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Diketahui tan17∘=247​ (nilai pembulatan). Tentukan nilai bentuk trigonometri berikut. a. cos17∘sin73∘−sin17∘tan73∘  b. sec287∘cosec163∘+tan343∘​

Pertanyaan

Diketahui tan space 17 degree equals 7 over 24 (nilai pembulatan). Tentukan nilai bentuk trigonometri berikut.

a. cos space 17 degree space sin space 73 degree minus sin space 17 degree space tan space 73 degree 

b. fraction numerator cosec space 163 degree plus tan space 343 degree over denominator sec space 287 degree end fraction 

Pembahasan Soal:

Ingat kembali definisi sincostan, dan : 

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

Ingat kembali 

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space open parentheses 90 degree minus straight alpha close parentheses end cell equals cell cos space straight alpha end cell row cell tan space open parentheses 90 degree minus straight alpha close parentheses end cell equals cell cotan space straight alpha end cell row cell tan space open parentheses 360 degree minus straight alpha close parentheses end cell equals cell negative tan space straight alpha end cell row cell cosec space open parentheses 180 degree minus straight alpha close parentheses end cell equals cell cosec space straight alpha end cell row cell sec space open parentheses 270 degree plus straight alpha close parentheses end cell equals cell cosec space straight alpha end cell end table  

Diketahui tan space 17 degree equals 7 over 24 dengan menggunakan teorema Pythagoras maka diperoleh sisi miring:

table attributes columnalign right center left columnspacing 0px end attributes row cell sisi space miring end cell equals cell square root of 24 squared plus 7 squared end root end cell row blank equals cell square root of 576 plus 49 end root end cell row blank equals cell square root of 625 end cell row blank equals 25 end table 

a. sehingga diperoleh 

cos space 17 degree space sin space 73 degree minus sin space 17 degree space tan space 73 degree equals cos space 17 degree space sin space open parentheses 90 degree minus space 17 degree close parentheses minus sin space 17 degree space tan space open parentheses 90 degree minus space 17 degree close parentheses equals cos space 17 degree space cos space 17 degree minus sin space 17 degree space cotan space 17 degree equals 24 over 25 times 24 over 25 minus 7 over 25 times 24 over 7 equals space 576 over 625 minus 24 over 25 equals fraction numerator 576 minus 600 over denominator 625 end fraction equals negative 24 over 625 

Dengan demikian nilai cos space 17 degree space sin space 73 degree minus sin space 17 degree space tan space 73 degree adalah negative 24 over 625.

b. sehingga diperoleh

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator cosec space 163 degree plus tan space 343 degree over denominator sec space 287 degree end fraction end cell equals cell fraction numerator cosec space open parentheses 180 degree minus 17 degree close parentheses plus tan space open parentheses 360 degree minus 17 degree close parentheses over denominator sec space open parentheses 270 degree plus 17 degree close parentheses end fraction end cell row blank equals cell fraction numerator cos e c space 17 degree plus open parentheses negative tan space 17 degree close parentheses over denominator cos e c space 17 degree end fraction end cell row blank equals cell fraction numerator begin display style 25 over 7 end style minus begin display style 7 over 24 end style over denominator begin display style 25 over 7 end style end fraction end cell row blank equals cell fraction numerator begin display style fraction numerator 600 minus 49 over denominator 168 end fraction end style over denominator begin display style 25 over 7 end style end fraction end cell row blank equals cell fraction numerator begin display style 551 over 168 end style over denominator begin display style 25 over 7 end style end fraction end cell row blank equals cell fraction numerator 551 over denominator up diagonal strike 168 end fraction cross times fraction numerator up diagonal strike 7 over denominator 25 end fraction end cell row blank equals cell 551 over 600 end cell end table end style 

Dengan demikian nilai dari fraction numerator cosec space 163 degree plus tan space 343 degree over denominator sec space 287 degree end fraction  adalah 551 over 600.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Amamah

Mahasiswa/Alumni Universitas Negeri Malang

Terakhir diupdate 24 September 2021

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Pertanyaan yang serupa

Jika θ berada di kuadran kedua dengan tanθ=−32​. Tunjukkan bahwa: cos(270∘+θ)+cos(360∘−θ)sin(90∘−θ)−cos(180∘−θ)​=6

