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Jika  dan  sudut lancip dari suatu segitiga siku-siku dan  maka  adalah ...

Pertanyaan

Jika begin mathsize 14px style alpha end style dan begin mathsize 14px style beta end style sudut lancip dari suatu segitiga siku-siku dan begin mathsize 14px style tan space alpha equals square root of 2 space sin space beta end style maka begin mathsize 14px style sin squared space alpha end style adalah ...

  1. begin mathsize 14px style 4 over 5 end style 

  2. begin mathsize 14px style 3 over 4 end style 

  3. begin mathsize 14px style 2 over 3 end style 

  4. begin mathsize 14px style 1 half end style 

  5. begin mathsize 14px style 1 third end style 

Pembahasan Soal:

Jika begin mathsize 14px style alpha end style dan begin mathsize 14px style beta end style sudut lancip dari suatu segitiga siku-siku, maka 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell alpha plus beta end cell equals 90 row beta equals cell 90 minus alpha end cell end table end style 

Sehingga, 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell tan space alpha end cell equals cell square root of 2 space sin space beta end cell row cell fraction numerator sin space alpha over denominator cos space alpha end fraction end cell equals cell square root of 2 space sin space open parentheses 90 degree minus alpha close parentheses end cell row cell fraction numerator sin space alpha over denominator cos space alpha end fraction end cell equals cell square root of 2 space cos space alpha end cell row cell sin space alpha end cell equals cell square root of 2 space cos space alpha times cos space alpha end cell row cell sin space alpha end cell equals cell square root of 2 space cos squared space alpha end cell row cell sin space alpha end cell equals cell square root of 2 open parentheses 1 minus sin squared space alpha close parentheses end cell row cell sin space alpha end cell equals cell square root of 2 minus square root of 2 space sin squared space alpha end cell row 0 equals cell square root of 2 space sin squared space alpha plus sin space alpha minus square root of 2 end cell row 0 equals cell open parentheses square root of 2 space sin space alpha minus 1 close parentheses open parentheses sin space alpha plus square root of 2 close parentheses end cell row cell sin space alpha end cell equals cell fraction numerator 1 over denominator square root of 2 end fraction end cell row cell sin space alpha end cell equals cell negative square root of 2 end cell row blank blank blank row blank blank blank end table end style   

Karena begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank sin end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank alpha end table end style tidak mungkin begin mathsize 14px style negative square root of 2 end style, maka begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank sin end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank alpha end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator 1 over denominator square root of 2 end fraction end cell end table end style.

Maka, begin mathsize 14px style sin squared space alpha equals open parentheses fraction numerator 1 over denominator square root of 2 end fraction close parentheses squared equals 1 half end style.

Jadi, jawaban yang tepat adalah D.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

F. Nur

Mahasiswa/Alumni Universitas Muhammadiyah Malang

Terakhir diupdate 29 Maret 2021

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Pertanyaan yang serupa

Nilai  adalah ....

Pembahasan Soal:

Ingat kembali sifat relasi kuadran pada perbandingan trigonometri dan identitas trigonometri berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell sin open parentheses 90 degree minus x close parentheses end cell equals cell cos space x end cell row cell sin squared x plus cos squared x end cell equals 1 end table

Sehingga, didapat perhitungan berikut ini.

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space x end cell equals cell sin open parentheses 90 degree minus x close parentheses end cell row cell cos space 1 degree end cell equals cell sin space 89 degree end cell row cell cos space 2 degree end cell equals cell sin space 88 degree end cell row cell cos space 89 degree end cell equals cell sin space 1 degree end cell end table

Maka,

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell cos squared space open parentheses 1 degree close parentheses plus cos squared space open parentheses 2 degree close parentheses plus cos squared space open parentheses 3 degree close parentheses plus... plus cos squared space open parentheses 90 degree close parentheses end cell row blank equals cell open parentheses cos squared space 1 degree plus cos squared space 89 degree close parentheses plus open parentheses cos squared space 2 degree plus cos squared space 88 degree close parentheses plus... plus open parentheses cos squared space 44 degree plus cos squared space 46 degree close parentheses plus cos squared space 45 degree plus cos squared space 90 degree end cell row blank equals cell open parentheses cos squared space 1 degree plus sin squared space 1 degree close parentheses plus cos squared space 45 degree plus cos squared space 90 degree end cell row blank equals cell 1 plus 1 plus... plus 1 plus open parentheses 1 half square root of 2 close parentheses squared plus 0 end cell row blank equals cell 1 cross times 44 plus 2 over 4 end cell row blank equals cell 44 comma 5 end cell end table

Dengan demikian, nilai cos squared space open parentheses 1 degree close parentheses plus cos squared space open parentheses 2 degree close parentheses plus cos squared space open parentheses 3 degree close parentheses plus... plus cos squared space open parentheses 90 degree close parentheses adalah 44,5.

