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Bentuk lain dari fraction numerator open parentheses sin space straight alpha minus sin space straight beta close parentheses open parentheses sin space straight alpha space plus sin space straight beta close parentheses over denominator left parenthesis cos space straight alpha space cos space straight beta right parenthesis squared end fraction

 

Pembahasan Soal:

Ingatlah aturan berikut.

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 end cell equals cell 1 minus cos squared x end cell row cell open parentheses a plus b close parentheses open parentheses a minus b close parentheses end cell equals cell a squared minus b squared end cell end table

Maka,

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator open parentheses sin space straight alpha minus sin space straight beta close parentheses open parentheses sin space straight alpha space plus sin space straight beta close parentheses over denominator left parenthesis cos space straight alpha space cos space straight beta right parenthesis squared end fraction end cell equals cell fraction numerator open parentheses 1 minus cos squared alpha close parentheses minus open parentheses 1 minus cos squared beta close parentheses over denominator cos squared alpha space cos squared beta end fraction end cell row blank equals cell fraction numerator open parentheses 1 minus cos space alpha squared close parentheses minus 1 plus cos squared beta over denominator cos squared alpha space cos squared beta end fraction end cell row blank equals cell fraction numerator cos squared beta minus cos squared alpha over denominator cos squared alpha space cos squared beta end fraction end cell row blank equals cell fraction numerator open parentheses cos space beta minus cos space alpha close parentheses open parentheses cos space beta plus cos space alpha close parentheses over denominator cos squared alpha space cos squared beta end fraction end cell end table


Jadi, bentuk lain dari dari fraction numerator open parentheses sin space straight alpha minus sin space straight beta close parentheses open parentheses sin space straight alpha space plus sin space straight beta close parentheses over denominator left parenthesis cos space straight alpha space cos space straight beta right parenthesis squared end fraction adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator open parentheses cos space beta minus cos space alpha close parentheses open parentheses cos space beta plus cos space alpha close parentheses over denominator cos squared alpha space cos squared beta end fraction end cell end table.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

G. Albiah

Mahasiswa/Alumni Universitas Galuh Ciamis

Terakhir diupdate 11 Juli 2021

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Pertanyaan yang serupa

Pada segitiga ABC diketahui 3 sin A + 4 cos B = 6 dan 3 cos A + 4 sin B = 1. Nilai sin C adalah...

Pembahasan Soal:

Misal:

3 sin A + 4 cos B = 6 ... (1)

3 cos A + 4 sin B = 1 ... (2)

Persamaan (1) dan (2) di kuadratkan lalu di jumlahkan, maka:

begin mathsize 12px style open parentheses 3 space sin space A plus 4 space cos space B equals 6 close parentheses squared left right arrow 9 space sin squared space A plus 16 space cos squared space B plus 24 space sin space A space cos space B equals 36  open parentheses 3 space cos space A plus 4 space sin space B equals 1 close parentheses squared left right arrow 9 space cos squared space A plus 16 space sin squared space B plus 24 space cos space A space sin space B equals 1    9 space sin squared space A plus 16 space cos squared space B plus 24 space sin space A space cos space B equals 36  bottom enclose 9 space cos squared space A plus 16 space sin squared space B plus 24 space cos space A space sin space B equals space space 1 end enclose space space plus  9 open parentheses sin to the power of 2 space end exponent A plus cos squared space A close parentheses plus 16 open parentheses sin squared space B plus cos squared space B close parentheses plus 24 open parentheses sin space A space cos space B plus cos space A space sin space B close parentheses equals 37 end style

 

Ingat!

begin mathsize 12px style sin squared space A plus cos squared space A equals 1  sin open parentheses A plus B close parentheses equals sin space A space cos space B plus cos space A space sin space B  sin open parentheses A minus B close parentheses equals sin space A space cos space B minus cos space A space sin space B end style

maka

size 12px 9 open parentheses size 12px sin to the power of size 12px 2 size 12px space end exponent size 12px A size 12px plus size 12px cos to the power of size 12px 2 size 12px space size 12px A close parentheses size 12px plus size 12px 16 open parentheses size 12px sin to the power of size 12px 2 size 12px space size 12px B size 12px plus size 12px cos to the power of size 12px 2 size 12px space size 12px B close parentheses size 12px plus size 12px 24 open parentheses size 12px sin size 12px space size 12px A size 12px space size 12px cos size 12px space size 12px B size 12px plus size 12px cos size 12px space size 12px A size 12px space size 12px sin size 12px space size 12px B close parentheses size 12px equals size 12px 37  size 12px 9 size 12px. size 12px 1 size 12px plus size 12px 16 size 12px. size 12px 1 size 12px plus size 12px 24 open parentheses size 12px sin open parentheses size 12px A size 12px plus size 12px B close parentheses close parentheses size 12px equals size 12px 37  size 12px 24 open parentheses size 12px sin open parentheses size 12px A size 12px plus size 12px B close parentheses close parentheses size 12px equals size 12px 12  size 12px sin open parentheses size 12px A size 12px plus size 12px B close parentheses size 12px equals size 12px 1 over size 12px 2