Pembahasan Soal:

Ingat kembali perbandingan trigonometri sudut berelasi berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space open parentheses 180 degree minus theta close parentheses end cell equals cell negative space cos theta end cell row cell sin space open parentheses 90 degree minus theta close parentheses end cell equals cell cos space theta end cell row cell cos space open parentheses 270 degree plus theta close parentheses end cell equals cell sin space theta end cell row cell cos space open parentheses 360 degree minus theta close parentheses end cell equals cell cos space theta end cell end table

Jika tan space theta equals negative 2 over 3theta berada di kuadran II.

tan space theta equals negative 2 over 3 equals de over sa

table attributes columnalign right center left columnspacing 0px end attributes row x equals cell square root of 3 squared plus 2 squared end root end cell row blank equals cell square root of 9 plus 4 end root end cell row blank equals cell square root of 13 end cell end table

Maka, 

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space theta end cell equals cell de over mi equals fraction numerator 2 over denominator square root of 13 end fraction end cell row cell cos space theta end cell equals cell sa over mi equals negative fraction numerator 3 over denominator square root of 13 end fraction end cell end table

Sehingga, diperoleh perhitungan:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator sin space open parentheses 90 degree minus theta close parentheses minus cos space open parentheses 180 degree minus theta close parentheses over denominator cos space open parentheses 270 degree plus theta close parentheses plus cos space open parentheses 360 degree minus theta close parentheses end fraction end cell equals 6 row cell fraction numerator cos space theta minus open parentheses negative cos space theta close parentheses over denominator sin space theta plus cos space theta end fraction end cell equals 6 row cell fraction numerator 2 space cos space theta over denominator sin space theta plus cos space theta end fraction end cell equals 6 row cell fraction numerator 2 times negative begin display style fraction numerator 3 over denominator square root of 13 end fraction end style over denominator begin display style fraction numerator 2 over denominator square root of 13 end fraction end style minus begin display style fraction numerator 3 over denominator square root of 13 end fraction end style end fraction cross times fraction numerator square root of 13 over denominator square root of 13 end fraction end cell equals 6 row cell fraction numerator negative 6 over denominator 2 minus 3 end fraction end cell equals 6 row cell fraction numerator negative 6 over denominator negative 1 end fraction end cell equals 6 row 6 equals cell 6 space open parentheses Terbukti close parentheses end cell end table

Jadi, terbukti bahwa fraction numerator sin space open parentheses 90 degree minus theta close parentheses minus cos space open parentheses 180 degree minus theta close parentheses over denominator cos space open parentheses 270 degree plus theta close parentheses plus cos space open parentheses 360 degree minus theta close parentheses end fraction adalah 6.

0

Roboguru

Jika θ berada di kuadran kedua dengan tanθ=−32​. Tunjukkan bahwa: sin(270∘+θ)−cot(−θ)tan(90∘−θ)+cos(180∘−θ)​=2−13​2+13​​

Pembahasan Soal:

Ingat kembali perbandingan trigonometri sudut berelasi berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell tan space open parentheses 90 degree minus theta close parentheses end cell equals cell cot space theta end cell row cell sin space open parentheses 270 degree plus theta close parentheses end cell equals cell negative cos space theta end cell row cell cos space open parentheses 180 degree minus theta close parentheses end cell equals cell negative cos space theta end cell end table

Jika tan space theta equals negative 2 over 3theta berada di kuadran II.

tan space theta equals negative 2 over 3 equals de over sa

table attributes columnalign right center left columnspacing 0px end attributes row x equals cell square root of 3 squared plus 2 squared end root end cell row blank equals cell square root of 9 plus 4 end root end cell row blank equals cell square root of 13 end cell end table

Maka, 

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space theta end cell equals cell de over mi equals fraction numerator 2 over denominator square root of 13 end fraction end cell row cell cos space theta end cell equals cell sa over mi equals negative fraction numerator 3 over denominator square root of 13 end fraction end cell end table

