Roboguru

Diketahui sinα=135​dansinβ=53​ Jika  sudut lancip α dan β sudut tumpul, maka  nilai dari cos(α+β)= ….

Pertanyaan

Diketahui sin invisible function application alpha equals 5 over 13 space d a n space sin invisible function application beta equals 3 over 5 Jika  sudut lancip alpha dan beta sudut tumpul, maka  nilai dari cos invisible function application left parenthesis alpha plus beta right parenthesis equals ….

  1. 16 over 65

  2. negative 63 over 65

  3. negative 16 over 63

  4. negative 56 over 65

  5. negative 33 over 65

Pembahasan Soal:

Dari sin invisible function application alpha equals 5 over 13  maka kita memperoleh segitiga berikut,

Sehingga cos invisible function application alpha equals 12 over 13 . Kemudian dari sin invisible function application beta equals 3 over 5 diperoleh segitiga

Sehingga cos invisible function application beta equals negative 4 over 5 (ingat, karena  sudut tumpul maka nilai cosinus bernilai negatif).

cos invisible function application left parenthesis alpha plus beta right parenthesis equals cos invisible function application alpha cos invisible function application beta minus sin invisible function application alpha sin invisible function application beta  equals open parentheses 12 over 13 close parentheses open parentheses negative 4 over 5 close parentheses minus open parentheses 5 over 13 close parentheses open parentheses 3 over 5 close parentheses  equals negative 48 over 65 minus 15 over 65  equals negative 63 over 65

 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

N. Ayu

Mahasiswa/Alumni Universitas Negeri Padang

Terakhir diupdate 04 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Misalkan, sudut pada segitiga ABC adalah A, B, dan C. Jika sinB+sinC=2sinA, maka nilai dari tan2B​⋅tan2C​ adalah ....

Pembahasan Soal:

Ingat kembali:

sin space open parentheses 180 minus a close parentheses equals sin space a

sin space a plus sin space b equals 2 space sin space open parentheses fraction numerator A plus B over denominator 2 end fraction close parentheses times cos space open parentheses fraction numerator A minus B over denominator 2 end fraction close parentheses  

cos space open parentheses a plus b close parentheses equals cos space a space cos space b minus sin space a space sin space b 

cos space open parentheses a minus b close parentheses equals cos space a space cos space b plus sin space a space sin space b

sin space a plus b equals 2 space sin space open parentheses fraction numerator A plus B over denominator 2 end fraction close parentheses times cos space open parentheses fraction numerator A plus B over denominator 2 end fraction close parentheses

tan space a equals fraction numerator sin space a over denominator cos space a end fraction

jumlah semua sudut pada segitiga adalah 180 degree, sehingga:

table attributes columnalign right center left columnspacing 0px end attributes row cell A plus B plus C end cell equals cell 180 degree end cell row A equals cell 180 minus open parentheses A plus B close parentheses end cell end table 

Sehingga diperoleh perhitungan:

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin space B plus sin space C end cell equals cell 2 space sin space A end cell row cell sin space B plus sin space C end cell equals cell 2 space sin space open parentheses 180 minus open parentheses B plus C close parentheses close parentheses end cell row cell sin space B plus sin space C end cell equals cell 2 space sin space open parentheses B plus C close parentheses end cell row cell 2 space sin space open parentheses fraction numerator B plus C over denominator 2 end fraction close parentheses times cos space open parentheses fraction numerator B minus C over denominator 2 end fraction close parentheses end cell equals cell 2 open parentheses 2 space sin space open parentheses fraction numerator B plus C over denominator 2 end fraction close parentheses times cos space open parentheses fraction numerator B plus C over denominator 2 end fraction close parentheses close parentheses end cell row cell up diagonal strike 2 space sin space open parentheses fraction numerator B plus C over denominator 2 end fraction close parentheses times end strike cos space open parentheses fraction numerator B minus C over denominator 2 end fraction close parentheses end cell equals cell 2 times up diagonal strike 2 space sin space open parentheses fraction numerator B plus C over denominator 2 end fraction close parentheses end strike times cos space open parentheses fraction numerator B plus C over denominator 2 end fraction close parentheses end cell row cell cos space open parentheses fraction numerator B minus C over denominator 2 end fraction close parentheses end cell equals cell 2 cos space open parentheses fraction numerator B plus C over denominator 2 end fraction close parentheses end cell row cell cos space 1 half B times cos space 1 half C plus sin space 1 half B times sin space 1 half C end cell equals cell 2 open parentheses cos space 1 half B times cos space 1 half C minus sin space 1 half B times sin space 1 half C close parentheses end cell row cell cos space 1 half B times cos space 1 half C plus sin space 1 half B times sin space 1 half C end cell equals cell 2 cos space 1 half B times cos space 1 half C minus 2 sin space 1 half B times sin space 1 half C end cell row cell 3 sin space 1 half B times sin space 1 half C end cell equals cell cos space 1 half B times cos space 1 half C end cell row cell fraction numerator sin space 1 half B times sin space 1 half C over denominator cos space 1 half B times cos space 1 half C end fraction end cell equals cell 1 third end cell row cell tan space 1 half B space tan space 1 half C end cell equals cell 1 third end cell end table end style

