Find the maximum and minimum values of the function y sinx 2 cosx 2

4 (root number 2 cannot be typed, replace it with a).

Let t=sinx+cosx=a*sin(x+45), so -a<=t<=a

t 2 = (sinx) 2 + (cosx) 2 + 2 (sinx) * (cosx) = 1 + 2 (sinx) * (cosx) so (sinx) * (cosx) = (t 2-1) 2

So y=(t 2-1) 2-2t+4=(1 2)t 2-2t+(7 2)=(1 2)*(t-2) 2+(11 2).

So find the maximum value of the quadratic function on [-a,a]) So, when t=-a, y has a maximum value, and when t=a, y has a minimum value.

t=2 ,f(x+2 )=cos[sin(x+2 )]=cos(sinx)=f(x) Let t=sinx, so t[-1,1].

y=cost decreases monotonically on t [-1,1] When t=1, the minimum value of y is cos1

Simplification: 2-(sin2) 2+cos4?Solution: Original formula = 2-(sin2) 2+1+1-2(sin2) 2

3-3(sin2) 2 = 3(1-(sin2) 2) = 3(cos2) 2

Since cos2=less than 0, =-cos2 is 3 under the root number

Find the tangent of a+b first and then the tangent of a+(a+b).

Auxiliary angle formula.

The quadratic secondary denominator is seen as the "1" numerator and denominator of the hidden row, and is divided by the square of the cosine.

cos(2x+2 pie 7} cos(x+7) 2-1 then recipe.

Raise the jackpot and then give you a detailed look! It's so stingy!

In addition, there are products on the stationmaster's group **, cheap ***.

When the 45° minimum formula pushes ab<=(a2+b2) 2, if and only if a=b holds the minimum approximate equal.

Let's do it with a universal formula, and the amount of calculation of horror is roughly equal to.

Categories: Electro Zen Brain Internet >> Programming >> Other Programming LanguagesProblem Description:

Find the maximum and minimum values of the function y=cos2x+sinx and say the process Thank you.

Analysis: y=1-2 (square of sinx) + sinx 2 (square of (sinx-1 4)) + 9 8

sinx is less than or equal to 1 than or equal to 1, and the noisy mountain rises to a small rise of y in equal to 9 8, greater than or equal to 2

So the minimum is -2 and the maximum is 9 8

This question is geometric.

y=sinx/(cosx+2)

It represents the slope from (sinx, cosx) to (-2,0) points.

And (sinx, cosx) represents the point on the circle x 2 + y 2 = 1.

Then it is actually the range of the slope from the point of the circle x 2+y 2=1 to (-2,0).

Draw a picture. Then in fact, when passing (-2,0) and tangent to the circle, the maximum and minimum slopes are obtained.

Let the straight line y=k(x-2), tangent to the circle.

Then (the distance to a straight line is radius 1.)

So:|k*(0-2)-0|1+k 2)=1 solution: k = 3 3 or k = - 3 3

So the maximum value is 3 3 and the minimum value is - 3 3 Proficiency in looking at algebra problems from a geometric perspective.

Let's answer with the knowledge of circles!

In order not to get confused, I use a instead of the function argument in the question.

Notice that the parametric equation for a circle with the center of the circle at the origin and a radius of 1 is x=cosa

y=sina

Isn't f(a)=(sina-1) (cosa-2) the slope of the line connecting the point p(cosa,sina) and q(2,1)?

q is a fixed point, and when p moves on a circle, it is easy to see that the slope of the pq line (i.e., f(a)) obtains the maximum and minimum values respectively when pq is tangent to the circle (in two cases), and finding the slope of the two tangents is the answer.

In fact, let the equation for pq be .

y-1=k(x-2) (it passes through (2,1) and the slope is k) i.e. kx-y-2k+1=0

Note that since pq is tangent to the circle, the distance from the origin to the straight line is 1, and the formula for the distance from the point to the straight line is there.

k*0-0-2k+1|[(k 2+1) (1 2)]=1 squared to obtain a quadratic equation.

3k^2-4k=0

Therefore k = 0,4 3

So the minimum value is 0 and the maximum value is 4 3

y=(2-sinx)/(2-cosx)

Both sides of the equation are multiplied by 2-cosx

y(2-cosx)=2-sinx

Place the equation to the left, and that is.

2y-ycosx=2-sinx

1+y )·sin(x- )=2-2y, where sin = y (1+y), cos =1 (1+y), sin(x- )=(2-2y) (1+y)

sin(x-θ)1,∴|2-2y)/√(1+y²)|1, get 3y -8y+3 0

The solution is (4-7) 3 y (4+ 7) 3, so the maximum value is (4+ 7) 3

Maximum:

Minimum: 2 2

Analysis: Derivative of the original function is obtained'= (cosx + sinx - 2)(cosx - sinx), both cosx and sinx value ranges are [ -1,1], so the sign of the first parenthesis must be negative, and when the second parenthesis is 0, there are two cases, one before and after the zero value.

First positive and then negative, a type of zero value before and after negative and then positive. So one case is the maximum, and the other is the minimum.

When both cosx and sinx take positive values of 2 2, it is the second case, and the minimum value is 2 2;

When both cosx and sinx take positive values of 2 2, it is the second case, and the minimum value is 2 2;

I've answered this before.

Answer: y=(2-sinx) (2-cosx) (both sides are multiplied by 2-cosx, 2-cosx≠0)2y-ycosx=2-sinx

sinx-ycosx=2-2y

1+y²)sin(x-∅)=2-2y

sin(x- )=(2-2y) (1+y)so |(2-2y)/√(1+y²)|1|(2-2y)|≤1+y²)

Squared 4-8y+4y 1+y

3y²-8y+3≤0

4+√7)/3≤y≤(4-√7)/3

So the maximum value (4+7) 3, the minimum value (4-7) 3

y=(2-sinx) (2-cosx) multiply both sides of the equation by 2-cosxy(2-cosx)=2-sinx and the left side of the equation gives 2y-ycosx=2-sinx

1+y )·sin(x- )=2-2y, where sin = y (1+y), cos =1 (1+y), sin(x- )=(2-2y) (1+y)

sin(x-θ)1,∴|2-2y)/√(1+y²)|1, get 3y -8y+3 0

The solution is (4-7) 3 y (4+ 7) 3, so the maximum value is (4+ 7) 3

Because 2-cosx is greater than 0

So multiply both sides of the equation by 2-cosx, and you get it.

2y-y*cosx=2-sinx, and the range of y can be solved by using the unity deformation to make sin(x-)=1.

y=cos2x-sinx

1-2sin²x-sinx

2(sinx+1 4) +9 8>=9 8 The maximum value is 9/8, and sinx=-1 4.

When sinx=1, the minimum value of -2*25 16+9 8=-1y=cos x-sinx is obtained

1-sin²x-sinx

(sinx+1 2) +5 4>=5 4 The maximum value is 5/4, and is obtained when sinx=-1 2.

When sinx=1, the minimum value of -9 4+5 4=-1 is obtained

(The square of sinx+ is equal to the square of the absolute value of sinx+, which is 0 to nine-quarters, and the range of positive and negative cannot be squared directly, and you should pronounce four-nineths ..........)