Sehingga, diperoleh perhitungan:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator tan space open parentheses 90 degree minus theta close parentheses plus cos space open parentheses 180 degree minus theta close parentheses over denominator sin space open parentheses 270 degree plus theta close parentheses minus cot space open parentheses negative theta close parentheses end fraction end cell equals cell fraction numerator 2 plus square root of 13 over denominator 2 minus square root of 13 end fraction end cell row cell fraction numerator cot space theta minus cos space theta over denominator negative space cos plus cot space theta end fraction end cell equals cell fraction numerator 2 plus square root of 13 over denominator 2 minus square root of 13 end fraction end cell row cell fraction numerator cot space theta minus cos space theta over denominator cot space theta minus space cos end fraction end cell equals cell fraction numerator 2 plus square root of 13 over denominator 2 minus square root of 13 end fraction end cell row 1 equals cell fraction numerator 2 plus square root of 13 over denominator 2 minus square root of 13 end fraction open parentheses Tidak space terbukti close parentheses end cell end table

Jadi, tidak terbukti bahwa fraction numerator tan space open parentheses 90 degree minus theta close parentheses plus cos space open parentheses 180 degree minus theta close parentheses over denominator sin space open parentheses 270 degree plus theta close parentheses minus cot space open parentheses negative theta close parentheses end fraction adalah fraction numerator 2 plus square root of 13 over denominator 2 minus square root of 13 end fraction.

0

Roboguru

Jika tana=−32​ dan a sudut dikuadran II, maka tan(270∘+a)+ctg(360∘−a)sin(90∘−a)−cos(180∘−a)​=…

Pembahasan Soal:

Berdasarkan konsep perbandingan trigonometri pada sudut begin mathsize 14px style open parentheses 90 degree minus x close parentheses end style, sudut begin mathsize 14px style open parentheses 180 degree minus x close parentheses end style, sudut begin mathsize 14px style open parentheses 270 degree plus x close parentheses end style, dan sudut begin mathsize 14px style open parentheses 360 degree minus x close parentheses end style, maka diperoleh bahwa:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin space open parentheses 90 degree minus a close parentheses end cell equals cell cos space a end cell row cell cos space open parentheses 180 degree minus a close parentheses end cell equals cell negative cos space a end cell row cell tan space left parenthesis 270 plus a right parenthesis end cell equals cell negative ctg space a end cell row cell ctg space left parenthesis 360 minus a right parenthesis end cell equals cell negative ctg space a end cell end table end style 

Sehingga penyederhanaan operasi trigonometri tersebut adalahsebagai berikut:

  begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator sin space open parentheses 90 degree minus a close parentheses minus cos space open parentheses 180 degree minus a close parentheses over denominator tan space open parentheses 270 degree plus a close parentheses plus ctg space open parentheses 360 degree minus a close parentheses end fraction end cell equals cell fraction numerator cos space a minus open parentheses negative cos space a close parentheses over denominator negative ctg space a plus open parentheses negative ctg space a close parentheses end fraction end cell row blank equals cell fraction numerator 2 times cos space a over denominator negative 2 times ctg space a end fraction end cell row blank equals cell fraction numerator cos space a over denominator negative ctg space a end fraction end cell row blank equals cell negative fraction numerator cos space a over denominator begin display style fraction numerator cos space a over denominator sin space a end fraction end style end fraction end cell row blank equals cell negative sin space a end cell end table end style 

Misalkan begin mathsize 14px style b end style adalah suatu sudut di kuadran I yang memenuhi persamaan berikut:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis 180 minus b right parenthesis end cell equals a end table end style 

Berdasarkan konsep perbandingan trigonometri sudut begin mathsize 14px style open parentheses 180 degree minus x close parentheses end style, diperoleh bahwa:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell tan space left parenthesis 180 minus b right parenthesis end cell equals cell negative tan space b end cell row cell tan space a end cell equals cell negative tan space b end cell row cell negative 2 over 3 end cell equals cell negative tan space b end cell row cell tan space b end cell equals cell 2 over 3 end cell end table end style  