Dengan demikian, tan space B over 2 times tan space C over 2 adalah 1 third

Jadi, jawaban yang tepat adalah A

0

Roboguru

Himpunan penyelesaian dari persamaan sin(3x+225)∘+sin(3x−225)∘=−21​2​ untuk 0∘≤x≤180∘ adalah ....

Pembahasan Soal:

Ingat rumus penjumlahan sinus berikut untuk penyelesaian soal tersebut, yaitu:

sin open parentheses A plus B close parentheses plus sin open parentheses A minus B close parentheses equals 2 space sin space A space cos space B 

Dari soal diketahui bahwa A = 3x dan B = 225degree. Sehingga diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space open parentheses 3 x plus 225 close parentheses degree plus sin space open parentheses 3 x minus 225 close parentheses degree end cell equals cell negative 1 half square root of 2 end cell row cell 2 space sin space 3 x space cos space 225 degree end cell equals cell negative 1 half square root of 2 end cell end table 

Kita cari nilai cos space 225 degree dengan mengaplikasikan rumus penjumlahan dua sudut pada cosinus, sebagai berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space 225 degree end cell equals cell space cos space open parentheses 180 degree plus 45 degree close parentheses end cell row blank equals cell cos space 180 degree space cos space 45 degree minus sin space 180 degree space sin space 45 degree end cell row blank equals cell open parentheses negative 1 close parentheses open parentheses 1 half square root of 2 close parentheses minus 0 end cell row blank equals cell negative 1 half square root of 2 end cell end table 

Substitusi kembali nilai cos space 225 degree, sehingga diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 space sin space 3 x space cos space 225 degree end cell equals cell negative 1 half square root of 2 end cell row cell 2 space sin space 3 x space open parentheses up diagonal strike negative 1 half square root of 2 end strike close parentheses end cell equals cell up diagonal strike negative 1 half square root of 2 end strike end cell row cell sin space 3 x end cell equals cell 1 half end cell row cell sin space 3 x end cell equals cell sin space 30 degree end cell end table 

Berdasarkan rumus dasar trigonometri untuk sinus yaitu:

sin space x equals sin space A left right double arrow x equals A plus n times 360 space atau space x equals open parentheses 180 minus A close parentheses plus n times 360 

Maka:

 table attributes columnalign right center left columnspacing 0px end attributes row cell 3 x end cell equals cell 30 plus n times 360 end cell row x equals cell fraction numerator 30 plus n times 360 over denominator 3 end fraction end cell row x equals cell 10 plus n times 120 end cell end table atau table attributes columnalign right center left columnspacing 0px end attributes row cell 3 x end cell equals cell open parentheses 180 minus 30 close parentheses plus n times 360 end cell row cell 3 x end cell equals cell 150 plus n times 360 end cell row x equals cell fraction numerator 150 plus n times 360 over denominator 3 end fraction end cell row x equals cell 50 plus n times 120 end cell end table 

 

table row cell n equals 0 end cell rightwards double arrow cell x subscript 1 equals 10 plus 0 equals 10 end cell row blank blank cell x subscript 2 equals 50 plus 0 equals 50 end cell row cell n equals 1 end cell rightwards double arrow cell x subscript 3 equals 10 plus 120 equals 130 end cell row blank blank cell x subscript 4 equals 50 plus 120 equals 170 end cell row blank blank cell x subscript 5 equals 10 plus 240 equals 250 space open parentheses TM close parentheses end cell end table 

Dengan demikian, himpunan penyelesaian adalah HP = open curly brackets 10 degree comma space 50 degree comma space 130 degree comma space 170 degree close curly brackets.

