over 3 end style over denominator 1 minus open parentheses begin display style 2 over 3 end style close parentheses squared end fraction end style end fraction end cell row blank equals cell fraction numerator 1 over denominator begin display style fraction numerator begin display style 4 over 3 end style over denominator 1 minus begin display style 4 over 9 end style end fraction end style end fraction end cell row blank equals cell fraction numerator 1 over denominator begin display style fraction numerator begin display style 4 over 3 end style over denominator begin display style fraction numerator 9 minus 4 over denominator 9 end fraction end style end fraction end style end fraction end cell row blank equals cell fraction numerator 1 over denominator begin display style fraction numerator begin display style 4 over 3 end style over denominator begin display style 5 over 9 end style end fraction end style end fraction end cell row blank equals cell fraction numerator 1 over denominator begin display style 4 over 3 end style cross times begin display style 9 over 5 end style end fraction end cell row blank equals cell fraction numerator 1 over denominator begin display style 36 over 15 end style end fraction end cell row blank equals cell 1 cross times 15 over 36 end cell row blank equals cell 15 over 36 end cell row blank equals cell 5 over 12 end cell end table 

Jadi, nilai cot space 2 straight theta equals 5 over 12.

Roboguru

Jika  dan , hitunglah nilai dari: c.

Pembahasan Soal:

Sudut Rangkap pada Tangen

tan space 2 alpha equals fraction numerator 2 space tan space alpha over denominator 1 minus tan squared space alpha end fraction     

Pada rumus sudut rangkap, 2 alpha merupakan 2 kali dari alpha.

Diketahui tan space alpha equals negative 1 half dan straight pi over 2 less than alpha less than straight pi, maka letak sudut alpha adalah

 table attributes columnalign right center left columnspacing 0px end attributes row cell straight pi over 2 end cell less than cell alpha less than straight pi end cell row cell fraction numerator 180 degree over denominator 2 end fraction end cell less than cell alpha less than 180 degree end cell row cell 90 degree end cell less than cell alpha less than 180 degree space open parentheses kuadran space II close parentheses end cell end table 

Pada kuadran II, tangen dan cosinus bernilai negatif sedangkan sinus bernilai positif.    

Sehingga, tan space 4 alpha adalah:

table attributes columnalign right center left columnspacing 0px end attributes row cell tan space 4 alpha end cell equals cell fraction numerator 2 space tan space 2 alpha over denominator 1 minus tan squared space 2 alpha end fraction end cell row blank equals cell fraction numerator 2 times fraction numerator 2 space tan space alpha over denominator 1 minus tan squared space alpha end fraction over denominator 1 minus open parentheses fraction numerator 2 space tan space alpha over denominator 1 minus tan squared space alpha end fraction close parentheses squared end fraction end cell row blank equals cell fraction numerator fraction numerator 4 times open parentheses negative begin display style 1 half end style close parentheses over denominator 1 minus open parentheses negative 1 half close parentheses squared end fraction over denominator 1 minus open parentheses fraction numerator 2 times open parentheses negative 1 half close parentheses over denominator 1 minus open parentheses negative 1 half close parentheses squared end fraction close parentheses squared end fraction end cell row blank equals cell fraction numerator begin display style fraction numerator negative 2 over denominator 1 minus begin display style 1 fourth end style end fraction end style over denominator 1 minus begin display style open parentheses negative 1 close parentheses squared over open parentheses begin display style 3 over 4 end style close parentheses squared end style end fraction end cell row blank equals cell fraction numerator negative 2 cross times begin display style 4 over 3 end style over denominator 1 minus open parentheses 1 cross times begin display style 16 over 9 end style close parentheses end fraction end cell row blank equals cell fraction numerator negative begin display style 8 over 3 end style over denominator begin display style 9 over 9 end style minus begin display style 16 over 9 end style end fraction end cell row blank equals cell fraction numerator negative begin display style 8 over 3 end style over denominator begin display style negative 7 over 9 end style end fraction end cell row blank equals cell negative fraction numerator 8 over denominator up diagonal strike 3 end fraction cross times open parentheses negative fraction numerator up diagonal strike 9 over denominator 7 end fraction close parentheses end cell row blank equals cell fraction numerator 8 cross times 3 over denominator 7 end fraction end cell row blank equals cell 24 over 7 end cell end table      

Jadi, table attributes columnalign right center left columnspacing 0px end attributes row blank blank tan end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 4 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank alpha end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 24 over 7 end cell end table.

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Diketahui , dengan . Nilai ...

