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Hasil dari [tex]\displaystyle \csc^2 \frac{\pi}{7}+\csc^2 \frac{2\pi}{7}+\csc^2 \frac{3\pi}{7}[/tex] yaitu ...

Berikut ini adalah pertanyaan dari syakhayaz pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Hasil dari \displaystyle \csc^2 \frac{\pi}{7}+\csc^2 \frac{2\pi}{7}+\csc^2 \frac{3\pi}{7} yaitu ...

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Jawab:
= 8
(Ya, delapan.)

Penjelasan dengan langkah-langkah:
Soal ini bikin puyeng kepalaku
\displaystyle\tt csc^2(x)=\frac{1}{sin^2(x)}=\frac{2}{2sin^2(x)}\\\\=\frac{2}{sin^2(x)+sin^2(x)}\\\\=\frac{2}{sin^2(x)+cos^2(x)-cos^2(x)+sin^2(x)}\\\\=\frac{2}{sin^2(x)+cos^2(x)-(cos^2(x)-sin^2(x))}\\\\\because sin^2(x)+cos^2(x)=1,~~cos^2(x)-sin^2(x)=cos(2x)\therefore\\\\=\frac{2}{1-cos(2x)}

\displaystyle\tt x=\frac{\pi}{7}~,~\frac{2\pi}{7}~,~\frac{3\pi}{7}\\\\2x=\frac{2\pi}{7}~,~\frac{4\pi}{7}~,~\frac{6\pi}{7}

\displaystyle \tt p=cos\left(\frac{2\pi}{7}\right)\\\\q=cos\left(\frac{4\pi}{7}\right)\\\\r=cos\left(\frac{6\pi}{7}\right)

Oleh karena itu hasil dari
\displaystyle\tt csc^2\left(\frac{\pi}{7}\right)+csc^2\left(\frac{2\pi}{7}\right)+csc^2\left(\frac{3\pi}{7}\right)\\\\=\frac{2}{1-cos(2\left(\frac{\pi}{7}\right))}+\frac{2}{1-cos(2\left(\frac{2\pi}{7}\right))}+\frac{2}{1-cos(2\left(3\frac{\pi}{7}\right))}\\\\=\frac{2}{1-cos(\frac{2\pi}{7})}+\frac{2}{1-cos(\frac{4\pi}{7})}+\frac{2}{1-cos(\frac{6\pi}{7})}\\\\=\frac{2}{1-p}+\frac{2}{1-q}+\frac{2}{1-r}\\\\=2\left(\frac{1}{1-p}+\frac{1}{1-q}+\frac{1}{1-r}\right)
Maka sesuai KPK penyebut, disederhanakan
\displaystyle\tt=2\left(\frac{-2(p+q+r)+(pq+pr+qr)+3}{1-(p+q+r)+(pq+pr+qr)-pqr}\right)

Masih kuat? Masih!
Cari nilai pq + pr + qr
\tt =cos\left(\frac{2\pi}{7}\right)cos\left(\frac{4\pi}{7}\right)+cos\left(\frac{2\pi}{7}\right)cos\left(\frac{6\pi}{7}\right)+cos\left(\frac{4\pi}{7}\right)cos\left(\frac{6\pi}{7}\right)\\\\\because cos(a)cos(b)=\frac{1}{2}(cos(a-b)+(cos(a+b))\therefore\\\\=\frac{1}{2}(cos\left(\frac{2\pi}{7}-\frac{4\pi}{7}\right)+cos\left(\frac{2\pi}{7}+\frac{4\pi}{7}\right)+cos\left(\frac{2\pi}{7}-\frac{6\pi}{7}\right)+cos\left(\frac{2\pi}{7}+\frac{6\pi}{7}\right)
\tt cos\left(\frac{4\pi}{7}-\frac{6\pi}{7}\right)+cos\left(\frac{4\pi}{7}+\frac{6\pi}{7}\right)\\=\frac{1}{2}(cos(\frac{2\pi}{7})+cos(\frac{6\pi}{7})+cos(\frac{4\pi}{7})+cos(\frac{8\pi}{7})+cos(\frac{2\pi}{7})+cos(\frac{10\pi}{7}))\\=\frac{1}{2}(cos(\frac{2\pi}{7})+cos(\frac{6\pi}{7})+cos(\frac{4\pi}{7})+cos(2\pi-\frac{8\pi}{7})+cos(\frac{2\pi}{7})+cos(2\pi-\frac{10\pi}{7}))
\tt =\frac{1}{2}(cos(\frac{2\pi}{7})+cos(\frac{6\pi}{7})+cos(\frac{4\pi}{7})+cos(\frac{14\pi-8\pi}{7})+cos(\frac{2\pi}{7})+cos(\frac{14\pi-10\pi}{7}))\\=\frac{1}{2}(cos(\frac{2\pi}{7})+cos(\frac{6\pi}{7})+cos(\frac{4\pi}{7})+cos(\frac{6\pi}{7})+cos(\frac{2\pi}{7})+cos(\frac{4\pi}{7}))\\=\frac{1}{2}(2(cos(\frac{2\pi}{7})+cos(\frac{4\pi}{7})+cos(\frac{6\pi}{7})))\\=\frac{1}{2}(2(p+q+r))=p+q+r
pq + pr + qr = p + q + r

