Jika ente lagi mencari jawaban atas soal Tentukan nilai dari
[tex]\displaystyle \frac\sin^4 2022^\circ+\cos^4 2022^\circ+3\sin^2 2022^\circ \cos^2 2022^\circ\sin^6 2022^\circ+\cos^6 2022^\circ+4\sin^2 2022^\circ \cos^2 2022^\circ[/tex] ente berada di website yang tepat.
Kami mempunyai 1 jawaban mengenai Tentukan nilai dari
[tex]\displaystyle \frac\sin^4 2022^\circ+\cos^4 2022^\circ+3\sin^2 2022^\circ \cos^2 2022^\circ\sin^6 2022^\circ+\cos^6 2022^\circ+4\sin^2 2022^\circ \cos^2 2022^\circ[/tex]. Silakan baca jawaban lebih lanjut disini:

Tentukan Nilai Dari
[tex]\displaystyle \frac\sin^4 2022^\circ+\cos^4 2022^\circ+3\sin^2 2022^\circ \cos^2 2022^\circ\sin^6 2022^\circ+\cos^6 2022^\circ+4\sin^2 2022^\circ \cos^2 2022^\circ[/tex]
Jawaban: #1:
Hasil dari [tex]\displaystyle\fracsin^42022^0+cos^42022^0+3sin^22022^0cos^22022^0sin^62022^0+cos^62022^0+4sin^22022^0cos^22022^0[/tex] adalah 1.
PEMBAHASAN
Trigonometri merupakan ilmu matematika yang mempelajari hubungan antara panjang dan sudut pada segitiga. Dalam trigonometri terdapat banyak rumus identitas, antara lain :
[tex]sin^2\theta+cos^2\theta=1[/tex]
[tex]sin2\theta=2sin\theta cos\theta[/tex]
[tex]cos2\theta=cos^2\theta-sin^2\theta[/tex]
.
DIKETAHUI
[tex]\displaystyle\fracsin^42022^0+cos^42022^0+3sin^22022^0cos^22022^0sin^62022^0+cos^62022^0+4sin^22022^0cos^22022^0=[/tex]
.
DITANYA
Tentukan hasilnya.
.
PENYELESAIAN
Misal a = 2022⁰
[tex]\displaystyle\fracsin^42022^0+cos^42022^0+3sin^22022^0cos^22022^0sin^62022^0+cos^62022^0+4sin^22022^0cos^22022^0[/tex]
[tex]\displaystyle=\fracsin^4a+cos^4a+3sin^2acos^2asin^6a+cos^6a+4sin^2acos^2a[/tex]
[tex]\displaystyle=\frac(sin^2a)^2+(cos^2a)^2+3sin^2acos^2a(sin^2a)^3+(cos^2a)^3+4sin^2acos^2a[/tex]
[tex]---------------[/tex]
Gunakan rumus :
[tex](x+y)^2=x^2+2xy+y^2[/tex]
[tex]x^2+y^2=(x+y)^2-2xy[/tex]
.
[tex](x+y)^3=x^3+3x^2y+3xy^2+y^3[/tex]
[tex](x+y)^3=x^3+y^3+3xy(x+y)[/tex]
[tex]x^3+y^3=(x+y)^3-3xy(x+y)[/tex]
[tex]---------------[/tex]
[tex]\displaystyle=\frac[(sin^2a+cos^2a)^2-2sin^2acos^2a]+3sin^2acos^2a[(sin^2a+cos^2a)^3-3sin^2acos^2a(sin^2a+cos^2a)]+4sin^2acos^2a[/tex]
[tex]\displaystyle=\frac[(1)^2-2sin^2acos^2a]+3sin^2acos^2a[(1)^3-3sin^2acos^2a(1)]+4sin^2acos^2a[/tex]
[tex]\displaystyle=\frac1+sin^2acos^2a1+sin^2acos^2a[/tex]
[tex]\displaystyle=1[/tex]
.
KESIMPULAN
Hasil dari [tex]\displaystyle\fracsin^42022^0+cos^42022^0+3sin^22022^0cos^22022^0sin^62022^0+cos^62022^0+4sin^22022^0cos^22022^0[/tex] adalah 1.
.
PELAJARI LEBIH LANJUT
- Membuktikan identitas trigonometri : https://brainly.co.id/tugas/29135063
- Perbandingan trigonometri : https://brainly.co.id/tugas/29090996
- Persamaan trigonometri : https://brainly.co.id/tugas/34382463
.
DETAIL JAWABAN
Kelas : 10
Mapel: Matematika
Bab : Trigonometri
Kode Kategorisasi: 10.2.7

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