Trigonometry 的問題 ......

由 0 至 90o
所有函數都是正
由 90o 至 180o
sin 函數都是正
由 180o 至 270o
tan 函數都是正
由 270o 至 360o
cos 函數都是正

1,Given that "tan a" = 5/6 and 0o < a < 360o , find "sin a" and "cos a"" ;
tan a = 5/6
以一個直角三角形考慮，對邊是 5，鄰邊是 6
斜邊是√(52+62) = √61
所以
sin a = 5 / √61 = (5√61)/61 或 -(5√61)/61
cos a = 6 / √61 = (6√61)/61 或 -(6√61)/61

2.Given that "sin a" = k and 180o < a < 270o , express "cos a" and "tan a" in terms of k .
因 180o < a < 270o ，sin 函數是負值，所以 k < 0.
以一個直角三角形考慮，對邊是 k，斜邊是 1
鄰邊是√(1 + k2)
cos a = -√(1 + k2) / 1
tan a = -k / √(1 + k2) = -k√(1 + k2) / √(1 + k2)

3.If "cos a" = 2/3 where 0度 < a < 360度 , find the value of (sin a *tan a)
以一個直角三角形考慮，鄰邊是 2，斜邊是 3
對邊是 = √(32 - 22) =√5
sin a * tan a = (√5 / 3)(√5 / 2)
= 5/6 或 -5/6

1. If a lies on the first quadrant, both sin a and cos a are positive.
therefore sin a = 5/(sqrt[5^2+6^2]) = 5/sqrt[61]
cos a = 6/sqrt[61].
If a lies on the third quadrant, both sin a and cos a are negative.
so sin a = -5/(sqrt[5^2+6^2]) = -5/sqrt[61]
cos a = -6/sqrt[61].

2. similarly, sin a = k, and k is negative.
cos a is also negative. therefore cos a = sqrt[1-k^2]
however, tan a is postive in this case. tan a should equal to -k/sqrt[1-k^2].

3. sin a * tan a = (sin a)^2/cos a = (1- (cos a)^2)/cos a = 5/6

2007-03-08 20:21:54 補充：
CAST diagram can help remember. S | A---------- T | C C means cosine is positiveA means All is positiveS means sine is positiveT means tangent is positiveWe can remember it as &quot;All Students Teach Chinese.&quot;