Oleh karena itu, jawaban yang benar adalah D.

0

Roboguru

tan15∘+tan75∘=...

Pembahasan Soal:

Dengan menggunakan rumus sinus penjumlahan dua sudut, rumus cosinus penjumlahan & pengurangan dua sudut. Maka penjumlahan & pengurangan tangen dirumuskan sebagai berikut.

open parentheses straight i close parentheses space tan space alpha plus tan space beta equals fraction numerator 2 space sin space open parentheses alpha plus beta close parentheses over denominator cos space open parentheses alpha plus beta close parentheses plus cos space open parentheses alpha minus beta close parentheses end fraction left parenthesis ii right parenthesis space tan space alpha minus tan space beta equals fraction numerator 2 space sin space open parentheses alpha minus beta close parentheses over denominator cos space open parentheses alpha plus beta close parentheses plus cos space open parentheses alpha minus beta close parentheses end fraction

Berdasarkan rumus no (i) di atas, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell tan space 15 degree plus tan space 75 degree end cell equals cell fraction numerator 2 space sin space open parentheses 15 degree plus 75 degree close parentheses over denominator cos space open parentheses 15 degree plus 75 degree close parentheses plus cos space open parentheses 15 degree minus 75 degree close parentheses end fraction end cell row blank equals cell fraction numerator 2 space sin space 90 degree over denominator cos space 90 degree plus cos space open parentheses negative 60 degree close parentheses end fraction end cell row blank equals cell fraction numerator 2 space sin space 90 degree over denominator cos space 90 degree plus cos space open parentheses 60 degree close parentheses end fraction end cell row blank equals cell fraction numerator 2 times 1 over denominator 0 plus begin display style 1 half end style end fraction end cell row blank equals cell fraction numerator 2 over denominator begin display style 1 half end style end fraction end cell row blank equals cell 2 times 2 end cell row blank equals 4 end table

Dengan demikian, hasil dari tan space 15 degree plus tan space 75 degree adalah 4.

Oleh karena itu, jawaban yang benar adalah D.

0

Roboguru

Tentukan himpunan penyelesaian tiap persamaan berikut untuk 0∘≤x≤360∘. b. cos(2x+30∘)+cos(4x−60∘)=0

Pembahasan Soal:

Ingat bahwa:

  • cos space x plus cos space y equals 2 space cos space 1 half left parenthesis x plus y right parenthesis space cos space 1 half left parenthesis x minus y right parenthesis  

Mencari himpunan penyelesaian:

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space left parenthesis 2 x plus 30 degree right parenthesis plus cos space left parenthesis 4 x minus 60 degree right parenthesis end cell equals 0 row cell 2 cos space open parentheses fraction numerator 6 x minus 30 degree over denominator 2 end fraction close parentheses cos space open parentheses fraction numerator negative 2 x plus 90 degree over denominator 2 end fraction close parentheses end cell equals 0 row cell 2 cos space open parentheses 3 x minus 15 degree close parentheses cos space open parentheses negative x plus 45 degree close parentheses end cell equals 0 row cell cos space open parentheses 3 x minus 15 degree close parentheses cos space open parentheses negative x plus 45 degree close parentheses end cell equals 0 end table  

table attributes columnalign right center left columnspacing 0px end attributes row blank rightwards arrow cell cos space open parentheses 3 x minus 15 degree close parentheses equals 0 end cell row blank blank blank row blank rightwards arrow cell 3 x minus 15 degree equals 90 degree plus 360 degree n comma thin space end cell row x equals cell fraction numerator 360 degree n over denominator 3 end fraction plus 35 degree end cell row blank blank atau row cell 3 x minus 15 degree end cell equals cell 270 degree plus 360 degree n end cell row x equals cell fraction numerator 360 degree n over denominator 3 end fraction plus 95 degree end cell end table   

table attributes columnalign right center left columnspacing 0px end attributes row blank rightwards arrow cell cos open parentheses negative x plus 45 degree thin space close parentheses equals 0 end cell row blank blank blank row blank rightwards arrow cell negative x plus 45 degree equals 90 degree plus 360 degree n end cell row x equals cell negative 360 degree n minus 45 degree end cell row blank blank atau row cell negative x plus 45 degree end cell equals cell 270 degree plus 360 degree n end cell row x equals cell negative 360 degree n minus 225 degree end cell end table  

