Pertanyaan

$begin mathsize 14px style fraction numerator sin space 45 degree plus cos space 45 degree over denominator tan space 60 degree end fraction equals.... end style$

Pembahasan Soal:

Kita gunakan nilai perbandingan trigonometri untuk sudut-sudut istimewa.

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator sin space 45 degree plus cos space 45 degree over denominator tan space 60 degree end fraction end cell equals cell fraction numerator begin display style 1 half end style square root of 2 plus begin display style 1 half end style square root of 2 over denominator square root of 3 end fraction end cell row blank equals cell fraction numerator square root of 2 over denominator square root of 3 end fraction times fraction numerator square root of 3 over denominator square root of 3 end fraction end cell row blank equals cell 1 third square root of 6 end cell end table end style$

Jadi, nilai dari $begin mathsize 14px style fraction numerator sin space 45 degree plus cos space 45 degree over denominator tan space 60 degree end fraction equals 1 third square root of 6 end style$ .

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

F. Kartikasari

Mahasiswa/Alumni Universitas Negeri Surabaya

Terakhir diupdate 28 Maret 2021

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Pembahasan Soal:

$begin mathsize 14px style fraction numerator 2 left parenthesis cos space 60 degree right parenthesis squared plus 4 left parenthesis sec space 30 degree right parenthesis to the power of 2 end exponent minus left parenthesis tan space 45 degree right parenthesis squared over denominator left parenthesis sin space 30 degree right parenthesis squared plus left parenthesis cos space 30 degree right parenthesis squared end fraction equals fraction numerator 2 open parentheses cos blank 60 degree close parentheses squared plus 4 open parentheses fraction numerator 1 over denominator cos space 30 degree end fraction close parentheses squared minus open parentheses tan blank 45 degree close parentheses squared over denominator open parentheses sin blank 30 degree close parentheses squared plus open parentheses cos blank 30 degree close parentheses squared end fraction equals fraction numerator 2 open parentheses 1 half close parentheses squared plus 4 open parentheses fraction numerator 2 square root of 3 over denominator 3 end fraction close parentheses squared minus open parentheses 1 close parentheses squared over denominator open parentheses 1 half close parentheses squared plus open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses squared end fraction equals fraction numerator 2 open parentheses 1 fourth close parentheses plus 4 open parentheses 12 over 9 close parentheses minus 1 over denominator 1 fourth plus 3 over 4 end fraction equals fraction numerator open parentheses 1 half close parentheses plus 4 open parentheses 4 over 3 close parentheses minus 1 over denominator 4 over 4 end fraction equals fraction numerator 1 half plus 16 over 3 minus 1 over denominator 1 end fraction equals 3 over 6 plus 32 over 6 minus 6 over 6 equals 29 over 6 equals 4 5 over 6 end style$

Jadi,

Roboguru

Nilai dari

Pembahasan Soal:

Dengan menggunakan konsep sudut berelasi diperoleh

$table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell tan space 1.860 degree plus sin space 1.485 degree end cell row blank equals cell tan space open parentheses 5 times 360 degree plus 60 degree close parentheses plus sin space open parentheses 4 times 360 degree plus 45 degree close parentheses end cell row blank equals cell tan space 60 degree plus sin space 45 degree end cell row blank equals cell square root of 3 plus 1 half square root of 2 end cell row blank equals cell 1 half open parentheses 2 square root of 3 plus square root of 2 close parentheses end cell end table$

Dengan demikian nilai dari $tan space 1.860 degree plus sin space 1.485 degree equals 1 half open parentheses 2 square root of 3 plus square root of 2 close parentheses$ .

Roboguru

Tentukan nilai fungsi berikut: untuk .

Pembahasan Soal:

Untuk mencari nilai fungsi $begin mathsize 14px style f open parentheses 15 degree close parentheses equals tan space 15 degree end style$ , kita dapat mengubah sudut $begin mathsize 14px style 15 degree end style$ menjadi pengurangan dua sudut istimewa terlebih dahulu, yaitu $begin mathsize 14px style 15 degree equals 45 degree minus 30 degree end style$ .

