High school trigonometric function questions about the advancement of speed

Depending on the conditions, f(x) is f(x) 0 at -1 x 1 and f(x) 0 at 1 x 3

So f(1)=1+b+c=0, b+c=-1f(0)=c 0

f(x)=x 2+(-1-c)x+c=(x-1)(x-c)If 0 c 1, at c x 1, f(x) 0, contradictory, so c 1, at 1 x 3 f(x) 0

So c 3f(x) = (x-1)(x-c), and c 3, so the maximum value of f(x) at -1 x 1 is f(-1) = 2(c+1) = 8

c=3,b=-4

1) It is known that no matter what value m, n goes, f(sin m)>=0, f(2+cos n)<=0.

When the value of f(x) is greater than or equal to 0 on the interval [-1,1] and less than or equal to 0 on [1,3], because the opening is upward.

So the abscissa of the vertex must be greater than 1, so on the interval [-1,1], f(x) decreases, f(1)=0, brings in the function, b+c=-1

2) If the value interval of the abscissa of the vertex is (1,2), let the abscissa of the vertex be 1+t,00, but from the value of no matter what value n goes to meet f(2+cos n)<=0, we can know that f(3)<=0, contradictory.

So the abscissa of the vertex is greater than or equal to 2, so b<=-4, known by b+c=-1, c>=3

3) So f(-1)=8, substitute the original formula, and then get c=3, b=4

Proof that: 1) The value range of sinm is [-1,1], and the range of values of 2+cosn is [1,3], because for any m, n is true, I choose the appropriate m so that sinm=1, then f(1)>=0;

Choose the appropriate n so that cosn+2=1, then f(1)<=0 then f(1)=0, so 1+b+c=0, so b+c=-12) choose the appropriate n, so that cosn+2=3, f(3)<=09+3b+c<=0, because b=-1-c, substitution will get c>=33) this parabola opening upwards, you draw a sketch, you can know that the function is monotonically decreasing on [, so f(-1) is the maximum, so f(-1)=1- b+c=8, b+c=-1, so b=3, c=-4

The quadratic function f(x)=x 2+bx+c(b,c belongs to r), no matter what value m,n goes, it satisfies f(sin m)>=0, f(2+cos n)<=0

It can be seen that when sinm=1 and cosn=-1, f(1) 0; f(1) 0 gives f(1)=0, b+c=-1

f(x)=(x-1)(x-c)

f(2+cos n) 0 so f(3) 0,c 3x<1.

f'(x)<0

The maximum value of f(sin m) is 8, i.e., f(-1)=81-b+c=8

Get. b=-4c=3

1<=sinm<=1

1<=2+cosn<=3.

So: f(x)=x 2+bx+c is 0 on [-1,1], 0 on [1,3].

f(1)≥0,f(1)≤0

f(1)=0

1+b+c=0

b+c=-1.

f(3)≤0

9+3b+c≤0

b=-1-c

9-3-3c+c≤0

c The opening is upward, 0 on [-1,1], 0 on [1,3].

So the axis of symmetry is to the right of x=1!!

So on [-1,1] is a subtractive function.

f(sin m) maximum = f(-1)=1-b+c=8b+c=7b+c=-1b=

It's not going to be an exam, is it? Don't do it, let's talk about this attitude.

The attitude is not good, no matter how high the score is, I will not do it.

4sin^2(b+c)/2-cos2a=7/2.It is possible to push out cos2a=-1 2a=60 degrees.

Multiply the equation (b+c) a by 2r according to the sinusoidal theorem to get sinb+sinc sina

sina = 60 degrees So the range of (b+c) a is equal to the range of 2 root 3 (sinb+sinc) 3.

Then solve at the same angle.

Using the method of "a532012866" will do, and the result is (1,2).

f(x)= 3sin(wx+fai)-cos(wx+fai)=2sin(wx+fai-pie 6).

is an odd function. then there is f(0)=0

FAI can be found, and the axis of symmetry stool for f(x) is.

wx+fai-paitan town 6=pai 2

The distance between the two adjacent axes of symmetry of the x=(7pai 6-fai) wf(x) image can be found in bold and bold way.

Pai 2 can find w

sin (a-b) + cos(a-b) = 1a-b in the second quadrant.

So cos(a-b)<0

So cos(a-b)=-4 5

Is it sin(a+b)=-3 5?

The same goes for cos(a+b)=4 5

So cos2b

cos[(a+b)-(a-b)]

cos(a+b)cos(a-b)+sin(a+b)sin(a-b)=-16/25-9/25

1 If it is sin(a+b)=-5 13

That's the same way.

First of all, the title is wrong. sin(a+b)=-3 5, it can't be -5 3.

Then, according to -3 5, the result is -1

Process: cos(a+b)=4 5 , cos(a-b)=-4 5 ;

cos2b=[(a+b)-(a-b)]=cos(a+b)cos(a-b)+sin(a+b)sin(a-b)=-1.

Hope it helps!

cos(a-b)=root(1-sin(a-b))=-4 5cos(a+b)=root(1-sin(a+b))=4 5cos2b=cos[(a+b)-(a-b)]=cos(a+b)cos(a-b)-sin(a+b)sin(a-b)=4 5*(-4 5)-(3 5*3 5)=-7 25

If there is an error, please let it know!!

Thank you!!

π/2 ≤2x+π/4≤π/2

3π/8≤x≤π/8

Are you sure the answer is okay?

Solution: d -

According to the cosine theorem: ac = ab +bc -2ab*bccos = ad +dc -2ad*dccos ( -

That is, 4 +3 -24cos = 5 +6 -60cos ( - get cos = -3 7

ac = 4 + 3 -24cos = 25 + 72 7 = 247 7 i.e. ac = 247 7