Compulsory 4 Mathematics Solve a problem

When is the first quadrant, it is positive. When is the second quadrant, it is negative. Physics should have been taught, in physics it was the special angles.

0-360 can be 37 degrees, or 143 degrees. That is, a three-sided right triangle in mathematics. It is the corner corresponding to the side length of 3.

Feel free to ask. If satisfied. Yours is my biggest motivation. ^_

Of course, sin 3 5 may be an acute angle or an obtuse angle.

When is an acute angle, the (2) in sin(2) falls in the second quadrant, which is positive.

So, sin( 2) 4 5

When is an obtuse angle, 2 falls in the third quadrant, and sin(2) is negative.

So, sin( 2) 4 5

Of course, it is a positive and negative value, and cos is found to be a square drop, which must be positive or negative, and in most cases it is obtained by substituting the second article. I forgot the formula, you can't figure it out, paste the formula, I'll show you the calculation.

Probably consider the periodicity of +2k as it can also be seen on the image.

When the angle alpha is in the third quadrant.

It may be that the angle may be less than 90, and if 90 is added, it will be less than 180, which is +

1 All 0 + 0 + 0-1 + (-1) = -2

Well, if you exceed 2, you can directly subtract 2 to do it, because 2 is a cycle, so if you exceed 4, you can subtract 4

Then it can be transformed into sin0 + cos (3 2 ) + tan -sin (1 2 ) + cos

This equation is to be calculated separately. Let x=sin and note that x -1 gives -2 -3x2 +x to destroy 0

Divide the inequality of the tan kernel into two unary quadratic inequalities.

The first of these inequalities solves for x-2, 3, or x1

The second inequality -3x2+x0 solves x 0 or the remainder x 1 3 and combines the obtained results with -1,1 to get the answer in the figure.

Solution: It is easy to know from the meaning of the question w=2

Shift the image of y=f(x) to the left by |φ|unit length.

Then the function becomes f(x)=sin(2(x+|.)φ4)=f(x)=sin(2x+2|φ|4)

This image is symmetrical with respect to the y-axis of Rule 2|φ|4=k + 2 so |k 2 + 8, k is the number of the whole brigade.

Dismantle the stupid simple liquid, so one of the values is Just take the value of k and correspond to it.

The unit year is split into a circle, sin -1 2, and the angle that conforms to the range is below the straight jujube line y=-1 2, that is, between -5 6——-6 is written as a set is [-5 6+2k, -6+2k] (k is an integer).

The sinusoidal image below the red line is also compliant, and it is also [-5 6+2k ,-6+2k ] (k is an integer).

丨sina丨=-sina, get sina<=0丨cosa丨=-cosa, get cosa<=0, so a is in the third quadrant. And sina·cosa≠0, i.e. not on the coordinate axis.

Therefore tana>0

Then the point p (tana, sina) is in the fourth quadrant.

The thing under the root number 1+2cosx》0,cosx》-1 2,x belongs to [2k, 2k+2 3, 2k+4 3, 2k+

The thing in lg must be greater than 0, as above, we can get sinx>1 2, x belongs to [2k + 6, 2k + 5 6 ].

Intersect and get x belongs to [2k + 6, 2k + 2 3 ] I won't put the diagram on it.

1+2cosx≥0，2sinx-1>0。Simplification: cosx -1 2, sinx>1 2.

Therefore, the range of the value of x, that is, the defined domain is (2k + 6, 2k + 2 3] k z). The image is a periodic function, and the period is 2, which only needs to be drawn in (6,2,3). Change, extremum, inflection point, asymptote.

1: a=,, get a=

aub=a, substituting x=1 into x 2-ax+a+1=0, a has no solution.

Substituting x=2 into x 2-ax+a+1=0, a=3a c=c, substituting x=2 into x 2-mx+2=0, yielding m=32: x 2+ax+b=2x, the root is 2, which is the double root.

x^2+(a-2)x+b=0

x-2)^2=x^2-4x+4

The contrast factor is:

4=a-24=b

i.e.: a=-2, b=4

3:1, a can be reduced to a=} b= a is included in b, then, a -4 (draw through the number axis).

2. cub= and because a is contained in cub, there is -2 If 1 (1-a) belongs to s, then 1 (1-1 (1-a)) belongs to s Simplification 1 (1-1 (1-a)) is 1 1 a.

2) 2 s, according to nature 2, there is.

1 (1-2) s, i.e., -1 s

1/[1-(-1)]∈s，1/2∈s

1/(1-1/2)∈s，2∈s

The other two elements are -1, 1 2

This question can be expanded into another sub-question, I don't know if you will understand, mention it again.

Verification: There are at least three different elements in the set.

If 1-1 a belongs to s, then 1 (1-(1-1 a)) belongs to s;

Simplification 1 (1-(1-1 a))=a;

So this set contains at least three elements: a, 1 (1-a, 1 a) (it can be proved that these three elements are unequal, because the discriminants of these three equations are less than 0 and have no real roots).

PS: The trick is to do more questions.

These questions are so easy, but it's too cumbersome to type with a pen, I can write them down on paper and show them to you!

Circle x 2 + y 2 = 25 center of the circle (0,0) radius r=5 The length of the chord is 8, then the distance from the center of the circle to the straight line = 3

1.Straight line x=-3 distance from the center of the circle to the straight line =3 half chord = 4 chord length is 8 so straight line x=-3 is a 2The other is set to have a slope of k

The equation for a straight line is y+2 3=kx+3k

The distance from the center of the circle to the straight line d=|3k-2/3|/(1+k^2)=3k=-8/9

The equation for a straight line is 8x-9y+18=0

Radius 5 chord length is 8 and half chord length is 4

The distance from the center of the circle to the chord is 3 (5 square - 4 square and open roots).

Straight line x=-3

The equation for a straight line is y+2 3 = k(x+3).

That is, y-kx +2 3-3k=0

If the radius of the circle is 5 and the length of the chord is 8, then the distance from the center of the circle to the straight line where the chord is located is sqrt(5 2-(8 2) 2) = 3

Whereas the point-to-straight distance is.

2/3-3k|sqrt(1+k 2) = 3, which is 9(1+k 2) = (2 3-3k) 29k 2 +9 = 4 9 - 4k + 9k 24k = 4 9-9 = -77 9

k = -77/36

Substituting the k-line equation yields!

Since the length of the truncated chord is 8, the distance from the center of the circle o to the straight line is d = root number [25-(8 2) 2] = 3

1) When the straight line is perpendicular to the x-axis, the condition is obviously satisfied, and the equation for the straight line is x=-3;

2) When the straight line is not perpendicular to the x-axis, let the linear line equation be y=k(x+3)-2 3, then |3k-2/3|/√(k^2+1)=3

The solution is k=-77 36

..Later, annoying, don't write.

One is x 3 and the other is x 3 with respect to the symmetrical straight line of op, which should be pretty easy to calculate.