✔ 最佳答案
1. cos23"- cos67" / tan23" ("<--呢2點係度)
cos23-cos67/tan23
=cos23-sin(90-67)/(sin23/cos23)
=cos23-sin23/(sin23/cos23)
=cos23-cos23
=0
2. -sin² 48"- sin² 42"
=-sin²48-cos²(90-42)
=-sin²48-cos²48
=-1
3. tan3θ= 1 / tan45"
tan3θ= 1 / 1
3θ=180n+45
θ=60n+15 (15, 75, 135, 195, 255, 315, etc)
4. 90">θ > 0", which is/are correct?
a. 1 >/= cos θ>/= 0 (what is >/=? "greater than or equal to"? OR "not greater than or equal to"? anyway this is not correct)
b. 1 > sin²θ> 0 (correct)
c. tanθ> 0 (correct)
2009-06-16 22:32:53 補充:
任何角度A, (sinA)^2+(cosA)^2 一定等於一,這是畢氏定理
相對直角三角型
sinA=對邊/斜邊
cosA=鄰邊/斜邊
另 對邊^2+鄰邊^2=斜邊^2 (畢氏定理)
因此 (對邊/斜邊)^2+(鄰邊/斜邊)^2=1
(sinA)^2+(cosA)^2=1
所以-sin²48-cos²48 = -1 *(sin²48+cos²48)=-1
2009-06-16 22:34:53 補充:
如果是初中問題,相信角度會在90度內.
tan3θ= 1 / 1
3θ = 45
θ = 15
2009-06-16 22:38:57 補充:
cos0=1
cos90=0
因90">θ > 0"不包含0及90這兩個角度,所以
1 > cos θ> 0
並不包含"="部分