Pertanyaan

Jika x di kuadran II dan tan x = –a, maka sin x = ....

$fraction numerator a over denominator square root of left parenthesis a squared plus 1 right parenthesis end root end fraction$
$fraction numerator negative a over denominator square root of left parenthesis a squared plus 1 right parenthesis end root end fraction$
$fraction numerator 1 over denominator square root of left parenthesis a squared plus 1 right parenthesis end root end fraction$
$fraction numerator 1 over denominator a square root of left parenthesis a squared plus 1 right parenthesis end root end fraction$
$fraction numerator negative 1 over denominator a square root of left parenthesis a squared plus 1 right parenthesis end root end fraction$

I. Roy

Master Teacher

Mahasiswa/Alumni Universitas Negeri Surabaya

Jawaban terverifikasi

Pembahasan

Diketahui: tan x = –a dan berada di kuadran II, maka: Untuk mencari sisi miring sudut x = r, kita gunakana teorema Pythagoras, seperti berikut :

Diketahui: tan x = –a dan berada di kuadran II, maka:

$begin mathsize 12px style tan space x equals negative a equals negative a over 1 equals fraction numerator d e p a n space left parenthesis y right parenthesis over denominator s a m p i n g space left parenthesis x right parenthesis end fraction end style$

Untuk mencari sisi miring sudut x = r, kita gunakana teorema Pythagoras, seperti berikut :

$begin mathsize 12px style r to the power of 2 space end exponent equals space y squared space plus space x squared r equals square root of a squared plus left parenthesis negative 1 right parenthesis squared end root equals square root of a squared plus 1 end root J a d i comma space sin space x equals fraction numerator d e p a n space left parenthesis y right parenthesis over denominator m i r i n g space left parenthesis r right parenthesis end fraction equals fraction numerator a over denominator square root of a squared plus 1 end root end fraction end style$