Maka
\displaystyle\tt2\left(\frac{-2(p+q+r)+(pq+pr+qr)+3}{1-(p+q+r)+(pq+pr+qr)-pqr}\right)\\\\=2\left(\frac{-2(p+q+r)+(p+q+r)+3}{1-(p+q+r)+(p+q+r)-pqr}\right)\\\\=\large\boxed{\bf 2\left(\frac{-(p+q+r)+3}{1-pqr}\right)}~~yg~~dicari~~nilainya

Cari nilai p + q + r

\tt \displaystyle =cos\left(\frac{2\pi}{7}\right)+cos\left(\frac{4\pi}{7}\right)+cos\left(\frac{6\pi}{7}\right)\\\\=\left(cos\left(\frac{2\pi}{7}\right)+cos\left(\frac{4\pi}{7}\right)+cos\left(\frac{6\pi}{7}\right)\right)\left(\frac{2sin\left(\frac{\pi}{7}\right)}{2sin\left(\frac{\pi}{7}\right)}\right)
Karena 2sin(a)cos(b) = sin(a+b)+sin(a-b) maka
\displaystyle\tt=\frac{sin\left(\not \frac{3\pi}{7}\right)+sin\left(-\frac{\pi}{7}\right)+\not sin\left(\frac{5\pi}{7}\right)+\not sin\left(-\frac{3\pi}{7}\right)+sin\left(\frac{\not7\pi}{\not7}\right)+\not sin\left(-\frac{5\pi}{7}\right)}{2sin\left(\frac{\pi}{7}\right)}
sin π = sin 0 = 0
\displaystyle\tt=\frac{-sin\left(\frac{\pi}{7}\right)}{2sin\left(\frac{\pi}{7}\right)}=-\frac{1}{2}
\large\boxed{\tt~p+q+r=-\frac{1}{2}~}
Sedangkan jumlah karakter maksimum adalah 5000
jadi lanjutannya ada di gambar . . .

