几道高中数学三角函数选择题,求解
1.已知△ABC中,tanA=-5/12, cosA=( ) A.12/13 B.5/13 C.-5/13 D.-12/13 2.已知sinα=√5/5,则(sinα)^4-(cosα)^4的值为多少( ) A.-1/5 B.-3/5 C.1/5 D.3/5 3.将函数y=sin2x+cos2x的图像向左平移π/4个单位,所得图像的解析式为( ) A.y=cos2x+sin2x B.y=cos2x-sin2x C.y=sin2x-cos2x D.y=cosxsinx 希望可以写出具体步骤
1。解:tanA=sinA/cosA=-5/12 得:sinA=(-5/12)cosA sin²A+cos²A=[(-5/12)cosA]²+cos²A=(169/144)cos²A=1 那么:cosA=±12/13 而:tanA<0 ∴cosA=-12/13 选D。
2。(sinα)^4-(cosα)^4 =(sin²α+cos²α)(sin²α-cos²α) =sin²α-cos²α =sin²α-(1-sin²α) =2sin²α-1 =2×(√5/5)²-1 =-3/5。
选B。 3。将函数y=sin2x+cos2x的图像向左平移π/4个单位 y=sin2(x+π/4)+cos2(x+π/4) =sin(2x+π/2)+cos(2x+π/2) =cos2x-sin2x 选B 。
1 D tanA=sinA/cosA sinA和cosA的平方和为1 所以选D 2 (sinα)^4-(cosα)^4=((sinα)^2+(cosα)^2)((sinα)^2-(cosα)^2)=(sinα)^2-(cosα)^2 sinα=√5/5 (sinα)^2=1/5 (cosα)^2=4/5 答案B 3 y=sin2x+cos2x=√2sin(2x+π/4) 向左平移π/4个单位 y=√2sin(2(x+π/4)+π/4) =√2sin(2x+π/2+π/4) = √2cos(2x+π/4) 答案 A
答:选A∵sinx=-1/√5>-1/2,π<x<7π/6,2π<2x<7π/3∴cosx=-2/√ n2x=2sinxcosx=4/s2x=3/5.xin(2x-...详情>>
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