Roboguru

Nilai dari  cos40∘cos50∘cos10∘​  adalah ...

Pertanyaan

Nilai dari  fraction numerator cos space 10 degree over denominator cos space 40 degree space cos space 50 degree end fraction  adalah ...space space 

  1. 3space space 

  2. 2space space 

  3. 1space space 

  4. 1 halfspace space 

  5. 1 fourthspace space 

Pembahasan Soal:

Ingat,

Perkalian Trigonometri Cosinus dengan Cosinus

2 space cos space straight A space cos space straight B equals cos space left parenthesis straight A plus straight B right parenthesis space plus space cos space left parenthesis straight A – straight B right parenthesis

Sudut Berelasi

cos space left parenthesis negative straight A right parenthesis space equals space cos space straight A

Berdasarkan rumus tersebut, diperoleh sebagai berikut

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator cos space 10 degree over denominator cos space 40 degree space cos space 50 degree end fraction end cell equals cell fraction numerator cos space 10 degree over denominator begin display style 1 half end style open parentheses cos space open parentheses 40 degree plus 50 degree close parentheses plus cos space open parentheses 40 degree minus 50 degree close parentheses close parentheses end fraction end cell row blank equals cell fraction numerator 2 space cos space 10 degree over denominator begin display style open parentheses cos space open parentheses 90 degree close parentheses plus cos space open parentheses negative 10 degree close parentheses close parentheses end style end fraction end cell row blank equals cell fraction numerator 2 space cos space 10 degree over denominator begin display style open parentheses 0 plus cos space open parentheses 10 degree close parentheses close parentheses end style end fraction end cell row blank equals cell fraction numerator 2 space cos space 10 degree over denominator begin display style cos space 10 degree end style end fraction end cell row blank equals 2 end table

Dengan demikian, nilai dari  fraction numerator cos space 10 degree over denominator cos space 40 degree space cos space 50 degree end fraction  adalah  2

Oleh karena itu, jawaban yang benar adalah B.space space 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Acfreelance

Terakhir diupdate 15 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

Pertanyaan yang serupa

Tentukan nilai dari: sin35∘cos5∘−cos35∘cos85∘.

Pembahasan Soal:

Gunakan konsep rumus perkalian trigonometri dan sudut relasi open parentheses 90 degree minus x close parentheses.

table attributes columnalign right center left columnspacing 2px end attributes row cell 2 space sin space straight A space cos space straight B end cell equals cell sin space open parentheses straight A plus straight B close parentheses plus sin space open parentheses straight A minus straight B close parentheses end cell row atau blank blank row cell sin space straight A space cos space straight B end cell equals cell 1 half open parentheses sin space open parentheses straight A plus straight B close parentheses plus sin space open parentheses straight A minus straight B close parentheses close parentheses end cell row blank blank blank row cell 2 space cos space straight A space cos space straight B end cell equals cell cos space open parentheses straight A plus straight B close parentheses plus cos space open parentheses straight A minus straight B close parentheses end cell row atau blank blank row cell cos space straight A space cos space straight B end cell equals cell 1 half open parentheses cos space open parentheses straight A plus straight B close parentheses plus cos space open parentheses straight A minus straight B close parentheses close parentheses end cell row blank blank blank row cell sin space open parentheses 90 degree minus alpha close parentheses end cell equals cell cos space alpha end cell row cell cos space open parentheses negative alpha close parentheses end cell equals cell cos space alpha end cell end table

Ingat kembali nilai trigonometri sudut istimewa 30 degree dan sudut istimewa di kuadran II 120 degree.

table attributes columnalign right center left columnspacing 2px end attributes row cell sin space 30 degree end cell equals cell 1 half end cell row cell cos space 120 degree end cell equals cell negative 1 half end cell end table

Perhatikan perhitungan berikut.

table attributes columnalign right center left columnspacing 2px end attributes row blank blank cell sin space 35 degree space cos space 5 degree minus cos space 35 degree space cos space 85 degree end cell row blank equals cell 1 half open parentheses sin space open parentheses 35 degree plus 5 degree close parentheses plus sin space open parentheses 35 degree minus 5 degree close parentheses close parentheses minus end cell row blank blank cell 1 half open parentheses cos space open parentheses 35 degree plus 85 degree close parentheses plus cos space open parentheses 35 degree minus 85 degree close parentheses close parentheses end cell row blank equals cell 1 half open parentheses sin space 40 degree plus sin space 30 degree close parentheses minus 1 half open parentheses cos space 120 degree plus cos space open parentheses negative 50 degree close parentheses close parentheses end cell row blank equals cell 1 half open parentheses sin space open parentheses 90 degree minus 50 degree close parentheses plus 1 half close parentheses minus 1 half open parentheses negative 1 half plus cos space 50 degree close parentheses end cell row blank equals cell 1 half open parentheses cos space 50 degree plus 1 half close parentheses minus 1 half open parentheses negative 1 half plus cos space 50 degree close parentheses end cell row blank equals cell fraction numerator cos space 50 degree over denominator 2 end fraction plus 1 fourth plus 1 fourth minus fraction numerator cos space 50 degree over denominator 2 end fraction end cell row blank equals cell open parentheses 1 fourth plus 1 fourth close parentheses plus open parentheses fraction numerator cos space 50 degree over denominator 2 end fraction minus fraction numerator cos space 50 degree over denominator 2 end fraction close parentheses end cell row blank equals cell 2 over 4 plus 0 end cell row blank equals cell 1 half end cell end table 