 

Sudut dalam segitiga ABC, maka C = begin mathsize 12px style 180 degree minus open parentheses A plus B close parentheses end style

Sehingga, nilai dari sin C yaitu:

begin mathsize 12px style sin space C equals sin open parentheses 180 degree minus open parentheses A plus B close parentheses close parentheses  sin space C equals sin space 180 degree space cos open parentheses A plus B close parentheses minus cos space 180 degree space sin open parentheses A plus B close parentheses  sin space C equals sin open parentheses A plus B close parentheses  sin space C equals 1 half end style

0

Roboguru

Jika  maka

Pembahasan Soal:

Sifat penjumlahan dan pengurangan trigonometri :

sin space straight A plus sin space straight B equals 2 sin 1 half left parenthesis straight A plus straight B right parenthesis cos 1 half left parenthesis straight A minus straight B right parenthesis sin space straight A minus sin space straight B equals 2 cos 1 half left parenthesis straight A plus straight B right parenthesis sin 1 half left parenthesis straight A minus straight B right parenthesis cos space straight A plus cos space straight B equals 2 cos 1 half left parenthesis straight A plus straight B right parenthesis cos 1 half left parenthesis straight A minus straight B right parenthesis cos space straight A minus cos space straight B equals 2 sin 1 half left parenthesis straight A plus straight B right parenthesis sin 1 half left parenthesis straight A minus straight B right parenthesis 

Sifat perkalian trigonometri :

2 sin space straight A space cos space straight B equals sin left parenthesis straight A plus straight B right parenthesis plus sin left parenthesis straight A minus straight B right parenthesis 2 cos space straight A space sin space straight B equals sin left parenthesis straight A plus straight B right parenthesis minus sin left parenthesis straight A minus straight B right parenthesis 2 cos space straight A space cos space straight B equals cos left parenthesis straight A plus straight B right parenthesis plus cos left parenthesis straight A minus straight B right parenthesis minus 2 sin space straight A space sin space straight B equals cos left parenthesis straight A plus straight B right parenthesis plus cos left parenthesis straight A minus straight B right parenthesis 

Dengan menggunakan sifat tersebut, maka :

fraction numerator tan space 1 half open parentheses straight A plus straight B close parentheses over denominator tan space 1 half open parentheses straight A minus straight B close parentheses end fraction equals fraction numerator sin space 1 half open parentheses straight A plus straight B close parentheses over denominator cos space 1 half open parentheses straight A plus straight B close parentheses end fraction cross times fraction numerator cos space 1 half open parentheses straight A minus straight B close parentheses over denominator sin space 1 half open parentheses straight A minus straight B close parentheses end fraction equals fraction numerator 1 half open parentheses 2 sin space 1 half open parentheses straight A plus straight B close parentheses cos space 1 half open parentheses straight A minus straight B close parentheses close parentheses over denominator 1 half open parentheses 2 cos space 1 half open parentheses straight A plus straight B close parentheses sin space 1 half open parentheses straight A minus straight B close parentheses close parentheses end fraction equals fraction numerator 1 half open parentheses sin space straight A plus sin space straight B close parentheses over denominator 1 half open parentheses sin space straight A minus sin space straight B close parentheses end fraction equals fraction numerator 1 half open parentheses 3 plus 5 close parentheses over denominator 1 half open parentheses 3 minus 5 close parentheses end fraction equals fraction numerator 4 over denominator negative 1 end fraction equals negative 4     

Maka, fraction numerator tan space 1 half open parentheses straight A plus straight B close parentheses over denominator tan space 1 half open parentheses straight A minus straight B close parentheses end fraction equals negative 4

Oleh karena itu, jawaban yang benar adalah A.

1

Roboguru

sin2(8π​+2A​)−sin2(8π​−2A​)=...

