Iklan

Pertanyaan

cos 30 0 ∘ ⋅ sin 31 5 ∘ cos 18 0 ∘ ⋅ sin 22 5 ∘ − tan 31 5 ∘ ​ = ...

    

  1. begin mathsize 14px style negative 3 open parentheses 1 plus square root of 2 close parentheses end style 

  2. begin mathsize 14px style negative 2 open parentheses 1 plus square root of 2 close parentheses end style 

  3. begin mathsize 14px style negative open parentheses 1 plus square root of 2 close parentheses end style 

  4. begin mathsize 14px style negative 1 half open parentheses 1 plus square root of 2 close parentheses end style 

  5. begin mathsize 14px style negative 1 fourth open parentheses 1 plus square root of 2 close parentheses end style 

Ikuti Tryout SNBT & Menangkan E-Wallet 100rb

Habis dalam

02

:

15

:

15

:

03

Klaim

Iklan

S. Nur

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah B.

jawaban yang tepat adalah B.

Pembahasan

Dengan demikian, jawaban yang tepat adalah B.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator cos space 180 degree times sin space 225 degree minus tan space 315 degree over denominator cos space 300 degree times sin space 315 degree end fraction end cell row blank equals cell blank fraction numerator cos space left parenthesis 180 degree plus 0 degree right parenthesis times sin space open parentheses 180 degree plus 45 degree close parentheses minus tan space left parenthesis 360 degree minus 45 degree right parenthesis over denominator cos space open parentheses 360 degree minus 60 degree close parentheses times sin space open parentheses 360 degree minus 45 degree close parentheses end fraction end cell row blank equals cell fraction numerator left parenthesis negative cos space 0 degree right parenthesis times left parenthesis negative sin space 45 degree right parenthesis minus left parenthesis negative tan space 45 degree right parenthesis over denominator cos space 60 degree times open parentheses negative sin space 45 degree close parentheses end fraction end cell row blank equals cell fraction numerator open parentheses negative 1 close parentheses open parentheses negative 1 half square root of 2 close parentheses plus 1 over denominator open parentheses 1 half close parentheses open parentheses negative 1 half square root of 2 close parentheses end fraction end cell row blank equals cell fraction numerator 1 half square root of 2 plus 1 over denominator negative 1 fourth square root of 2 end fraction cross times fraction numerator 4 square root of 2 over denominator 4 square root of 2 end fraction end cell row blank equals cell fraction numerator 2 times 2 plus 4 square root of 2 over denominator negative 2 end fraction end cell row blank equals cell negative 2 minus 2 square root of 2 end cell row blank equals cell negative 2 open parentheses 1 plus square root of 2 close parentheses end cell end table end style 


Dengan demikian, jawaban yang tepat adalah B.

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

101

Iklan

Pertanyaan serupa

Hitunglah nilai dari: cos 30 0 ∘ + tan 33 0 ∘ + sin 36 0 ∘ sin 15 0 ∘ + tan 22 5 ∘ − cos 12 0 ∘ ​

58

5.0

Jawaban terverifikasi

RUANGGURU HQ

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

Coba GRATIS Aplikasi Roboguru

Coba GRATIS Aplikasi Ruangguru

Download di Google PlayDownload di AppstoreDownload di App Gallery

Produk Ruangguru

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