Question

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三角函數 39、40、41例題

39.設180°<f<270°,試化簡√sin²0+(1+sin⑥) = 【_1 1
(in ²10 +63²0) = 1
105²410-1-51²³0
7
40.設sin0+cosA= 且0為第四象限角,則
--
13
(1) sin 0= [
-
005²0-1=-51²³000s off = Sihve
5
1
13
(Sing + cos19) == "169
= sin ²0 + cos²³0+ 2 sin-cos= 12. sin. cos=
41. sina+sin B = 1 cos a+cos3=0
(2) tan(180°+0) = { 1 cos= Esin
5
12.
a tand=b
atang-b
7
69 169
169 - 169 = Sin + √125in ²
cos² a + cos² B = [
3
1
2
19-21
sin
45