The minimum value of y 1 sinx 2 mcosx is 4 to find the value of m

If your 2 is squared, you should do so.

by y>=-4

Then 2-(cosx) 2-mcosx>=-4 let a=cosx

then the inequality is.

2-a^2-ma>=-4

Sorted out. 0>=a^2+ma-6；Let f(a)=a2+ma-6f(a) be the derivative of a, and we can get it.

The minimum point of the function is a=-m 2

When -m 2<=-1, then f(a) is singled over the defined domain, and f(a)max=f(1)=m-5=0, resulting in m=5

When -1<=-m 2<=1, then f(a)max=max(f(-1),f(1)) is not equal to 0

When 1<=-m 2, then f(a) is subtracted on the defined domain, then f(a)max=f(-1)=-m-5 is not equal to 0

In summary, m=5

y=1+(sinx)^2-mcosx=-(cosx)^2-mcosx+2=-[(cosx)^2+mcosx+m^2/4]+2+m^2/4=-[cosx+m/2]^2+2+m^2/4>=-4

cosx+m/2]^2<=6+m^2/4

Then when m>0 and cosx=1, [cosx+m2]2 can be taken to the maximum, m2 4+m+1=6+m2 4, that is, m=5

When m<0 and cosx=-1, [cosx+m2]2 is possible to get the maximum value, m2 4-m+1=6+m2 4, that is, m=-5

The value of cosx is between -1 and 1, containing -1 and 1, because the maximum value of this equation is 5.

The maximum value of MCOSX = 3 and cosx = 3 m, X can be determined when the value of m is determined

Didn't you say the value of m?? If m is greater than 0, then cosx is equal to 1 and has a maximum, and conversely, cosx is equal to -1 and has a maximum!!

Summary. Sorry, to correct it, the correct answer is: x>0, y>0, x+y= 2sinx+siny=sinx+sin( 2-x)=sinx+cosx 2 root number(sinx*cosx) when sinx=x=.

or sinx=, when x= 2, y=0sinx+cosx, the minimum is 1, that is, sinx+siny, the minimum is 1

Knowing x>0,y>0,x+y= 2, find the minimum value of sinx+siny.

x>0,y>0,x+y= 2sinx+siny=sinx+sin( 2-x)=sinx+cosx 2 root number (sinx*cosx) when sinx=cosx= root limb key number 2 2, sinx+codx is the minimum bucket root hunger file number 2

The above is the detailed answer.

You don't have a problem with that.

Exactly. x>0,y>0,x+y= 2sinx+siny=sinx+sin( 2-x)=sinx+cosx 2 (sinx*cosx) when sinx=cosx= root 2 2, x=y= 4sinx+cosx, the minimum is the root nai xiaofengsheng and the number 2, that is, sinx+siny, the minimum is the root number 2

Not Chunlu is embarrassed, the front bucket has a little writing error, has been corrected, the correct answer is the pin: x>0, y>0, x+y= 2sinx+siny=sinx+sin( 2-x)=sinx+cosx 2 root number (sinx*cosx) when sinx=cosx=root number 2 2, x=y= 4sinx+cosx, the minimum value is the root number 2, that is, the minimum value of sinx+siny is the root number 2

Sorry, to correct it, the correct answer is: x>0, y>0, x+y= 2sinx+siny=sinx+sin( 2-x) = sinx+cosx 2 root number (sinx*cosx) when sinx=, x= the most abrasive small eggplant swim difference is sinx+siny, the minimum is 1

Sorry, the correct answer is: x>0, y>0, x+y= 2sinx+siny=sinx+sin( 2-x)=sinx+cosx 2 root number (sinx*cosx) when sinx=, x=the smallest of the field is 1, that is, sinx+siny, the smallest is 1

I'm sorry, but the correct answer is the trace: x>0,y>0,x+y= 2sinx+siny=sinx+sin( 2-x)=sinx+cosx 2 root number (sinx*cosx) when sinx=x=x=. or sinx=, when x= 2, y=0sinx+cosx, the minimum is 1, that is, sinx+siny, the minimum is 1

Summary. That is, when x takes 4/4 is that there is a minimum value of 0, known x>0, y>0, x+y= 2, find the minimum value of sinx+siny.

