急急急....sin cos tan

2009-05-04 4:06 am
唔該幫我計: (要步驟)

1. cos60°cos30°+sin60°sin30° /tan30°

2. 2cos^2 45°-tan45° /sin50°

3.sin31°tan59°-cos31°

4. (1+tan^2 55°)sin^2 35°+1

答案是:
1. 3/2
2.0
3.0
4.2

回答 (3)

2009-05-04 4:23 am
✔ 最佳答案
1.
cos60cos30+sin60sin30 /tan30
=[(1/2)(√3/2)+(1/2)(√3/2)]/(1/√3)
=(√3/2)/(1/√3)
=3/2

2.
2cos^2 45-tan45 /sin50
=[2(√2/2)^2-1] /sin50
=[2(1/2)-1] /sin50
=0

3.
sin31tan59-cos31
=sin31(1/tan31)-cos31
=sin31(cos31/sin31)-cos31
=cos31-cos31
=0

4.
(1+tan^2 55)sin^2 35+1
=[1+(sin^2 55/cos^2 55)]sin^2 35+1
=[(sin^2 55+cos^2 55)/cos^2 55]sin^2 35+1
=(1/cos^2 55)sin^2 35+1
=(1/sin^2 35)sin^2 35+1
=1+1
=2
2009-05-05 12:41 am
第一題同時可以係

(cos60°cos30°+sin60°sin30°) /tan30°
=cos(60°-30°)/tan30°
=cos 30°/tan 30°
=((sqrt3) /2)/(1/sqrt3)
=3/2
2009-05-04 4:26 am
錯題目
1. (cos60°cos30°+sin60°sin30°) /tan30先岩


(cos60°cos30°+sin60°sin30°) /tan30
=(1/2*√3/2+√3/2*1/2)/tan30
=(2√3/4)/tan30
=(2√3/4)*√3
=3/2.

因為你好似搞錯左題目.其他我廢是搞.你搞番好d題目.我再幫你
~~~~~~其得+( )~~~~~~~~~

2009-05-03 20:26:42 補充:
√=開方 *=x
參考: 我


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