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Anonymous users2024-02-16Because mn=0
So sina-2cosa=0
sina=2cosa
Divide the two sides by cosa to get tana = 2
f（x）=cos2x+2sinx=1-2sinx·sinx+2sinx=-2(sinx·sinx-sinx)+1
2[(sinx-1/2)^2-1/4]+1=-2(sinx-1/2)^2+3/2
And because |sinx|< = 1, push it step by step and naturally come out ,,,3,3 2].
The result is calculated by my mouth, I don't know if it is correct, you can see for yourself.
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Anonymous users2024-02-15(1) m*n = root number 3sina-cosa = 12sin (a-6) = 1
sin(a-π/6)=1/2
Because 0a = 3
2) f(x) = root number 3 2 * (1 + cos2x) + 1 2 * sin2x = sin (2x + 3) + root number 3 2
Because 0< = x< = 3
0<=2x<=2π/3
3<=2x+π/3<=π
So the root number 3 2 < = f(x)< = 1 + the root number 3 2
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Anonymous users2024-02-14(1) Because mn=0
So sina-2cosa=0
sina=2cosa
Divide the two sides by cosa to get tana = 2
2)f（x）=cos2x+2sinx=1-2sinx·sinx+2sinx=-2(sinx·sinx-sinx)+1
2[(sinx-1/2)^2-1/4]+1=-2(sinx-1/2)^2+3/2
And because |sinx|< = 1, push it step by step and naturally come out ,,,3,3 2].
The result is calculated by my mouth, I don't know if it is correct, you can see for yourself.
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Anonymous users2024-02-131) sina -2cosa=0
So tana=2
2)y=cos2x +2sinx =1-2(sinx)^2 +2sinx
2t^2 +2t +1 t =sinx , 1<=t<=1
2(t-1/2)^2 +3/2
When t=12, ymax=32
When t=-1, tmin=-3
The value range is [-3,3 2].
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Anonymous users2024-02-12I didn't finish the question......
The vectors are known m=(sina,cosa), vectors n=(traces1,-2), and mn=0
1) Find the value of tana.
Jujube Year 2) finds the range of the function f(x)=cos2x+tanasinx(x belongs to r).
m*n=sina-2cosa=0
tana=sina/cosa=2
f(x)=cos2x+tanasinx
cos2x+2sinx
1-2(sinx)^2+2sinx
3/2)-2(sinx - 1/2)^21<=sinx<=1
3 2 <=sinx - 1 2 "Ziyansan = 1 20<= sinx - 1 2) 2 <=9 4 So: (3 2)-2*(9 4)<=f(x)<=3 23<=f(x)<=3 2
Value range: [-3,3 2].
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Anonymous users2024-02-11m = (root number 3,1), n = (cosa+1,sina), and m n gives the equation cosa+1 = 3sina
Combined with cos a+sin a=1, sina=0 or sina = 3 2, since a (0, ), so sina = 3 2, which is obtained by the vector ab*vector ac=2.
c*b*cosa=2 knows that cosa>0 so cosa=1 2 thus bc=2 cosa=4 then sδ=1 2bcsina= 3 and cosa=(b +c -a) 2bc=1 2 of a =b +c -bc>=2bc-bc=bc=4, so a>=2
That is, the minimum value of BC is 2
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Anonymous users2024-02-10The vector m=(root number 3,1), n = (cosa+1, sina), and the vector m n,(cosa+1) 3=sina,3sina-cosa=1,sin(a-30°)=1 2,30°=bc=4, when b=c=2, take the equal sign, the minimum value of bc is 2
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