High One Trigonometric Function Help!! High School Trigonometric Problems! Solve

, squared (cos) 2+4sin cos +4(sin) 2=5

Note (cos) 2 + (sin) 2 = 1, then.

cosα)^2+4sinαcosα+4(sinα)^2=5[(cosα)^2+(sinα)^2]

That is, (sin) 2-4sin cos +4(cos) 2=0, (sin -2cos) 2=0, tan =2, choose b

2.(cos) 2+(sin) 2=1, then (cos) 2=1-(sin) 2

cos ) 2 = (1-sin) (1 + sin), 1-sin ) cos = cos (1 + sin), so), (sin -1) cos = -cos (1 + sin) = 2, choose c

cos = (k-1) (k-3), then cot = (k-1) (k+1).

Because (cos) 2+(sin) 2=1, [(k+1) (k-3)] 2+[(k-1) (k-3)] 2=1, i.e., (k+1) 2+(k-1) 2=(k-3) 2

k=1 or k=-7, then there is cot = 0 or 4 3, choose b

2.The numerator and denominator are multiplied by 1-sinx at the same time

cosx(1-sinx) 1+sinx(1-sinx)=-2 is simplified: 1-sinx cosx=-2

So: sinx-1 cosx=2

(1) 1+sinx+cosx is not equal to 0, and x is not equal to 2k k+

2) f(x)=1 + sin2x) (1+sinx +cosx), discuss the monotonicity of (sin2x) (1+(2)*sin(x+ 4)) using the derivative method (for fractions).

Because (cosx) 2-(sinx) 2=cos2x

From cos2x=2(cosx) 2-1, (cosx) 4=(1+cos2x) 2 4

Again(sinx) 2=(1-cos2x) 2

Substitution, can be simplified.

f(x)=[3(cos2x)^2+cos2x]/(2cos2x)=(3/2)*cos2x+1/2

From cos2x≠0, we get 2x≠k + 2, i.e., x≠k 2 + 4

Therefore y=3 2*cos2x+1 2 (x≠k 2+ 4).

The function definition domain is.

-1 cos2x 1 and cos2x≠0

The function range is =[-1,1 2) (1 2,2].

cosx) 4=(1+cos2x) 2 4 What does this mean?? How do you divide by 4 at the end? In the same way, why does (sinx) 2=(1-cos2x) 2 divide by 2 at the end??

Answer: This is the formula for double angles.

Because 1+cos2x=2(cosx) 2, (1+cos2x) 2=4(cosx) 4

Because 1-cos2x=2(sinx) 2, (1-cos2x) 2=(sinx) 2

2kπ+5π/6

2k+1)π-/6

The difference between him and 2k - 6 is that one of the coefficients is odd and the other is even, so they can be combined, which is an integer.

Similarly, 2k +7 6=(2k+1) + 6 and 2k + 6 are also combined into integers.

And the integer can be represented by k.

So the merger is [k - 6, k + 6].

or the union is [k - 6, k + 6].

This should be analyzed in combination with trigonometric lines: [- 6, 6] This range is rotated by 180 degrees, and it will be [5 6, 7 6] plus k to get [k - 6, k + 6].

And then there's that one.

Can you help me look at all the questions?

The result is three for the root number of two, and three for five times the root number of the two.

Two-to-three times. Black line] [black line].

Just send ** if you ask a question, just write it on paper, because I'm a high school student, and the teacher will say if the question is not written, it's troublesome, thank you.

Answer: Okay. sinxcos(x+π/4) =√2/2sinxcosx-√2/2sin^2x =√2/4sin2x-√2/4(1-cos2x) =√2/4sin2x+√2/4cos2x--√2/4 =√2/4sin(2x+π/4)--2/4 sin(2x+π/4)

The simplest is: 2sin (2x+4)- 2sin(2x+4)+4 root number 2.

Just bring it in.

The most difficult thing is to solve the problem.

Since the period of the function f(sin 2(x)) is , it can be replaced by a periodic transformation.

That is, 5 6 = - 6, 7 6 = + 6

Then the above formula can be changed after finishing.

The so-called definition domain is the range of x (independent variable) that makes the function meaningful, and the above can be found by the formula 0<=sin x<=1 4, 0<=(cos 2-1 2)<=1 4,.

Student: You can try to draw an image of these two parts in a planar Cartesian coordinate system.

You'll notice that they actually represent the same area.

The distance between adjacent maximums is half a period, so the period of the function is 2 w = 2 * (2 3 - 6), knowing w = 2.

The maximum value is 2, and a is 2.

Substituting the maximum point into the function gives y= 6

u=2sin^2(2x+π)acos(2x+π/2)+1

2sin^2(2x)-asin2x+1

When x [0, ], sin2x [-1,1], so that y=sin2x, then u= u(y)=2y 2-ay+1=2(y-a 4) 2+1-a 2 8,y [-1,1].

u is a quadratic function with the opening facing up of the independent variable y, the axis of symmetry is y=a 4, and the midpoint of the independent variable interval [-1,1] is 0, it is easy to know that when a 4 0, that is, a 0, umax=u(y)max=u(-1)=3+a, at this time, y=sin2x=-1, x=3 2

When a 4<0, i.e., a<0, umax=u(y)max=u(1)=3-a, at this time y=sin2x=1, x=2

Wait a minute, I'm done writing and sending **.

4cosx sin（x+6/π)-1

4cosxsinxcos6 +cosxsin6 )-1=4cosx[(3)2sinx+(1 2)cosx]-1=2(3)sinxcosx+2cosx-1=(3)sin2x+cos2x

2[(√3)/2sin2x+1/2cons2x]=2sin(2x+π/6)

So t=2 2=

x is in the range of 6/4 and 4/4

2x+6 at [-6/3].

f(x) belongs to [-2/1,1].

f(x)max=1 f(x)min= -1 2, I mean I'm tired of playing ==Ask for adoption.

With sinusoidal sin(x+y)=sinxcosy+cosxsiny, so sin(x+ 6)=sinxcos 6+cosxsin 6=(3) 2sinx+1 2cosx

So f(x)=4cosxsin(x+ 6)-1=4cos[sinxcos 6+cosxsin 6]-1=4cos[(3) 2sinx+1 2cosx]-1=2 3cosxsinx+2(cosx)squared-1= 3sin2x+cos2x

2sin(2x+π/6)

So the period t=2 2=

x is in the interval 6, 4

2x+ 6 at [- 6, 2 3].

The maximum value is f(x)=f(2)=2;

The minimum value is f(x)=f(6)=-1