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標題:
三條math 15分
發問:
10a) 4(cos330 - sin150) 10b) √3 sin120 + cos150 tan240 11) If tan⊕=3 and ⊕lies in quad.III ,find sin⊕+cos⊕ 12) IF sin⊕= -2/√5 and tan⊕<0 ,find cos ⊕ without using calculator 要有步驟
10a) 4(cos330 - sin150) 4(cos330 - sin150) =4(cos30-sin30) =4[(√3/2)-0.5] =2√3-2 10b) √3 sin120 + cos150 tan240 √3 sin120 + cos150 tan240 =√3 sin60 -cos30 tan60 =√3 (√3/2) -(√3/2) (√3) =0 11) If tan⊕=3 and ⊕lies in quad.III ,find sin⊕+cos⊕ tan⊕=3 =>sin⊕=3cos⊕ sin⊕+cos⊕ =3cos⊕+cos⊕ =4cos⊕ =-4/√(1+tan^2 ⊕) =-4/√(1+9) =-4/√10 12) IF sin⊕= -2/√5 and tan⊕<0 ,find cos ⊕ cos ⊕ =√(1-sin^2⊕) =√[1-(-2/√5)^2] =1/√5 2007-05-06 22:54:37 補充: 樓上第一題計錯la…
其他解答:
10a) 4(cos330-sin150) =4[(-cos30)-(sin30)] =4(-√3/2-?) =-2√3-2 =-2(√3+1) b) √3 sin120 + cos150 tan240 =√3 sin60+(-cos30)(tan60) =√3 (√3/2)+(-√3/2)(√3) =3/2-3/2 =0 11同12題係英文- - 睇吾明- -~
三條math 15分
發問:
10a) 4(cos330 - sin150) 10b) √3 sin120 + cos150 tan240 11) If tan⊕=3 and ⊕lies in quad.III ,find sin⊕+cos⊕ 12) IF sin⊕= -2/√5 and tan⊕<0 ,find cos ⊕ without using calculator 要有步驟
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最佳解答:10a) 4(cos330 - sin150) 4(cos330 - sin150) =4(cos30-sin30) =4[(√3/2)-0.5] =2√3-2 10b) √3 sin120 + cos150 tan240 √3 sin120 + cos150 tan240 =√3 sin60 -cos30 tan60 =√3 (√3/2) -(√3/2) (√3) =0 11) If tan⊕=3 and ⊕lies in quad.III ,find sin⊕+cos⊕ tan⊕=3 =>sin⊕=3cos⊕ sin⊕+cos⊕ =3cos⊕+cos⊕ =4cos⊕ =-4/√(1+tan^2 ⊕) =-4/√(1+9) =-4/√10 12) IF sin⊕= -2/√5 and tan⊕<0 ,find cos ⊕ cos ⊕ =√(1-sin^2⊕) =√[1-(-2/√5)^2] =1/√5 2007-05-06 22:54:37 補充: 樓上第一題計錯la…
其他解答:
10a) 4(cos330-sin150) =4[(-cos30)-(sin30)] =4(-√3/2-?) =-2√3-2 =-2(√3+1) b) √3 sin120 + cos150 tan240 =√3 sin60+(-cos30)(tan60) =√3 (√3/2)+(-√3/2)(√3) =3/2-3/2 =0 11同12題係英文- - 睇吾明- -~
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