Sehingga didapat penyelesaiannya adalah HP equals open curly brackets table attributes columnalign right center left columnspacing 0px end attributes row x equals cell negative 360 degree n minus 45 degree comma space atau end cell row x equals cell negative 360 degree n minus 225 degree comma space atau end cell row x equals cell fraction numerator 360 degree thin space n over denominator 3 end fraction plus 35 degree comma space atau end cell row x equals cell fraction numerator 360 degree n over denominator 3 end fraction plus 95 degree end cell end table close curly brackets.

Yang memenuhi 0 degree less or equal than x less or equal than 360 degree adalah:

table attributes columnalign center left center end attributes row blank cell x equals negative 360 degree n minus 45 degree end cell blank row cell x equals 0 rightwards arrow end cell cell x equals negative 45 degree end cell memenuhi row blank cell x equals negative 360 degree n minus 225 degree end cell blank row cell x equals 0 rightwards arrow end cell cell x equals negative 225 degree end cell memenuhi row blank cell x equals fraction numerator 360 degree n over denominator 3 end fraction plus 35 degree end cell blank row cell x equals 0 rightwards arrow end cell cell x equals 35 degree end cell memenuhi row cell x equals 1 rightwards arrow end cell cell x equals 155 degree end cell memenuhi row cell x equals 2 rightwards arrow end cell cell x equals 275 degree end cell memenuhi row blank cell x equals fraction numerator 360 degree n over denominator 3 end fraction plus 95 degree end cell blank row cell x equals 0 rightwards arrow end cell cell x equals 95 degree end cell memenuhi row cell x equals 1 rightwards arrow end cell cell x equals 215 degree end cell memenuhi row cell x equals 2 rightwards arrow end cell cell x equals 335 degree end cell memenuhi end table 

Didapat:

HP equals open curly brackets 35 degree comma space 95 degree comma space 155 degree comma space 215 degree comma space 275 degree comma space 335 degree close curly brackets 

Jadi, himpunan penyelesaiannya adalah HP equals open curly brackets 35 degree comma space 95 degree comma space 155 degree comma space 215 degree comma space 275 degree comma space 335 degree close curly brackets.

0

Roboguru

Jika tanα=1dantanβ=1/3, dimana α, β sudut lancip, maka nilai dari cos(α+β) adalah...

Pembahasan Soal:

Diketahui tan space alpha equals 1 space dan space tan space beta equals 1 divided by 3, dengan menggambarkan segitiga siku-siku dan mengingat definisi tangen yaitu perbandingan sisi depan dan samping sudut, seperti terlihat pada gambar berikut:

Berdasarkan gambar, maka

cos space straight alpha equals fraction numerator samping space over denominator miring end fraction equals fraction numerator 1 over denominator square root of 2 end fraction comma space sin space straight alpha equals depan over miring equals fraction numerator 1 over denominator square root of 2 end fraction cos space straight beta equals fraction numerator samping space over denominator miring end fraction equals fraction numerator 3 over denominator square root of 10 end fraction comma space sin space straight beta equals depan over miring equals fraction numerator 1 over denominator square root of 10 end fraction

Nilai trigonometri pada sudut lancip adalah positif. Selanjutnya,

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space open parentheses straight alpha plus straight beta close parentheses end cell equals cell cos space straight alpha space cos space straight beta minus sin space straight alpha space sin space straight beta end cell row blank equals cell fraction numerator 1 over denominator square root of 2 end fraction open parentheses fraction numerator 3 over denominator square root of 10 end fraction close parentheses minus fraction numerator 1 over denominator square root of 2 end fraction open parentheses fraction numerator 1 over denominator square root of 10 end fraction close parentheses end cell row blank equals cell fraction numerator 3 over denominator square root of 20 end fraction minus fraction numerator 1 over denominator square root of 20 end fraction end cell row blank equals cell fraction numerator 2 over denominator square root of 20 end fraction end cell row blank equals cell fraction numerator 2 over denominator 2 square root of 5 end fraction end cell row blank equals cell 1 fifth square root of 5 end cell end table

Jadi, diperoleh hasilnya adalah 1 fifth square root of 5.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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