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Nilai dari

Pertanyaan

Nilai dari fraction numerator sin space 70 degree space sec space 140 degree space tan space 50 degree over denominator cos space 20 degree space sec space 40 degree space tan space 130 degree end fraction equals space... 

  1. negative 1 

  2. 0 

  3. 1 

  4. tan space 20 degree 

  5. sec space 20 degree 

Pembahasan Soal:

Dengan menggunakan konsep sudut berelasi, diperoleh 

fraction numerator sin space 70 degree space sec space 140 degree space tan space 50 degree over denominator cos space 20 degree space sec space 40 degree space tan space 130 degree end fraction equals fraction numerator sin space open parentheses 90 degree minus 20 degree close parentheses space sec space open parentheses 180 degree minus 40 degree close parentheses space tan space 50 degree over denominator cos space 20 degree space sec space 40 degree space tan space open parentheses 180 minus 50 close parentheses end fraction equals fraction numerator cos space 20 degree space open parentheses negative sec space 40 degree close parentheses space tan space 50 degree over denominator cos space 20 degree space sec space 40 degree space open parentheses negative tan space 50 degree close parentheses end fraction equals 1 

Dengan demikian, jawaban yang tepat adalah C.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

L. Rante

Mahasiswa/Alumni Universitas Negeri Makassar

Terakhir diupdate 04 Juni 2021

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Pertanyaan yang serupa

Tentukan bentuk sederhana dari

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space 70 degree end cell equals cell sin open parentheses 90 minus 20 close parentheses degree end cell row blank equals cell cos space 20 end cell row cell sec space 140 degree end cell equals cell sec open parentheses 180 minus 40 close parentheses end cell row blank equals cell negative sec space 40 degree end cell row cell tan space 130 degree end cell equals cell tan open parentheses 180 minus 50 close parentheses end cell row blank equals cell negative tan space 50 degree end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator sin space 70 degree s e c space 140 degree space tan 50 degree over denominator cos space 20 degree s e c space 40 degree space tan space 130 degree end fraction end cell equals cell fraction numerator cos space 20 degree times negative s e c space 40 degree times tan space 50 degree over denominator cos space 20 degree times s e c space 40 degree times negative tan space 50 degree end fraction end cell row blank equals cell fraction numerator negative 1 over denominator negative 1 end fraction end cell row blank equals 1 end table

Jadi, bentuk sederhana dari fraction numerator sin space 70 degree s e c space 140 degree space tan 50 degree over denominator cos space 20 degree s e c space 40 degree space tan space 130 degree end fraction adalah 1

 

Roboguru

Buktikan bahwa a.

Pembahasan Soal:

Dengan menggunakan konsep sudut berelasi diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator sin open parentheses 270 plus p close parentheses times sin open parentheses 180 minus p close parentheses over denominator cos open parentheses 90 minus p close parentheses times cos open parentheses 180 minus p close parentheses end fraction end cell equals cell fraction numerator negative cos space p times space sin space p over denominator sin space p times negative cos space p end fraction end cell row blank equals cell fraction numerator negative sin space p space cos space p over denominator negative sin space p space cos space p space end fraction end cell row blank equals 1 end table

Dengan demikian terbukti bahwa fraction numerator sin open parentheses 270 plus p close parentheses times sin open parentheses 180 minus p close parentheses over denominator cos open parentheses 90 minus p close parentheses times cos open parentheses 180 minus p close parentheses end fraction equals 1.

Roboguru

Bentuk sederhana dari         adalah ....

Pembahasan Soal:

begin mathsize 14px style sin space left parenthesis 360 degree minus alpha right parenthesis plus cos space left parenthesis 90 degree minus alpha right parenthesis minus cos space left parenthesis 180 degree minus alpha right parenthesis plus sin space left parenthesis 270 degree minus alpha right parenthesis equals negative sin space alpha plus sin space alpha minus open parentheses negative cos space alpha close parentheses plus open parentheses negative cos space alpha close parentheses equals negative sin space alpha plus sin space alpha plus cos space alpha minus cos space alpha equals 0 end style 

jadi jawaban yang paling tepat adalah E

Roboguru

Nilai dari adalah ....

Pembahasan Soal:

begin mathsize 14px style fraction numerator sin 30 degree space tan 135 degree space cos 45 degree over denominator cos 60 degree space sin space 225 degree space tan 150 degree end fraction space equals fraction numerator sin 30 degree space open parentheses negative tan 45 degree close parentheses space cos 45 degree over denominator cos 60 degree space open parentheses negative sin space 45 degree close parentheses space open parentheses negative tan 30 degree close parentheses end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator 1 half. open parentheses negative 1 close parentheses. open parentheses 1 half square root of 2 close parentheses over denominator 1 half. open parentheses negative 1 half square root of 2 close parentheses. open parentheses negative 1 third square root of 3 close parentheses end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator negative 1 fourth square root of 2 over denominator 1 over 12 square root of 6 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals negative fraction numerator 3 over denominator square root of 3 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals negative square root of 3 end style  

Jadi, jawaban yang tepat adalah B.

Roboguru

Jika  berada di kuadran kedua dengan . Tunjukkan bahwa:

Pembahasan Soal:

Ingat kembali perbandingan trigonometri sudut berelasi berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell tan space open parentheses 90 degree minus theta close parentheses end cell equals cell cot space theta end cell row cell sin space open parentheses 270 degree plus theta close parentheses end cell equals cell negative cos space theta end cell row cell cos space open parentheses 180 degree minus theta close parentheses end cell equals cell negative cos space theta end cell end table

Jika tan space theta equals negative 2 over 3theta berada di kuadran II.

tan space theta equals negative 2 over 3 equals de over sa

table attributes columnalign right center left columnspacing 0px end attributes row x equals cell square root of 3 squared plus 2 squared end root end cell row blank equals cell square root of 9 plus 4 end root end cell row blank equals cell square root of 13 end cell end table

Maka, 

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space theta end cell equals cell de over mi equals fraction numerator 2 over denominator square root of 13 end fraction end cell row cell cos space theta end cell equals cell sa over mi equals negative fraction numerator 3 over denominator square root of 13 end fraction end cell end table

Sehingga, diperoleh perhitungan:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator tan space open parentheses 90 degree minus theta close parentheses plus cos space open parentheses 180 degree minus theta close parentheses over denominator sin space open parentheses 270 degree plus theta close parentheses minus cot space open parentheses negative theta close parentheses end fraction end cell equals cell fraction numerator 2 plus square root of 13 over denominator 2 minus square root of 13 end fraction end cell row cell fraction numerator cot space theta minus cos space theta over denominator negative space cos plus cot space theta end fraction end cell equals cell fraction numerator 2 plus square root of 13 over denominator 2 minus square root of 13 end fraction end cell row cell fraction numerator cot space theta minus cos space theta over denominator cot space theta minus space cos end fraction end cell equals cell fraction numerator 2 plus square root of 13 over denominator 2 minus square root of 13 end fraction end cell row 1 equals cell fraction numerator 2 plus square root of 13 over denominator 2 minus square root of 13 end fraction open parentheses Tidak space terbukti close parentheses end cell end table

Jadi, tidak terbukti bahwa fraction numerator tan space open parentheses 90 degree minus theta close parentheses plus cos space open parentheses 180 degree minus theta close parentheses over denominator sin space open parentheses 270 degree plus theta close parentheses minus cot space open parentheses negative theta close parentheses end fraction adalah fraction numerator 2 plus square root of 13 over denominator 2 minus square root of 13 end fraction.

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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