Jawab:= 8(Ya, delapan.) Penjelasan dengan langkah-langkah:Soal ini bikin puyeng kepalaku[tex]\displaystyle\tt csc^2(x)=\frac{1}{sin^2(x)}=\frac{2}{2sin^2(x)}\\\\=\frac{2}{sin^2(x)+sin^2(x)}\\\\=\frac{2}{sin^2(x)+cos^2(x)-cos^2(x)+sin^2(x)}\\\\=\frac{2}{sin^2(x)+cos^2(x)-(cos^2(x)-sin^2(x))}\\\\\because sin^2(x)+cos^2(x)=1,~~cos^2(x)-sin^2(x)=cos(2x)\therefore\\\\=\frac{2}{1-cos(2x)}[/tex][tex]\displaystyle\tt x=\frac{\pi}{7}~,~\frac{2\pi}{7}~,~\frac{3\pi}{7}\\\\2x=\frac{2\pi}{7}~,~\frac{4\pi}{7}~,~\frac{6\pi}{7}[/tex][tex]\displaystyle \tt p=cos\left(\frac{2\pi}{7}\right)\\\\q=cos\left(\frac{4\pi}{7}\right)\\\\r=cos\left(\frac{6\pi}{7}\right)[/tex]Oleh karena itu hasil dari[tex]\displaystyle\tt csc^2\left(\frac{\pi}{7}\right)+csc^2\left(\frac{2\pi}{7}\right)+csc^2\left(\frac{3\pi}{7}\right)\\\\=\frac{2}{1-cos(2\left(\frac{\pi}{7}\right))}+\frac{2}{1-cos(2\left(\frac{2\pi}{7}\right))}+\frac{2}{1-cos(2\left(3\frac{\pi}{7}\right))}\\\\=\frac{2}{1-cos(\frac{2\pi}{7})}+\frac{2}{1-cos(\frac{4\pi}{7})}+\frac{2}{1-cos(\frac{6\pi}{7})}\\\\=\frac{2}{1-p}+\frac{2}{1-q}+\frac{2}{1-r}\\\\=2\left(\frac{1}{1-p}+\frac{1}{1-q}+\frac{1}{1-r}\right)[/tex]Maka sesuai KPK penyebut, disederhanakan[tex]\displaystyle\tt=2\left(\frac{-2(p+q+r)+(pq+pr+qr)+3}{1-(p+q+r)+(pq+pr+qr)-pqr}\right)[/tex]Masih kuat? Masih!Cari nilai pq + pr + qr[tex]\tt =cos\left(\frac{2\pi}{7}\right)cos\left(\frac{4\pi}{7}\right)+cos\left(\frac{2\pi}{7}\right)cos\left(\frac{6\pi}{7}\right)+cos\left(\frac{4\pi}{7}\right)cos\left(\frac{6\pi}{7}\right)\\\\\because cos(a)cos(b)=\frac{1}{2}(cos(a-b)+(cos(a+b))\therefore\\\\=\frac{1}{2}(cos\left(\frac{2\pi}{7}-\frac{4\pi}{7}\right)+cos\left(\frac{2\pi}{7}+\frac{4\pi}{7}\right)+cos\left(\frac{2\pi}{7}-\frac{6\pi}{7}\right)+cos\left(\frac{2\pi}{7}+\frac{6\pi}{7}\right)[/tex][tex]\tt cos\left(\frac{4\pi}{7}-\frac{6\pi}{7}\right)+cos\left(\frac{4\pi}{7}+\frac{6\pi}{7}\right)\\=\frac{1}{2}(cos(\frac{2\pi}{7})+cos(\frac{6\pi}{7})+cos(\frac{4\pi}{7})+cos(\frac{8\pi}{7})+cos(\frac{2\pi}{7})+cos(\frac{10\pi}{7}))\\=\frac{1}{2}(cos(\frac{2\pi}{7})+cos(\frac{6\pi}{7})+cos(\frac{4\pi}{7})+cos(2\pi-\frac{8\pi}{7})+cos(\frac{2\pi}{7})+cos(2\pi-\frac{10\pi}{7}))[/tex][tex]\tt =\frac{1}{2}(cos(\frac{2\pi}{7})+cos(\frac{6\pi}{7})+cos(\frac{4\pi}{7})+cos(\frac{14\pi-8\pi}{7})+cos(\frac{2\pi}{7})+cos(\frac{14\pi-10\pi}{7}))\\=\frac{1}{2}(cos(\frac{2\pi}{7})+cos(\frac{6\pi}{7})+cos(\frac{4\pi}{7})+cos(\frac{6\pi}{7})+cos(\frac{2\pi}{7})+cos(\frac{4\pi}{7}))\\=\frac{1}{2}(2(cos(\frac{2\pi}{7})+cos(\frac{4\pi}{7})+cos(\frac{6\pi}{7})))\\=\frac{1}{2}(2(p+q+r))=p+q+r[/tex]pq + pr + qr = p + q + rMaka[tex]\displaystyle\tt2\left(\frac{-2(p+q+r)+(pq+pr+qr)+3}{1-(p+q+r)+(pq+pr+qr)-pqr}\right)\\\\=2\left(\frac{-2(p+q+r)+(p+q+r)+3}{1-(p+q+r)+(p+q+r)-pqr}\right)\\\\=\large\boxed{\bf 2\left(\frac{-(p+q+r)+3}{1-pqr}\right)}~~yg~~dicari~~nilainya[/tex]Cari nilai p + q + r[tex]\tt \displaystyle =cos\left(\frac{2\pi}{7}\right)+cos\left(\frac{4\pi}{7}\right)+cos\left(\frac{6\pi}{7}\right)\\\\=\left(cos\left(\frac{2\pi}{7}\right)+cos\left(\frac{4\pi}{7}\right)+cos\left(\frac{6\pi}{7}\right)\right)\left(\frac{2sin\left(\frac{\pi}{7}\right)}{2sin\left(\frac{\pi}{7}\right)}\right)[/tex]Karena 2sin(a)cos(b) = sin(a+b)+sin(a-b) maka[tex]\displaystyle\tt=\frac{sin\left(\not \frac{3\pi}{7}\right)+sin\left(-\frac{\pi}{7}\right)+\not sin\left(\frac{5\pi}{7}\right)+\not sin\left(-\frac{3\pi}{7}\right)+sin\left(\frac{\not7\pi}{\not7}\right)+\not sin\left(-\frac{5\pi}{7}\right)}{2sin\left(\frac{\pi}{7}\right)}[/tex]sin π = sin 0 = 0[tex]\displaystyle\tt=\frac{-sin\left(\frac{\pi}{7}\right)}{2sin\left(\frac{\pi}{7}\right)}=-\frac{1}{2}[/tex][tex]\large\boxed{\tt~p+q+r=-\frac{1}{2}~}[/tex]Sedangkan jumlah karakter maksimum adalah 5000jadi lanjutannya ada di gambar . . .

Semoga dengan pertanyaan yang sudah terjawab oleh xcvi dapat membantu memudahkan mengerjakan soal, tugas dan PR sekolah kalian.

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Last Update: Sat, 18 Mar 23
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