高一题目,急!
已知X+Y=3-cos4a,X-Y=4sin2a,求证X^1/2+Y^1/2=2
已知x+y=3-cos4a,x-y=4sin2a,求证x^1/2+y^1/2=2 证明: x+y=3-cos4a, x-y=4sin2a 两式相加、减,分别得 2x=3-cos4a+4sin2a 2y=3-cos4a-4sin2a 而 2x=3-cos4a+4sin2a =3-[1-2(sin2a)^2]+4sin2a =2(sin2a)^2+4sin2a+2 =(2sin2a+2)(sin2a+1) x=(sin2a+1)^2 √x=(sin2a+1) 2y=3-cos4a-4sin2a =(2sin2a-2)(sin2a-1) y=(sin2a-1)^2 √y=1-sin2a 所以 √x+√y =sin2a+1+1-sin2a =2
不想后面两位大哥比较聪明也,小弟也是这么想的,在这就不用写拉。
x+y=3-cos4a=3-(1-2sin^22a)=2(1+sin^22a)...(1),x-y=4sin2a....(2) (1)+(2),x=1+sin^22a+2sin2a=(1+sin2a)^2, (1)-(2),y=1+sin^22a-2sin2a=1-sin2a)^2, x^1/2+y^1/2=1+sin2a+1-sin2a=2
x+y=3-cos4a=3-(1-2sin^22a)=2(1+sin^22a)...(1),x-y=4sin2a....(2) (1)+(2),x=1+sin^22a+2sin2a=(1+sin2a)^2, (1)-(2),y=1+sin^22a-2sin2a=1-sin2a)^2, x^1/2+y^1/2=1+sin2a+1-sin2a=2
X + Y = 3-cos4a = 3-[1-2*(sin2a)^2] = 2 + 2*(sin2a)^2 X - Y = 4*sin2a ==> X = 1 + (sin2a)^2 + 2*sin2a = (1 + sin2a)^2, Y = 1 + (sin2a)^2 - 2*sin2a = (1 - sin2a)^2 因此: X^1/2 + Y^1/2 = [(1 + sin2a)^2]^1/2 + [(1 - sin2a)^2]^1/2 = (1 + sin2a) + (1 - sin2a) = 2
答:根据tan2a,可知分子要化简为sin2a,分母cos2a,选择适当的二倍角公式向这方面靠 1+sin4а-cos4а=1+2sin2a*cos2a-(1-2s...详情>>
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