Pembahasan Soal:

Ingat definisi sinus, cosinus, dan tangen:

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space alpha end cell equals cell fraction numerator sisi space depan over denominator sisi space miring end fraction end cell row cell cos space alpha end cell equals cell fraction numerator sisi space samping over denominator sisi space miring end fraction end cell row cell tan space alpha end cell equals cell fraction numerator sin space alpha over denominator cos space alpha end fraction end cell end table 

Ingat konsep sudut rangkap:

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space 2 alpha end cell equals cell 2 space sin space alpha space cos space alpha end cell row cell cos space 2 alpha end cell equals cell 1 minus sin squared alpha end cell row cell tan space 2 alpha end cell equals cell fraction numerator sin space 2 alpha over denominator cos space 2 alpha end fraction end cell end table  

Diketahui sin space A equals fraction numerator 2 over denominator square root of 5 end fraction, dengan 0 less than A less than 90 degree maka dengan menggunakan teorema Pythagoras diperoleh sisi samping:

table attributes columnalign right center left columnspacing 0px end attributes row cell sisi space depan end cell equals 2 row cell sisi space miring end cell equals cell square root of 5 end cell row blank blank blank row cell sisi space samping end cell equals cell square root of open parentheses square root of 5 close parentheses squared minus 2 squared end root end cell row blank equals cell square root of 5 minus 4 end root end cell row blank equals 1 end table

sehingga 

cos space straight A equals fraction numerator 1 over denominator square root of 5 end fraction

Nilai dari 2A, ada 3 kemungkinan maksud dari soal yaitu sin space 2 straight Acos space 2 straight A, atau tan space 2 straight A sehingga:

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space 2 A end cell equals cell 2 space sin space A times cos space A end cell row blank equals cell 2 open parentheses fraction numerator 2 over denominator square root of 5 end fraction close parentheses open parentheses fraction numerator 1 over denominator square root of 5 end fraction close parentheses end cell row blank equals cell 4 over 5 end cell row blank blank blank row cell cos space 2 A end cell equals cell 1 minus 2 space sin squared A end cell row blank equals cell 1 minus 2 open parentheses fraction numerator 2 over denominator square root of 5 end fraction close parentheses squared end cell row blank equals cell 1 minus 2 open parentheses 4 over 5 close parentheses end cell row blank equals cell 1 minus 8 over 5 end cell row blank equals cell 5 over 5 minus 8 over 5 end cell row blank equals cell negative 3 over 5 end cell row blank blank blank row cell tan space 2 A end cell equals cell fraction numerator sin space 2 A over denominator cos space 2 A end fraction end cell row blank equals cell fraction numerator begin display style 4 over 5 end style over denominator negative begin display style 3 over 5 end style end fraction end cell row blank equals cell 4 over 5 cross times open parentheses negative 5 over 3 close parentheses end cell row blank equals cell negative 4 over 3 end cell end table  

kemungkinan yang ditanyak pada soal nilai tan space 2 straight A adalah negative 4 over 3

Oleh karena itu, jawaban yang benar adalah B.

Roboguru

Jika dan x sudut lancip, maka nilai dari tan3x adalah...

Pembahasan Soal:

tanx equals 1 half  tan space 3 straight x equals fraction numerator 3 tanx space minus space tan 3 straight x over denominator 1 minus 3 tan squared straight x end fraction  equals fraction numerator 3 open parentheses begin display style 1 half end style close parentheses minus open parentheses begin display style 1 half end style close parentheses squared over denominator 1 minus 3 open parentheses begin display style 1 half end style close parentheses squared end fraction  equals fraction numerator begin display style 3 over 2 end style minus begin display style 1 over 8 end style over denominator 1 minus begin display style 3 over 4 end style end fraction equals fraction numerator begin display style fraction numerator 12 minus 1 over denominator 8 end fraction end style over denominator begin display style fraction numerator 4 minus 3 over denominator 4 end fraction end style end fraction equals fraction numerator begin display style 11 over 8 end style over denominator begin display style 1 fourth end style end fraction equals 11 over 8 straight x 4 over 1  equals 11 over 2 equals 5 1 half

Roboguru

Jika  dengan . Maka

Pembahasan Soal:

Diketahui tan space x equals m, maka 

table attributes columnalign right center left columnspacing 0px end attributes row cell tan space 2 x end cell equals cell fraction numerator 2 space tan space x over denominator 1 minus tan squared x end fraction end cell row blank equals cell fraction numerator 2 times m over denominator 1 minus m squared end fraction end cell row blank equals cell fraction numerator 2 m over denominator 1 minus m squared end fraction end cell end table 

Dengan menggunakan konsep perbandingan trigonometri maka 

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space 2 x end cell equals cell fraction numerator 2 m over denominator square root of open parentheses 2 m close parentheses squared plus open parentheses 1 minus m squared close parentheses squared end root end fraction end cell row blank equals cell fraction numerator 2 m over denominator square root of 4 m squared plus 1 minus 2 m squared plus m to the power of 4 end root end fraction end cell row blank equals cell fraction numerator 2 m over denominator square root of m to the power of 4 plus 2 m squared plus 1 end root end fraction end cell row blank equals cell fraction numerator 2 m over denominator m squared plus 1 end fraction end cell end table 

Dengan menggunakan konsep sudut rangkap sinus, diperoleh 

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space x times cos space x end cell equals cell 1 half sin space 2 x end cell row blank equals cell 1 half times fraction numerator 2 m over denominator open parentheses m squared plus 1 close parentheses end fraction end cell row blank equals cell fraction numerator m over denominator m squared plus 1 end fraction end cell end table 

Jadi nilai sin space x times cos space x equals fraction numerator m over denominator m squared plus 1 end fraction

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