Pertanyaan

Pada gambar di atas diketahui: luas luas△PQR = luas△PKR + luas△QKR . Gunakan rumus segitiga: L = 2 1 pq sin R untuk membuktikan bahwa sin ( a + b ) = sin a cos b + cos a sin b .

Pada gambar di atas diketahui:

luas $luas△PQR = luas △PKR + luas△QKR$ . Gunakan rumus segitiga: $L = \frac{1}{2} pq sin R$ untuk membuktikan bahwa $sin (a + b) = sin a cos b + cos a sin b$ .

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Klaim

R. Febrianti

Master Teacher

Mahasiswa/Alumni Universitas Negeri Malang

Jawaban terverifikasi

Jawaban

berdasarkan hitungan di atas terbukti bahwa .

berdasarkan hitungan di atas terbukti bahwa 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 plus b close parentheses equals sin space a space cos space b plus cos space a space sin space b

Pembahasan

Soal tersebut tidak dapat dikerjakan karena letak tidak diketahui. Soal tersebut bisa dikerjakan jika diasumsikan . Berdasarkan aturan sinus tentang mencari luas segitiga maka luas yaitu Berdasarkan aturan sinus tentang mencari luas segitiga maka luas yaitu Berdasarkan aturan sinus tentang mencari luas segitiga maka luas yaitu Berdasarkan gambar diperoleh bahwa Berdasarkan perbandingan sisi trigonometri maka pada diperoleh Berdasarkan perbandingan sisi trigonometri maka pada diperoleh Sehingga Jadi berdasarkan hitungan di atas terbukti bahwa .

Soal tersebut tidak dapat dikerjakan karena letak a space text dan end text space b tidak diketahui. Soal tersebut bisa dikerjakan jika diasumsikan text ∠PRK end text equals a text dan ∠QRK end text equals b .

Berdasarkan aturan sinus tentang mencari luas segitiga maka luas triangle text PRK end text yaitu

text L end text subscript 1 equals 1 half cross times q cross times text RK end text cross times sin space a

Berdasarkan aturan sinus tentang mencari luas segitiga maka luas triangle text QRK end text yaitu

text L end text subscript 2 equals 1 half cross times p cross times text RK end text cross times sin space b

Berdasarkan aturan sinus tentang mencari luas segitiga maka luas triangle text PQR end text yaitu

text L end text equals 1 half cross times p cross times q cross times sin space open parentheses a plus b close parentheses

Berdasarkan gambar diperoleh bahwa

$begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell L triangle P Q R end cell equals cell L triangle P R K plus L triangle Q R K end cell row cell up diagonal strike 1 half end strike cross times p cross times q cross times sin open parentheses a plus b close parentheses end cell equals cell up diagonal strike 1 half end strike cross times q cross times R K cross times sin space a plus up diagonal strike 1 half end strike cross times p cross times R K cross times sin space b end cell row cell p cross times q cross times sin open parentheses a plus b close parentheses end cell equals cell q cross times R K cross times sin space a plus p cross times R K cross times sin space b end cell row cell sin open parentheses a plus b close parentheses end cell equals cell fraction numerator q cross times R K cross times sin space a plus p cross times R K cross times sin space b over denominator p cross times q end fraction end cell row cell sin open parentheses a plus b close parentheses end cell equals cell fraction numerator up diagonal strike q cross times R K cross times sin space a over denominator p cross times up diagonal strike q end fraction plus fraction numerator up diagonal strike p cross times R K cross times sin space cross times b over denominator up diagonal strike p cross times q end fraction end cell row cell sin open parentheses a plus b close parentheses end cell equals cell fraction numerator R K cross times sin space a over denominator p end fraction plus fraction numerator R K cross times sin space b over denominator q end fraction end cell row cell sin open parentheses a plus b close parentheses end cell equals cell fraction numerator R K over denominator p end fraction cross times sin space a plus fraction numerator R K over denominator q end fraction cross times sin b end cell row blank blank blank end table end style$

Berdasarkan perbandingan sisi trigonometri maka pada text △PRK end text diperoleh

$table attributes columnalign right center left columnspacing 0px end attributes row cell cos space a end cell equals cell fraction numerator text samping end text over denominator text miring end text end fraction end cell row cell cos space a end cell equals cell fraction numerator text RK end text over denominator q end fraction end cell end table$

Berdasarkan perbandingan sisi trigonometri maka pada triangle text QRK end text diperoleh

$table attributes columnalign right center left columnspacing 0px end attributes row cell cos space b end cell equals cell fraction numerator text samping end text over denominator text miring end text end fraction end cell row cell cos space b end cell equals cell fraction numerator text RK end text over denominator p end fraction end cell end table$

Sehingga

$table attributes columnalign right center left columnspacing 0px end attributes row cell L triangle P Q R end cell equals cell L triangle P R K plus L triangle Q R K end cell row cell up diagonal strike 1 half end strike cross times p cross times q cross times sin open parentheses a plus b close parentheses end cell equals cell up diagonal strike 1 half end strike cross times q cross times text RK end text cross times sin space a plus up diagonal strike 1 half end strike cross times p cross times text RK end text cross times sin space b end cell row cell p cross times q cross times sin open parentheses a plus b close parentheses end cell equals cell q cross times text RK end text cross times sin space a plus p cross times text RK end text cross times sin space b end cell row cell sin open parentheses a plus b close parentheses end cell equals cell fraction numerator q cross times text RK end text cross times sin space a plus p cross times text RK end text cross times sin space b over denominator p cross times q end fraction end cell row cell sin open parentheses a plus b close parentheses end cell equals cell fraction numerator up diagonal strike q cross times text RK end text cross times sin space a over denominator p cross times up diagonal strike q end fraction plus fraction numerator up diagonal strike p cross times text RK end text cross times sin space cross times b over denominator up diagonal strike p cross times q end fraction end cell row cell sin open parentheses a plus b close parentheses end cell equals cell fraction numerator text RK end text cross times sin space a over denominator p end fraction plus fraction numerator text RK end text cross times sin space b over denominator q end fraction end cell row cell sin open parentheses a plus b close parentheses end cell equals cell fraction numerator text RK end text over denominator p end fraction cross times sin space a plus fraction numerator text RK end text over denominator q end fraction cross times sin b end cell row cell sin open parentheses a plus b close parentheses end cell equals cell cos space b cross times sin space a plus cos space a cross times sin b end cell row cell sin open parentheses a plus b close parentheses end cell equals cell sin space a space cos space b plus cos space a space sin space b end cell row blank blank blank end table$

Jadi berdasarkan hitungan di atas terbukti bahwa sin open parentheses a plus b close parentheses equals sin space a space cos space b plus cos space a space sin space b .