Pertanyaan

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

$fraction numerator negative a over denominator square root of open parentheses a squared space plus space 1 close parentheses end root end fraction$
$fraction numerator 1 over denominator square root of open parentheses a squared space plus space 1 close parentheses end root end fraction$
$fraction numerator 1 over denominator a square root of open parentheses a squared space plus space 1 close parentheses end root end fraction$
$fraction numerator negative 1 over denominator a square root of open parentheses a squared space plus space 1 close parentheses end root end fraction$

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Jawaban terverifikasi

Pembahasan

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

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

$tan space x equals negative a equals negative a over 1 equals fraction numerator d e p a n space y over denominator s a m p i n g space x space end fraction$ (lihat gambar di samping)

Untuk mencari sisi miring sudut x = r, kita gunakana teorema Pythagoras, seperti berikut :
$r squared space equals space y squared space plus space x squared r equals square root of a squared plus open parentheses negative 1 close parentheses squared end root equals square root of a squared plus 1 space end root J a d i comma space sin space x equals fraction numerator d e p a n space y over denominator m i r i n g space r end fraction equals fraction numerator a space over denominator square root of a squared plus 1 end root end fraction$