Nilai begin mathsize 14px style sin space b end style dapat diperoleh melalui perbandingan sisi segitiga siku-siku dengan salah sudutnya adalah begin mathsize 14px style b end style, panjang sisi depan sudut tersebut adalah 2 satuan, dan panjang sisi sampingnya adalah 3 satuan. Dengan begitu, panjang sisi miring segitiga tersebut adalah sebagai berikut:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sisi space miring end cell equals cell square root of 3 squared plus 2 squared end root end cell row blank equals cell square root of 9 plus 4 end root end cell row blank equals cell square root of 13 space satuan end cell end table end style  

Sehingga, 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin space b end cell equals cell fraction numerator sisi space depan over denominator sisi space miring end fraction end cell row blank equals cell fraction numerator 2 over denominator square root of 13 end fraction end cell row blank equals cell 2 over 13 square root of 13 space end cell row blank blank blank end table end style  

Dari sini diperoleh:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell negative sin space a end cell equals cell negative sin space left parenthesis 180 degree minus b right parenthesis end cell row blank equals cell negative sin space b end cell row blank equals cell negative 2 over 13 square root of 13 end cell end table end style 

Jadi, begin mathsize 14px style fraction numerator sin space open parentheses 90 degree minus a close parentheses minus cos space open parentheses 180 degree minus a close parentheses over denominator tan space open parentheses 270 degree plus a close parentheses plus ctg space open parentheses 360 degree minus a close parentheses end fraction equals negative 2 over 13 square root of 13 end style

0

Roboguru

Nyatakan koordinat kutub dalam koordinat kartesius. 4. D(20,330∘)

Pembahasan Soal:

Jika diketahui korrdinat kutub begin mathsize 14px style open parentheses straight r comma straight alpha close parentheses end style, maka koordinat kartesiusnya:

begin mathsize 14px style straight x equals straight r space cos space straight alpha straight y equals straight r space sin space straight alpha end style 

Dengan demikian, koordinat kartesiusnya:

xy==========20cos33020cos(36030)20cos3020×21310320sin33020sin(36030)20sin3020×2110    

Sehingga, koordinat kartesiusnya adalah (103,10).

 

0

Roboguru

Diketahui tan17∘=247​ (nilai pembulatan). Tentukan nilai bentuk trigonometri berikut. b. sec287∘cosec163∘+tan343∘​

Pembahasan Soal:

Dari soal tersebut, pertama-tama tentukan panjang sisi terhadap sumbu x dan y di sudut tersebut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell tan space 17 degree end cell equals cell 7 over 24 end cell row cell y over x end cell equals cell 7 over 24 end cell row y equals cell space 7 end cell row x equals cell space 24 end cell end table end style

Setelah itu, tentukan panjang sisi miring dari sudut tersebut dengan menggunakan panjang x dan y yang telah ditemukan.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row s equals cell square root of x squared plus y squared end root end cell row blank equals cell square root of 24 squared plus 7 squared end root end cell row blank equals 25 end table end style

Dengan menggunakan nilai tersebut nilai trigonometri pada soal bisa diperoleh.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator cos e c space 163 degree plus tan space 343 degree over denominator s e c space 287 degree end fraction end cell equals blank row cell fraction numerator begin display style fraction numerator 1 over denominator sin space left parenthesis 180 degree minus 17 degree right parenthesis end fraction end style plus tan space left parenthesis 360 degree minus 17 degree right parenthesis over denominator begin display style fraction numerator 1 over denominator cos space left parenthesis 270 degree plus 17 degree right parenthesis end fraction end style end fraction end cell equals blank row cell fraction numerator begin display style fraction numerator 1 over denominator sin space 17 degree end fraction end style minus space tan space 17 degree over denominator begin display style fraction numerator 1 over denominator sin space 17 degree end fraction end style end fraction end cell equals blank row cell fraction numerator begin display style s over y end style minus begin display style y over x end style over denominator begin display style y over s end style end fraction end cell equals blank row cell fraction numerator begin display style 25 over 7 end style minus begin display style 7 over 24 end style over denominator begin display style 25 over 7 end style end fraction end cell equals blank row cell 551 over 600 end cell equals blank end table end style

1

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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