Roboguru
SD
SMP
SMA
UTBK/SNBT

Iklan

Iklan

Pertanyaan

Jika sin(2x + 30 o ) = a dan sin(x + 45 o ) = b, maka sin(3x + 75 o )sin(x - 15 o ) = ....

Jika sin(2x + 30o) = a dan sin(x + 45o) = b, maka sin(3x + 75o)sin(x - 15o) = ....

  1. begin mathsize 14px style open parentheses a minus b close parentheses squared end style

  2. begin mathsize 14px style a squared minus b squared end style

  3. begin mathsize 14px style fraction numerator a squared over denominator square root of 2 end fraction minus fraction numerator begin display style b squared end style over denominator begin display style square root of 2 end style end fraction end style

  4. begin mathsize 14px style a squared minus fraction numerator 2 a b over denominator square root of 6 end fraction plus b squared end style

  5. begin mathsize 14px style 2 a b minus fraction numerator 1 over denominator square root of 6 end fraction end style

Ke roboguru plus

Iklan

A. Rizky

Master Teacher

Mahasiswa/Alumni Universitas Negeri Malang

Jawaban terverifikasi

Iklan

Pembahasan

begin mathsize 14px style text i) Tentukan nilai cos dari tiap bentuk end text text Ingat! end text c o s space alpha equals square root of 1 minus s i n squared space alpha end root text Maka end text c o s open parentheses 2 x plus 30 to the power of o close parentheses equals square root of 1 minus s i n squared open parentheses 2 x plus 30 to the power of o close parentheses end root equals square root of 1 minus a squared end root c o s open parentheses x plus 45 to the power of o close parentheses equals square root of 1 minus s i n squared open parentheses x plus 45 to the power of o close parentheses end root equals square root of 1 minus b squared end root text ii) end text text Perhatikan rumus untuk fungsi sin end text text Ingat! end text s i n open parentheses alpha plus beta close parentheses equals s i n space alpha space c o s space beta plus c o s space alpha space s i n space beta s i n open parentheses alpha minus beta close parentheses equals s i n space alpha space c o s space beta minus c o s space alpha space s i n space beta text Maka end text sin open parentheses 3 x plus 75 to the power of o close parentheses equals sin open parentheses open parentheses 2 x plus 30 to the power of o close parentheses plus open parentheses x plus 45 to the power of o close parentheses close parentheses equals sin open parentheses 2 x plus 30 to the power of o close parentheses cos open parentheses x plus 45 to the power of o close parentheses plus cos open parentheses 2 x plus 30 to the power of o close parentheses sin open parentheses x plus 45 to the power of o close parentheses equals a bullet square root of 1 minus b squared end root plus square root of 1 minus a squared end root bullet b text dan end text sin open parentheses x minus 15 to the power of o close parentheses equals sin open parentheses open parentheses 2 x plus 30 to the power of o close parentheses minus open parentheses x plus 45 to the power of o close parentheses close parentheses equals sin open parentheses 2 x plus 30 to the power of o close parentheses cos open parentheses x plus 45 to the power of o close parentheses minus cos open parentheses 2 x plus 30 to the power of o close parentheses sin open parentheses x plus 45 to the power of o close parentheses equals a bullet square root of 1 minus b squared end root minus square root of 1 minus a squared end root bullet b text sehingga end text sin open parentheses 3 x plus 75 to the power of o close parentheses sin open parentheses x minus 15 to the power of o close parentheses equals open parentheses a bullet square root of 1 minus b squared end root plus square root of 1 minus a squared end root bullet b close parentheses open parentheses a bullet square root of 1 minus b squared end root minus square root of 1 minus a squared end root bullet b close parentheses equals open parentheses a bullet square root of 1 minus b squared end root close parentheses squared minus open parentheses square root of 1 minus a squared end root bullet b close parentheses squared equals a squared open parentheses 1 minus b squared close parentheses minus open parentheses 1 minus a squared close parentheses b squared equals a squared minus a squared b squared minus b squared plus a squared b squared equals a squared minus b squared end style

Perdalam pemahamanmu bersama Master Teacher
di sesi Live Teaching, GRATIS!

3

Iklan

Iklan

Gold Icon

Klaim Gold gratis sekarang!

Dengan Gold kamu bisa tanya soal ke Forum sepuasnya, lho.

Pertanyaan serupa

Buktikan: b. sin ( x + y ) ⋅ sin ( x − y ) = sin 2 x − sin 2 y .

3

5.0

Jawaban terverifikasi

Dalam suatu segitiga ABC diketahui cos ∠ A = 5 3 ​ dan cos ∠ B = 13 5 ​ . Nilai sin ∠ C = ...

4

5.0

Jawaban terverifikasi

Iklan

Iklan

Dalam suatu segitiga ABC diketahui cos ∠ A = 5 3 ​ dan cos ∠ B = 13 5 ​ . Nilai sin ∠ C = ...

4

5.0

Jawaban terverifikasi

Buktikan: b. sin ( x + y ) ⋅ sin ( x − y ) = sin 2 x − sin 2 y .

3

5.0

Jawaban terverifikasi

Pada gambar berikut, diketahui AB = 10 cm , BC = 5 cm , ∠ BAP = x ∘ , dan BC tegak lurus AB . Proyeksi titik B dan C ke garis AP berturut-turut adalah titik D dan E . a. Tunjukkanlah ( i )...

2

0.0

Jawaban terverifikasi
Ruangguru

RUANGGURU HQ

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

Coba GRATIS Aplikasi Roboguru

Download di Google Play

Coba GRATIS Aplikasi Ruangguru

Download di Google PlayDownload di AppstoreDownload di App Gallery

Produk Ruangguru

Produk Lainnya

Bantuan & Panduan

Hubungi Kami

Ruangguru WhatsApp

+62 815-7441-0000

Email info@ruangguru.com

[email protected]

Contact 02140008000

02140008000

Ikuti Kami

©2024 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia