標題:
「數學」-直角三角形(三個函數)
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發問:
麻煩幫幫手呀~~~ 1.不用計算機簡化sin2次30度 cos2次60度tan45度。
最佳解答:
sin2次30度 = (1/2)^2 = 1/4 cos2次60度 = (1/2)^2 = 1/4 tan45度 = 1
其他解答:
1st right angle (other 2 angles =30oand 60o) triangle with: adjacent side (of angle 30o) = sqrt root 3 opposite side (of angle 30o) = 1 By using the pyth. Theorem (a^2+b^2=c^2), hypo = 2 2nd right angle (other 2 angle =45o) triangle with: adjacent side (of angle 45o) = 1 opposite side (of angle 45o) = 1 By using the pyth. Theorem (a^2+b^2=c^2), hypo = sqrt root2 Here are some of the basic triangle values you need to memorize by heart: From the 1st triangle: Sin 30o= cos 60o=1/2 Tan30o=1/sqrt root 3 Tan60o=sqrt root3 From the 2nd triangle: Cos45o=sin45o=1/sqrt root2 Tan45o=1 Thus the solution as follow: 1. (sin30o)^2=(1/2)^2=1/4 2. (cos60o)^2*tan45o=(1/2)^2*1 = 1/4|||||sin2 次30度 cos2次60度 都係等於 2分之1 tan 45 =1 而 1」2 + 1」2 + 1 = 2 或者 而cos2次θ + sin 2次θ =1 tan 45 = 1 so, cos2次θ + sin 2次 tan 45 = 1 1 = 2 2008-04-09 10:30:59 補充： 或者之前個段係唔要的. 或者之後個到係跟據條公式的.
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