Roboguru

Jika 0≤(α,B)≤3π​,cos(α+β)⋅cos(α−β)=0. dan sin(2π​−α)cosα=43​, maka α2−β2=…

Pertanyaan

Jika 0(α,B)3π,cos(α+β)cos(αβ)=0. dan sin(2πα)cosα=43, maka α2β2=

  1. 12π2 

  2. 6π2    

  3. 0    

  4. 12π2   

  5. 6π2     

Pembahasan Soal:

Ingat bahwa :

Sudut berelasi

sin(90α)=cosα

Rumus perkalian trigonometri

cosAcosB=21[cos(A+B)+cos(AB)]

dari soal diketahui

sin(2πα)cosαcosαcosαcos2αcosαcosαcosαα=======43434343213cos6π6π

cos(α+β)cos(αβ)21[cos(α+β+αβ)+cos(α+βα+β)]21[cos2α+cos2β]21[2cos2α1+2cos2β1]21[2cos2α+2cos2β2]cos2α+cos2β143+cos2β1431+cos2βcos2βcosβcosββ============000000004121cos3π3π

Sehingga

α2β2====(6π+3π)(6π3π)(6π+2π)(6π2π)363π2121π2

Jadi, jawaban yang tepat adalah A

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Roy

Mahasiswa/Alumni Universitas Negeri Surabaya

Terakhir diupdate 13 September 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: .

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

Buktikan identitas berikut. c.

Pembahasan Soal:

Ingat kembali:

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 space cos space straight alpha space cos space straight beta end cell equals cell cos space open parentheses straight alpha plus straight beta close parentheses plus cos space open parentheses straight alpha minus straight beta close parentheses end cell row cell cos space open parentheses 90 degree minus straight A close parentheses end cell equals cell sin space straight A end cell end table 

Akan dibuktikan 2 space cos space open parentheses straight theta plus straight pi over 4 close parentheses times cos space open parentheses fraction numerator 3 straight pi over denominator 4 end fraction minus straight theta close parentheses plus 1 equals sin space 2 straight theta. Maka:

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell Ruas space kiri end cell equals cell 2 space cos space open parentheses straight theta plus straight pi over 4 close parentheses times cos space open parentheses fraction numerator 3 straight pi over denominator 4 end fraction minus straight theta close parentheses plus 1 end cell row blank equals cell open square brackets cos space open parentheses open parentheses straight theta plus straight pi over 4 close parentheses plus open parentheses fraction numerator 3 straight pi over denominator 4 end fraction minus straight theta close parentheses close parentheses plus cos space open parentheses open parentheses straight theta plus straight pi over 4 close parentheses minus open parentheses fraction numerator 3 straight pi over denominator 4 end fraction minus straight theta close parentheses close parentheses close square brackets plus 1 end cell row blank equals cell open square brackets cos space open parentheses up diagonal strike straight theta plus straight pi over 4 plus fraction numerator 3 straight pi over denominator 4 end fraction up diagonal strike negative straight theta end strike close parentheses plus cos space open parentheses straight theta plus straight pi over 4 minus fraction numerator 3 straight pi over denominator 4 end fraction plus straight theta close parentheses close square brackets plus 1 end cell row blank equals cell open square brackets cos space open parentheses 4 over 4 straight pi close parentheses plus cos space open parentheses 2 straight theta minus 2 over 4 straight pi close parentheses close square brackets plus 1 end cell row blank equals cell open parentheses cos space 180 degree plus cos space open parentheses 2 straight theta minus 90 degree close parentheses close parentheses plus 1 end cell row blank equals cell open parentheses negative 1 plus sin space 2 straight theta close parentheses plus 1 end cell row blank equals cell up diagonal strike negative 1 end strike plus sin space 2 straight theta up diagonal strike plus 1 end strike end cell row blank equals cell sin space 2 straight theta end cell end table end style 

Jadi, terbukti bahwa 2 space cos space open parentheses straight theta plus straight pi over 4 close parentheses times cos space open parentheses fraction numerator 3 straight pi over denominator 4 end fraction minus straight theta close parentheses plus 1 equals sin space 2 straight theta.