Jadi, diperoleh nilai sin space 35 degree space cos space 5 degree minus cos space 35 degree space cos space 85 degree equals 1 half.

0

Roboguru

Getaran sebuah kawat ditentukan oleh persamaan: Tunjukkan bahwa persamaan tersebut ekuivalen dengan persamaan: y=4AcosT2πt​⋅cosλ2πx​

Pembahasan Soal:

Dengan menggunakan rumus trigonometri, diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row y equals cell 2 A open parentheses cos open square brackets 2 straight pi begin italic style left parenthesis t over T minus x over lambda right parenthesis end style close square brackets plus cos open square brackets 2 straight pi open parentheses t over T italic plus x over lambda close parentheses close square brackets close parentheses end cell row blank equals cell 2 A open parentheses 2 cos space 1 half open square brackets 2 straight pi open parentheses t over T italic minus x over lambda close parentheses plus 2 straight pi open parentheses straight t over straight T italic plus straight x over straight lambda close parentheses close square brackets times cos space 1 half open square brackets 2 straight pi open parentheses t over T italic minus x over lambda close parentheses minus 2 straight pi open parentheses straight t over straight T italic plus straight x over straight lambda close parentheses close square brackets close parentheses end cell row blank equals cell 2 A open parentheses 2 cos space 1 half open square brackets fraction numerator 2 straight pi t over denominator T end fraction minus fraction numerator 2 pi x over denominator straight lambda end fraction plus fraction numerator 2 pi t over denominator T end fraction plus fraction numerator 2 pi x over denominator straight lambda end fraction close square brackets times cos space 1 half open square brackets fraction numerator 2 straight pi t over denominator T end fraction minus fraction numerator 2 pi x over denominator straight lambda end fraction minus fraction numerator 2 pi t over denominator T end fraction minus fraction numerator 2 pi x over denominator straight lambda end fraction close square brackets close parentheses end cell row blank equals cell 2 A open parentheses 2 cos space open parentheses 1 half times fraction numerator 4 straight pi t over denominator T end fraction close parentheses times cos space open parentheses 1 half times fraction numerator negative 4 πx over denominator straight lambda end fraction close parentheses close parentheses end cell row blank equals cell 2 A open parentheses 2 cos fraction numerator 2 straight pi t over denominator T end fraction times cos fraction numerator 2 straight pi x over denominator straight lambda end fraction close parentheses end cell row blank equals cell 4 A cos fraction numerator 2 straight pi t over denominator T end fraction times cos fraction numerator 2 straight pi x over denominator straight lambda end fraction end cell row blank blank blank end table

Dengan demikian, terbukti bahwa

2 A open parentheses cos open square brackets 2 straight pi open parentheses straight t over straight T italic minus straight x over straight lambda close parentheses close square brackets plus cos open square brackets 2 straight pi open parentheses t over T italic plus x over lambda close parentheses close square brackets close parentheses equals 4 A cos space fraction numerator 2 pi t over denominator T end fraction times cos fraction numerator 2 pi x over denominator lambda end fraction

0

Roboguru

Nilai dari sin15∘1​+sin75∘1​=...

Pembahasan Soal:

Ingat kembali rumus-rumus penjumlahan sinus dan perkalian sinus sebagai berikut.

  1. sin space alpha plus sin space beta equals 2 space sin space open parentheses fraction numerator alpha plus beta over denominator 2 end fraction close parentheses space cos space open parentheses fraction numerator alpha minus beta over denominator 2 end fraction close parentheses
  2. 2 space sin space alpha space sin space beta equals cos space open parentheses alpha minus beta close parentheses minus cos space open parentheses alpha plus beta close parentheses

