f.2 maths

1) 4sinA/3sinA-cosA

=4sinA/3sinA-4sinA/cosA

=4/3-4tanA

=4/3-4(5/3)

=4/3-20/3

=4-20/3

= -16/3

2)cos(40-3a)=sin(7a+2)

sin(90-(40-3a))=sin(7a+2)

50+3a=7a+2

48=4a

a=12

3)sin19tan71-cos19

=sin19(sin71/cos71)-cos19

=sin19(cos19/sin19)-cos19

=cos19-cos19

=0

1).if tanA = 5/3, find the value of 4sinA/3sinA-cosA by using trigonometric identities.

tanA = 5/3
sinA/cosA=5/3
3sinA/5=cosA

=>4sinA/(3sinA-cosA )
=4sinA/(3sinA-3sinA/5)
=4/(12/5)
=5/3

2). cos(40度-3a) = sin(7a+2度)

(a=?)
cos(40°-3a) = sin(7a+2°)
cos(40°-3a)=cos(90°-7a-2°)
40°-3a=90°-7a-2°
4a=48°
a=12°


3). (sin19度tan71度-cos19度)
(sin19°tan71°-cos19°)
=(sin19°sin71°/cos71°)-cos19°
=(sin19°sin71°-cos19°cos71°)/(cos71°cos19°)
=(cos19°cos71°-cos19°cos71°)/(cos71°cos19°)
=0


2007-05-02 19:36:59 補充：
樓上第一題計錯了。

tanA = 5/3
sinA/cosA=5/3
so sinA=5,cosA=3

4sinA/3sinA-cosA
=4(5)/3(3)-3
= -7/9//

cos(40-3a)=sin(7a+2)
cos(40-3a)=sin(90-(7a+2)
40-3a=90-7a-2
4a=48
a=12//

sin19 tan71-cos19
=sin19 tan(90-19)-cos19
=sin19/tan19-cos19
=sin19/(sin19/cos19)-cos19
=cos19-cos19
=0//