0

Roboguru

Pembahasan Soal:

Rumus perkalian sinus dan kosinus adalah sebagai berikut.

open parentheses straight i close parentheses space space 2 space sin space A space cos space B equals sin space open parentheses A plus B close parentheses plus sin space open parentheses A minus B close parentheses left parenthesis ii right parenthesis space space 2 space cos space A space sin space B equals sin space open parentheses A plus B close parentheses minus sin space open parentheses A minus B close parentheses left parenthesis iii right parenthesis space 2 space cos space A space cos space B equals cos space open parentheses A plus B close parentheses plus cos space open parentheses A minus B close parentheses left parenthesis iv right parenthesis space 2 space sin space A space sin space B equals cos space open parentheses A minus B close parentheses minus cos space open parentheses A plus B close parentheses

Berdasarkan rumus no (i) dan (ii) di atas, maka:

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses sin space 75 degree space cos space 15 degree close parentheses minus open parentheses cos space 75 degree space sin space 15 degree close parentheses end cell row blank equals cell 1 half open parentheses sin space open parentheses 75 degree plus 15 degree close parentheses plus sin space open parentheses 75 degree minus 15 degree close parentheses close parentheses end cell row blank blank cell negative 1 half space open parentheses sin space open parentheses 75 degree plus 15 degree close parentheses minus sin space open parentheses 75 degree minus 15 degree close parentheses close parentheses end cell row blank equals cell 1 half open parentheses space sin space 90 degree plus sin space 60 degree close parentheses minus 1 half open parentheses sin space 90 degree minus sin space 60 degree close parentheses end cell row blank equals cell 1 half open parentheses 1 plus 1 half square root of 3 close parentheses minus 1 half open parentheses 1 minus 1 half square root of 3 close parentheses end cell row blank equals cell 1 half plus 1 fourth square root of 3 minus 1 half plus 1 fourth square root of 3 end cell row blank equals cell 2 over 4 square root of 3 end cell row blank equals cell 1 half square root of 3 end cell end table

Dengan demikian, hasil dari open parentheses sin space 75 degree space cos space 15 degree close parentheses minus open parentheses cos space 75 degree space sin space 15 degree close parentheses adalah 1 half square root of 3.

Oleh karena itu, jawaban yang benar adalah D.

0

Roboguru

Tunjukkanlah bahwa :

Pembahasan Soal:

Sifat penjumlahan dan pengurangan trigonometri :

sin space straight A plus sin space straight B equals 2 sin 1 half left parenthesis straight A plus straight B right parenthesis cos 1 half left parenthesis straight A minus straight B right parenthesis sin space straight A minus sin space straight B equals 2 cos 1 half left parenthesis straight A plus straight B right parenthesis sin 1 half left parenthesis straight A minus straight B right parenthesis cos space straight A plus cos space straight B equals 2 cos 1 half left parenthesis straight A plus straight B right parenthesis cos 1 half left parenthesis straight A minus straight B right parenthesis cos space straight A minus cos space straight B equals 2 sin 1 half left parenthesis straight A plus straight B right parenthesis sin 1 half left parenthesis straight A minus straight B right parenthesis 

Sifat perkalian trigonometri :

2 sin space straight A space cos space straight B equals sin left parenthesis straight A plus straight B right parenthesis plus sin left parenthesis straight A minus straight B right parenthesis 2 cos space straight A space sin space straight B equals sin left parenthesis straight A plus straight B right parenthesis minus sin left parenthesis straight A minus straight B right parenthesis 2 cos space straight A space cos space straight B equals cos left parenthesis straight A plus straight B right parenthesis plus cos left parenthesis straight A minus straight B right parenthesis minus 2 sin space straight A space sin space straight B equals cos left parenthesis straight A plus straight B right parenthesis plus cos left parenthesis straight A minus straight B right parenthesis 

Dengan menggunakan sifat tersebut, maka :

sin space 22 comma 5 degree cos space 22 comma 5 degree minus cos squared space 22 comma 5 degree equals 1 half open parentheses 2 sin space 22 comma 5 degree cos space 22 comma 5 degree close parentheses minus 1 half open parentheses 2 cos squared space 22 comma 5 degree close parentheses equals 1 half open parentheses sin space 45 degree close parentheses minus 1 half open parentheses 2 space cos space 22 comma 5 degree cos space 22 comma 5 degree close parentheses equals 1 half cross times 1 half square root of 2 minus 1 half open parentheses cos space 45 degree plus 1 close parentheses equals 1 half cross times 1 half square root of 2 minus 1 half cross times 1 half square root of 2 minus 1 half equals negative 1 half  

Maka terbukti bahwa sin space 22 comma 5 degree cos space 22 comma 5 degree minus cos squared space 22 comma 5 degree equals negative 1 half

0

Roboguru

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

Pembahasan Soal:

Ingat,

Perkalian Trigonometri Cosinus dengan Cosinus

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

Sudut Berelasi

cos(A)=cosA

Berdasarkan rumus tersebut, diperoleh sebagai berikut

cos40cos50cos10=====21(cos(40+50)+cos(4050))cos10(cos(90)+cos(10))2cos10(0+cos(10))2cos10cos102cos102

Dengan demikian, nilai dari  cos40cos50cos10  adalah  2

Oleh karena itu, jawaban yang benar adalah B. 

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