Berdasarkan kedua rumus di atas, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 1 over denominator sin space 15 degree end fraction plus fraction numerator 1 over denominator sin space 75 degree end fraction end cell equals cell fraction numerator sin space 75 degree plus sin space 15 degree over denominator sin space 15 degree space sin space 75 degree end fraction end cell row blank equals cell fraction numerator 2 space sin space open parentheses begin display style fraction numerator 90 degree over denominator 2 end fraction end style close parentheses space cos space open parentheses begin display style fraction numerator 60 degree over denominator 2 end fraction end style close parentheses over denominator begin display style 1 half end style space open parentheses cos space open parentheses negative 60 degree close parentheses minus cos space 90 degree space close parentheses end fraction end cell row blank equals cell 4 space fraction numerator sin space 45 degree space cos space 30 degree over denominator begin display style cos space open parentheses negative 60 degree close parentheses minus cos space 90 degree space end style end fraction end cell row blank equals cell 4 fraction numerator begin display style 1 half end style square root of 2 times begin display style 1 half end style square root of 3 over denominator begin display style 1 half end style minus 0 end fraction end cell row blank equals cell 4 fraction numerator begin display style 1 fourth square root of 6 end style over denominator begin display style 1 half end style end fraction end cell row blank equals cell 4 times 1 half square root of 6 end cell row blank equals cell 2 square root of 6 end cell end table

Dengan demikian, hasil dari fraction numerator 1 over denominator sin space 15 degree end fraction plus fraction numerator 1 over denominator sin space 75 degree end fraction adalah 2 square root of 6.

Oleh karena itu, jawaban yang benar adalah A.

0

Roboguru

cos45∘cos15∘=...

Pembahasan Soal:

Untuk menyelesaikan soal tersebut, kita bisa menggunakan rumus perkalian trigonometri:

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 space cos space A space cos space B end cell equals cell cos space open parentheses A plus B close parentheses plus space cos space open parentheses A minus B close parentheses end cell row cell cos space A space cos space B end cell equals cell fraction numerator cos space open parentheses A plus B close parentheses plus space cos space open parentheses A minus B close parentheses over denominator 2 end fraction end cell end table

Tabel sudut istimewa:

Pembahasan:

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space 45 degree space cos space 15 degree end cell equals cell fraction numerator cos space open parentheses 45 degree plus 15 degree close parentheses plus space cos space open parentheses 45 degree minus 15 degree close parentheses over denominator 2 end fraction end cell row blank equals cell fraction numerator cos space 60 degree plus cos space 30 degree over denominator 2 end fraction end cell row blank equals cell fraction numerator begin display style 1 half end style plus begin display style 1 half end style square root of 3 over denominator 2 end fraction end cell row blank equals cell 1 half open parentheses 1 plus square root of 3 close parentheses open parentheses 1 half close parentheses end cell row blank equals cell 1 fourth open parentheses 1 plus square root of 3 close parentheses end cell row blank equals cell 1 fourth plus 1 fourth square root of 3 end cell end table

Oleh karena itu, jawaban yang benar adalah C.

0

Roboguru

Nilai dari (tan10∘+tan70∘−tan50∘) adalah ....

Pembahasan Soal:

Ingat rumus jumlah dan selisih trigonometri berikut ini:

2sinAcosB=sin(A+B)+sin(AB)

2cosAsinB=sin(A+B)sin(AB)

2cosAcosB=cos(A+B)+cos(AB)

sinAsinB=2cos21(A+B)sin21(AB)

cosA+cosB=2cos21(A+B)cos21(AB)

Dengan menggunakan konsep di atas, diperoleh hasil:

tan10+tan70tan50=(cos10sin10+cos70sin70)cos50sin50=(cos70cos10cos70sin10+sin70cos10)cos50sin50=22(cos70cos10cos70sin10+sin70cos10)cos50sin50=(2cos70cos102cos70sin10+2sin70cos10)cos50sin50=(cos(70+10)+cos(7010)(sin(70+10)sin(7010))+(sin(70+10)+sin(7010)))cos50sin50=(cos80+cos60(sin80sin60)+(sin80+sin60))cos50sin50=cos80+212sin80cos50sin50=cos80cos50+21cos502sin80cos50sin50(cos80+21)(samakanpenyebut)=cos80cos50+21cos502sin80cos50cos80sin5021sin50=cos80cos50+21cos502sin80cos50cos80sin5021sin5022=2cos80cos50+cos5022sin80cos502cos80sin50sin50=cos(80+50)+cos(8050)+cos502(sin(80+50)+sin(8050))(sin(80+50)sin(8050))sin50=cos130+cos30+cos502(sin130+sin30)(sin130sin30)sin50=cos130+213+cos502(sin130+21)(sin13021)sin50=213+cos130+cos502sin130+1sin130+21sin50=213+cos130+cos5023+sin130sin50=213+(2cos21(130+50)cos21(13050))23+(2cos21(130+50)sin21(13050))=213+(2cos21(180)cos21(80))23+(2cos21(180)sin21(80))=213+(2cos90cos40)23+(2cos90sin40)=213+(20cos40)23+(20sin40)=213+023+0=21323=23÷23=23×32=33=33×33=333=3

Jadi, tan10+tan70tan50=3.

Oleh karena itu, jawaban yang benar adalah D.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved