maths1``問!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!help.......

2008-05-14 3:35 am

1. cos(40°－3﹫)=sin(7﹫+2°)
2. cos37°sin53°+cos53°sin37°
3. (sin19°tan71°－cos19°)2次
4. (cos﹫+1)(cos﹫－1)≡﹣sin﹫cos﹫tan﹫
5. cos(90°－A)分之cos2次A≡sin(90°－A)tan(90°－A)
6.Given that sin﹫=25分之7,find the value of 3tan﹫－cos﹫分之1by using Pythagoras' theorem.

﹫＝theta

回答 (1)

用戶138635

2008-05-14 3:40 am

✔ 最佳答案

1. cos(40°－3﹫)=sin(7﹫+2°)
sin(90°-40°+3@)=sin(7@+2°)
50+3@=7@+2
4@=48
@=12

2008-05-13 19:41:40 補充：
2. cos37°sin53°+cos53°sin37°
=sin(90-37)sin53+cos53cos(90-37)
=sin53sin53+cos53cos53
=(sin53)^2+(cos53)^2
=1

2008-05-13 19:42:00 補充：
^2=2次

2008-05-13 19:54:47 補充：
(sin19°tan71°－cos19°)^2
=(sin19X(cos19÷sin19)-cos19)^2
=0^2
=0

2008-05-13 20:00:15 補充：
(cos﹫+1)(cos﹫－1)≡﹣sin﹫cos﹫tan﹫

R.H.S.
=﹣sin﹫cos﹫tan﹫
=-sin@cos@(sin@÷cos@)
=-(sin@)^2
=-(1-(cos@)^2))
=(cos@)^2)-1
=(cos﹫+1)(cos﹫－1) <---------------x^2-y^2=(x+y)(x-y)
=L.H.S.

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