Lalu, kita gunakan rumus pengurangan selisih sudut pada tangen

$begin mathsize 14px style tan space open parentheses A minus B close parentheses equals fraction numerator tan space A minus tan space B over denominator 1 plus tan space A space tan space B end fraction end style$

sehingga diperoleh

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell tan space open parentheses 45 degree minus 30 degree close parentheses end cell equals cell fraction numerator tan space 45 degree minus tan space 30 degree over denominator 1 plus tan space 45 degree space tan space 30 degree end fraction end cell row blank equals cell fraction numerator 1 minus begin display style 1 third end style square root of 3 over denominator 1 plus 1 space cross times begin display style 1 third end style square root of 3 end fraction end cell row blank equals cell fraction numerator 1 minus begin display style 1 third end style square root of 3 over denominator 1 plus begin display style 1 third end style square root of 3 end fraction cross times fraction numerator 1 minus 1 third square root of 3 over denominator 1 minus 1 third square root of 3 end fraction end cell row blank equals cell fraction numerator open parentheses 1 minus 1 third square root of 3 close parentheses cross times open parentheses 1 minus 1 third square root of 3 close parentheses over denominator 1 minus begin display style 3 over 9 end style end fraction end cell row blank equals cell fraction numerator 1 minus begin display style 2 over 3 end style square root of 3 plus begin display style 1 third end style over denominator begin display style 6 over 9 end style end fraction end cell row blank equals cell fraction numerator begin display style 4 over 3 end style minus begin display style 2 over 3 end style square root of 3 over denominator begin display style 2 over 3 end style end fraction end cell row blank equals cell fraction numerator begin display style 4 minus 2 square root of 3 end style over denominator begin display style 2 end style end fraction end cell row blank equals cell 2 minus square root of 3 end cell end table end style$

Oleh karena itu, diperoleh $begin mathsize 14px style tan space 15 degree equals 2 minus square root of 3 end style$ .

Jadi, nilai fungsi $begin mathsize 14px style f left parenthesis x right parenthesis equals tan space x end style$ untuk $begin mathsize 14px style x equals 15 degree end style$ adalah $begin mathsize 14px style 2 minus square root of 3 end style$ . $space$

Roboguru

Hitunglah: c.

Pembahasan Soal:

Dengan menggunakan nilai trigonometri sudut istimewa diperoleh

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator sin left parenthesis 30 degree right parenthesis plus tan left parenthesis 45 degree right parenthesis plus cos left parenthesis 60 degree right parenthesis over denominator cosec left parenthesis 30 degree right parenthesis cotan left parenthesis 45 degree right parenthesis end fraction end cell equals cell fraction numerator begin display style 1 half end style plus 1 plus begin display style 1 half end style over denominator 2 times 1 end fraction end cell row blank equals cell 2 over 2 end cell row blank equals 1 end table end style$

Dengan demikian nilai dari $begin mathsize 14px style fraction numerator sin left parenthesis 30 degree right parenthesis plus tan left parenthesis 45 degree right parenthesis plus cos left parenthesis 60 degree right parenthesis over denominator cosec left parenthesis 30 degree right parenthesis cotan left parenthesis 45 degree right parenthesis end fraction end style$ adalah $begin mathsize 14px style 1 end style$ .

Roboguru

Nilai dari dikurangi dengan

Pembahasan Soal:

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator sin space 30 degree plus cos space 60 degree over denominator tan space 45 degree end fraction minus fraction numerator sin squared 45 degree over denominator cos squared 45 degree end fraction end cell equals cell fraction numerator begin display style 1 half end style plus begin display style 1 half end style over denominator 1 end fraction minus open parentheses begin display style 1 half end style square root of 2 close parentheses squared over open parentheses begin display style 1 half end style square root of 2 close parentheses squared end cell row blank equals cell 1 minus fraction numerator begin display style 1 half end style over denominator begin display style 1 half end style end fraction end cell row blank equals cell 1 minus 1 end cell row blank equals 0 end table end style$

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