Pembahasan Soal:

Ingat rumus jumlah dan selisih trigonometri berikut ini:

sinA+sinB=2sin21(A+B)cos21(AB)

sinAsinB=2cos21(A+B)sin21(AB)

Dengan menggunakan konsep di atas, diperoleh hasil:

======sin(8π+2A)+sin(8π2A)2sin21[(8π+2A)+(8π2A)]cos21[(8π+2A)(8π2A)]2sin21(82π)cos21(22A)2sin(8π)cos(2A)sin(8π+2A)sin(8π2A)2cos21[(8π+2A)+(8π2A)]sin21[(8π+2A)(8π2A)]2cos21(82π)sin21(22A)2cos(8π)sin(2A)

Maka

=====sin2(8π+2A)sin2(8π2A)[sin(8π+2A)+sin(8π2A)]×[sin(8π+2A)sin(8π2A)]2sin(8π)cos(2A)×2cos(8π)sin(2A)2sin(8π)cos(8π)×2sin(2A)cos(2A)sin(82π)sinA21sinA

Jadi, sin2(8π+2A)sin2(8π2A)=21sinA.

Oleh karena itu, jawaban yang tepat adalah D.

0

Roboguru

Buktikan  sinA−sinBsinA+sinB​=tan(2A−B​)tan(2A+B​)​

Pembahasan Soal:

Ingat bahwa :

Rumus jumlah dan selisih Trigonometri yaitu

sinA+sinB=2sin21(A+B)cos21(AB)sinAsinB=2cos21(A+B)sin21(AB)

Sehingga 

sinAsinBsinA+sinB====tan(2AB)tan(2A+B)2cos21(A+B)sin21(AB)2sin21(A+B)cos21(AB)tan(2A+B)tan(2AB)1tan(2AB)tan(2A+B)(terbukti)

Dengan demikian terbukti bahwa sinAsinBsinA+sinB=tan(2AB)tan(2A+B)

 

0

Roboguru

Pada segitiga ABC lancip diketahui  dan , maka

Pembahasan Soal:

Diketahui:

 table attributes columnalign right center left columnspacing 0px end attributes row cell sin space B end cell equals cell 12 over 13 end cell row cell cos space A end cell equals cell 4 over 5 end cell end table

Karena segitiga ABC lancip maka besar ketiga sudutnya angle 90 degree (kuadran I) sehingga nilai fungsi trigonometrinya selalu positif. Selanjutnya dapat ditentukan cos space B space dan space sin space italic A dengan identitas trigonometri sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row 1 equals cell sin squared space B plus cos squared space B end cell row cell cos italic space B end cell equals cell square root of 1 minus sign sin squared space B end root end cell row blank equals cell square root of 1 minus sign begin italic style left parenthesis 12 over 13 right parenthesis end style squared end root end cell row blank equals cell square root of 1 minus sign 144 over 169 end root end cell row blank equals cell square root of fraction numerator 169 minus sign 144 over denominator 169 end fraction end root end cell row blank equals cell square root of 25 over 169 end root end cell row blank equals cell 5 over 13 end cell row blank blank blank row 1 equals cell sin squared space italic A plus cos squared space italic A end cell row cell sin italic space italic A end cell equals cell square root of 1 minus sign cos squared space italic A end root end cell row blank equals cell square root of 1 minus sign begin italic style left parenthesis 4 over 5 right parenthesis end style squared end root end cell row blank equals cell square root of 1 minus sign 16 over 25 end root end cell row blank equals cell square root of fraction numerator 25 minus sign 16 over denominator 25 end fraction end root end cell row blank equals cell square root of 9 over 25 end root end cell row blank equals cell 3 over 5 end cell end table  

Karena jumlah besar sudut dalam suatu segitiga adalah 180 degree maka C equals 180 degree minus sign open parentheses italic A and B close parentheses. Sehingga dapat ditentukan nilai dari sin space C sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space C end cell equals cell sin space open parentheses 180 degree minus sign open parentheses italic A and B close parentheses close parentheses end cell row blank equals cell sin space 180 degree middle dot cos space left parenthesis italic A and B right parenthesis minus sign sin space left parenthesis italic A and B right parenthesis middle dot cos space 180 degree end cell row blank equals cell 0 middle dot cos space left parenthesis italic A and B right parenthesis minus sign sin space left parenthesis italic A and B right parenthesis middle dot open parentheses negative sign 1 close parentheses end cell row blank equals cell sin space left parenthesis italic A and B right parenthesis end cell row blank equals cell sin space italic A middle dot cos space B and sin space B middle dot cos space italic A end cell row blank equals cell 3 over 5 middle dot 5 over 13 plus 12 over 13 middle dot 4 over 5 end cell row blank equals cell fraction numerator 15 plus 48 over denominator 65 end fraction end cell row blank equals cell 63 over 65 end cell end table  

Dengan demikian nilai dari sin space C equals 63 over 65.

2

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