Roboguru

Buktikan bahwa csc6x(1−cos6x)−1=3cot2xcsc2x

Pertanyaan

Buktikan bahwa c s c to the power of 6 x left parenthesis 1 minus cos to the power of 6 x right parenthesis minus 1 equals 3 space c o t squared x space c s c squared x

Pembahasan Soal:

Ingat!

table attributes columnalign right center left columnspacing 0px end attributes row cell c o t space x end cell equals cell fraction numerator cos space x over denominator sin space x end fraction end cell row cell c s c space x end cell equals cell fraction numerator 1 over denominator sin space x end fraction end cell row cell sin squared x plus cos squared x end cell equals 1 end table

Akan dibuktikan: c s c to the power of 6 x left parenthesis 1 minus cos to the power of 6 x right parenthesis minus 1 equals 3 space c o t squared x space c s c squared x

Bukti: 

table attributes columnalign right center left columnspacing 0px end attributes row cell c s c to the power of 6 x left parenthesis 1 minus cos to the power of 6 x right parenthesis minus 1 end cell equals cell c s c to the power of 6 x minus cos to the power of 6 x space c s c to the power of 6 x minus 1 end cell row blank equals cell fraction numerator c s c to the power of 6 x minus cos to the power of 6 x over denominator sin to the power of 6 x minus 1 end fraction end cell row blank equals cell fraction numerator 1 over denominator sin to the power of 6 x end fraction minus fraction numerator cos to the power of 6 x over denominator sin to the power of 6 x end fraction minus fraction numerator sin to the power of 6 x over denominator sin to the power of 6 x end fraction end cell row blank equals cell fraction numerator 1 minus cos to the power of 6 x minus sin to the power of 6 x over denominator sin to the power of 6 x end fraction end cell row blank equals cell fraction numerator 1 minus left parenthesis cos to the power of 6 x plus sin to the power of 6 x right parenthesis over denominator sin to the power of 6 x end fraction end cell row blank equals cell fraction numerator 1 minus left parenthesis left parenthesis cos squared x plus sin squared x right parenthesis cubed minus 3 space cos to the power of 4 end exponent x space sin squared x minus 3 space cos squared x space sin to the power of 4 x right parenthesis over denominator sin to the power of 6 x end fraction end cell row blank equals cell fraction numerator 1 minus left parenthesis 1 cubed minus 3 space cos to the power of 4 x space sin squared x minus 3 space cos squared x space sin to the power of 4 x right parenthesis over denominator sin to the power of 6 x end fraction end cell row blank equals cell fraction numerator 3 space cos to the power of 4 x space sin squared x plus 3 space cos squared x space sin to the power of 4 x over denominator sin to the power of 6 x end fraction end cell row blank equals cell fraction numerator 3 space cos squared x space sin squared x left parenthesis cos squared x plus sin squared x right parenthesis over denominator sin to the power of 6 x end fraction end cell row blank equals cell fraction numerator 3 space cos squared x space sin squared x left parenthesis 1 right parenthesis over denominator sin to the power of 6 x end fraction end cell row blank equals cell fraction numerator 3 space cos squared x over denominator sin to the power of 4 x end fraction end cell row blank equals cell fraction numerator 3 space cos squared x over denominator sin squared x space sin squared x end fraction end cell row blank equals cell 3 fraction numerator cos squared x over denominator sin squared x end fraction times fraction numerator 1 over denominator sin squared x end fraction end cell row blank equals cell 3 space c o t squared x space c s c squared x end cell end table

Dengan demikian, terbukti bahwa c s c to the power of 6 x left parenthesis 1 minus cos to the power of 6 x right parenthesis minus 1 equals 3 space c o t squared x space c s c squared x.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Indah

Mahasiswa/Alumni Universitas Lampung

Terakhir diupdate 06 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

Pertanyaan yang serupa

secx+tanxsecx−tanx​​=....

1

Roboguru

(1+cotx−cscx)(1+tanx+secx)=...

0

Roboguru

Diketahui sin(A+B)=21​3​, tan(A−B)=31​3​, 0∘≤(A+B)≤90∘, dan A>B. Hitunglah sinA, cosA, tanA, secA, cosecA, dan cotanA.

0

Roboguru

(1−cos2A)cotan2A=...

0

Roboguru

Buktikan identitas-identitas trigonometri berikut! a. tanα+cotα=secαcscα

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved