中三 sin cos tan 數 (3條)

2007-06-10 6:41 am
已知答案 但唔知步驟


1. (cos^2 45°)(sin 30°) / tan^2 60° ans=1/12

2. cos30° cos45° + sin60° sin45° ans=√6/2

3. find θ ,

tan60° / sin(θ/3 +35°) ans: θ=75°
更新1:

打錯了第三題...sorry 3. find θ , tan60° / sin(θ/3 +35°)=√8 cos 45° ans: θ=75°

回答 (4)

2007-06-10 7:08 am
✔ 最佳答案
1.(cos^2 45°)(sin 30°) / tan^2 60° ans=1/12
=(1/√2)^2*(1/2)/3
=(1/2*1/2)/3
=(1/4)/3
=1/12
2.cos30° cos45° + sin60° sin45° ans=√6/2
=sin(90-30)sin(90-45)+sin60sin45
=sin60sin45+sin60sin45
=2(√3/2)(√2/2)
=√6/2
3.沒有可以解的方法
因為不是個方程
參考: me
2007-06-10 7:27 am
1, [(√2/2)^2](1/2)/(√3)^2
 =(2/4)(1/2)/3
 =2/24
 =1/12

2, (√3/2)(√2/2)+(√3/2)(√2/2)
 =2(√6/4)
 =√6\2

3, tan60°/sin(θ/3+35°)=√8cos 45°
   √3/sin(θ/3+35°)=2√2(√2/2)
   √3/sin(θ/3+35°)=2
     sin(θ/3+35°)=√3/2
     sin(θ/3+35°)=sin60°
       θ/3+35°=180°n+(-1)n*60°   (where n is integer)
         θ/3=180°n+(-1)n*60°-25°
          θ=540n+(-1)n*180°-75°
          θ=75°   (if 0°<θ<90°)
2007-06-10 7:02 am
1.(cos^2 45°)(sin 30°) / tan^2 60°

=[(√2 /2)^2*(1/2)]/(√3)^2

=[(2/4)*(1/2)]/3

=(2/8)/3

=(2/8)*(1/3)

=2/24

=1/12

2.cos30° cos45° + sin60° sin45°

=(√3 /2)*(√2 /2)+(√3 /2)*(√2 /2)

=√6/4+√6/4

=2√6/4

=√6/2

3.你好似有D野冇打落去。
2007-06-10 6:51 am
1.(√2/2)^2 (1/2)/(√3)^2
=(2/4)(1/2)/3
=2/24
=1/12

2. (√3/2)(√2/2)+(√3/2)(√2/2)
=2(√6/4)
=√6\2

3.我覺得佢等號另一邊應該有野比你先計到


收錄日期: 2021-04-25 21:47:41
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