Oleh karena itu, jawaban yang benar adalah B.

0

Roboguru

Jika sin x = 0,8, maka nilai dari  adalah ....

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell sin open parentheses straight pi over 2 minus x close parentheses end cell equals cell sin straight pi over 2 times cos space x minus cos straight pi over 2 times sin space x end cell row blank equals cell cos space x minus 0 end cell row blank equals cell cos space x end cell end table 

table attributes columnalign right center left columnspacing 0px end attributes row cell cos open parentheses straight pi plus straight x close parentheses end cell equals cell cos space straight pi times cos space straight x minus sin space straight pi times space sin space straight x end cell row blank equals cell negative 1 times cos space straight x minus 0 end cell row blank equals cell negative cos space straight x end cell end table 

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 space sin space open parentheses straight pi over 2 minus x close parentheses plus cos space open parentheses straight pi plus straight x close parentheses end cell equals cell 2 space cos space x minus cos space x end cell row blank equals cell cos space x end cell end table 

table attributes columnalign right center left columnspacing 0px end attributes row cell sin squared x plus cos squared x end cell equals 1 row cell cos squared x end cell equals cell 1 minus sin squared x end cell row cell cos space x end cell equals cell square root of 1 minus sin squared x end root end cell row blank equals cell square root of 1 minus open parentheses 0 comma 8 close parentheses squared end root end cell row blank equals cell square root of 1 minus 0 comma 64 end root end cell row blank equals cell square root of 0 comma 36 end root end cell row blank equals cell 0 comma 6 end cell end table 

Oleh karena itu, jawaban yang benar adalah B.

0

Roboguru

Bentuk cos75∘cosec75∘sin15∘sec15∘​ setara dengan ...

Pembahasan Soal:

Ingat kembali identitas trigonometri:

secθcosecθ==cosθ1sinθ1 

dan ingat kembali pada kuadran I berlaku:

sin(90α)cos(90α)==cosαsinα 

sehingga

cos75cosec75sin15sec15=======cos75sin751sin15cos151sin75cos75cos15sin15cos15sin15×cos75sin75cos15cos75sin15sin75cos15cos75sin(9075)sin(9015)cos15cos75cos75cos151  

Oleh karena itu, jawaban yang benar adalah E. 

0

Roboguru

Nilai  adalah ...

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell cos squared space open parentheses 15 to the power of o close parentheses plus cos squared open parentheses 35 to the power of o close parentheses plus cos squared open parentheses 55 to the power of o close parentheses plus cos squared open parentheses 75 to the power of o close parentheses end cell row blank equals cell cos squared open parentheses 15 to the power of o close parentheses plus cos squared open parentheses 35 to the power of o close parentheses plus cos squared open parentheses 90 to the power of o minus 35 to the power of o close parentheses plus cos squared open parentheses 90 to the power of o minus 15 to the power of o close parentheses end cell row blank equals cell cos squared open parentheses 15 to the power of o close parentheses plus cos squared open parentheses 35 to the power of o close parentheses plus sin squared open parentheses 35 to the power of o close parentheses plus sin squared open parentheses 15 to the power of o close parentheses end cell row blank equals cell open curly brackets cos squared open parentheses 15 to the power of o close parentheses plus sin squared open parentheses 15 to the power of o close parentheses close curly brackets plus open curly brackets cos squared open parentheses 35 to the power of o close parentheses plus sin squared open parentheses 35 to the power of o close parentheses close curly brackets end cell row blank equals cell 1 plus 1 end cell row blank equals 2 end table end style 


Dengan demikian, jawaban yang tepat adalah E.

0

Roboguru

Pembahasan Soal:

Ingat identitas trigonometri yaitu

sin squared x plus cos squared x equals 1

Maka diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell sin squared x plus cos squared x end cell equals 1 row cell sin squared x plus cos squared x minus 1 end cell equals 0 row cell sin squared x minus 1 end cell equals cell negative cos squared x end cell end table

Jadi berdasarkan hitungan di atas diperoleh bahwa table attributes columnalign right center left columnspacing 0px end attributes row cell sin squared x minus 1 end cell equals cell negative cos squared x end cell end table.

Oleh karena itu, jawaban yang benar adalah D.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

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