Question

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三角函數 1、3

1. 設 為銳角,且
tan 0+2
5
==
tan 0-1 2
Sin
cast2
S-1
?
5
2
Ecos-sin = [-
sih tz
Siny ==
√10
1
5
5 (sih-1) = 2(Sih+₂)
2. 試求sin30°xcos-45°-tan²30°xcos60°=【六】
=+*+-
3
x =
= = = = = = 7² = + = tv snofcos 0 = 1, caso = 1-sing
b
C
A BI
,
4cos.
jsin-5=25in14
3 sin = 9 sin = 3
=>
100+4=1200-5
1- (33²
3. AABCZA: ZB: 2C=1:2:3 sin Ax cos B+ tan² B = [.
13
Ao FX
was the
3
1
2
cas
1
2
ww
-
15
55
9 = cos
9
2x is

Answer 1

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可知

1、
設 x = tanθ，則原方程式交叉相乘得

2x+4 = 5x–5
x=3，又θ為銳角，畫個直角三角形就可以知道
sinθ = 3/√10, cosθ=1/√10

cosθ – sinθ = –2/√10 = –√10/5

3. 令比率是r, 則三角形內角和
180 = r+2r+3r=6r, r=30
所以ㄥA=30°, ㄥB=60°, ㄥC=90°
sinA = 1/2, cosB =1/2, tanB =√3
所求 = 1/4 + 3 =13/4。