Hurry up. Right away.

Answer: root number 2, the answer is as shown in the figure above.

Please look at the next question, it is to find the minimum value.

Good. The minimum is 0

Are you sure? Don't help me do anything wrong.

Someone else did it for me just now, and there is also a negative root number 2

Teacher! Is it still there?

Negative root number 2 cannot be taken

You look at the range. The maximum is 3/4

If you want to get -root number 2 then it is greater than in the third quadrant.

Isn't sin(2) equal to 1?

Equal to 0cos 2 is 1

That is, when x takes 4/4 is there a minimum value of 0

Summary. Hello, because f(x)=sinx+cosx= 2sin(x+ 4)Question 1 t=2 1=2 Question 2 When sin(x+ 4)=1, it is the maximum, that is, when f(x)= 2sin(x+ 4)=-1, it is the minimum value, i.e., f(x)=-2 Question 3 Because f(a)=3 4, i.e. 2sin(a+ 4)=3 4 2sinacos 4+ 2sin 4cosa=3 4sina+cosa=3 4 so (sina+cosa) = 9 16sin a + 2 sinacosa + cos = 9 161 + 2 sinacosa = 9 162 sinacosa = -7 16 so sin2a = 2sinacosa = -7 16

Knowing x>0,y>0,x+y= 2, find the minimum value of sinx+siny.

Hello! Hello, because f(x)=sinx+cosx= 2sin(x+ 4) the first question t=2 1=2 second difference and question When sin(x+ 4)=1, it is the maximum value, that is, f(x)= 2sin(x+ 4)=-1 time difference celebration, which is the minimum value, that is, f(x)=-2 Question 3 Because f(a)=3 4, that is, 2sin(a+ 4)=3 4 2sin(a+ 4)=3 4 2sinacos 4+ 2sin 4cosa=3 4sina+cosa=3 4So(imaginary beam sina + cosa) =9 16sin a + 2sinacosa + cos =9 161 + 2sinacosa = 9 162sinacosa = -7 16 so sin2a = 2sinacosa = -7 16

Knowing x>0,y>0,x+y= 2, find the minimum value of sinx+siny.

That's what I'm asking for! Teacher!

Hello, when Huigai 0 x good draft, sinx >0, y=sinx +1 sinx 2 (sinx *1 sinx)=2 The current socks judgment is only when sinx = 1 sinx takes the equal sign, (sinx) 2=1, sinx =1, that is, when x = 2, y takes the minimum value of 2

Teacher! Are you right?

Yes, dear.

x+y= 2, y= 2-x, sinx+siny=sinx+sin( 2-x)=sinx+cosx= 2sin(x+ 4), by x>0, y= 2-x>0, 0 is obtained as 4 "x+ 4<3 4, the kernel is obtained as 2 2 sin(x+ 4) 1, so the minimum value of 1 is 1

From x+y= 2, we get y= 2-x, sinx+siny=sinx+sin( 2-x)=sinx+cosx= 2sin(x+ 4), by He Changqing x>0, y= 2-x> Zen Lu 0, 0 is 4, so the minimum value sought by Xunyin does not exist, but there is a lower bound 1.

sinx+siny=sinx+sin(π/2-x)=sinx+cosx≥2√（sinxcosx）

When sinx=cosx, and the leakage height is as high as x= n4, the inequality is taken as "=".

So the minimum value of sinx+siny is restart2.

Solution: sinx ten siny = 2sin ( Suiyin ten y) 2cos( a y) 2 by the known > quiet mountain 0,y>0, ten y= 2 generations to get sin ten siny=2sin 4cos( 2一2y) 2= 2cos( 4-y) and y= 2- >0 and >o 0 "y< 2 y in the Qi people has no minimum value.

y=(1+sinx)(1+cosx) y=1+sinx+cosx+sinxcosx y=1+(sinx+cosx)+[sinx+cosx) 2-1] 2 Let sinx+cosx=t, so -1 t pure vertical 1 so y=1+t+t 2 2-1 2=(t 2+2t+1) slow match 2=(t+1) 2 since, -1 t 1, which is greater, 0 y 2 The minimum value is 0, The maximum value is 2....