Ask two high school trigonometric questions to solve two high school math trigonometric problems

1)=sin(a-π)/cos(π-a)*sin(a)/sin(a-π)cos(a)/sin(-a)

sin(a)/cos(a)*sin(a)/[-sin(a)]*cos(a)/[-sin(a)]

2) f（x)=sin(nπ-x）cos(nx+x)/cos[(n+1)π-x]×tan(x-nπ）cot（nπ/2+x)

When n is even: f(x)=sin(-x)cos(x) cos( -x)*tan(x)*cot(x).

sinx when n is odd: f(s) = sin( -x) cos( +x) cos(x)*tan(x)*cot( 2+x).

sin(x)[-cos(x)]/cos(x)*tan(x)*[tan(x)]

sin(x)[tan(x)]^2

So f(7 6) = -1 2 (n is even).

1。.Solve using an induction formula.

sin(a-5π）/cos(3π-a）×cos(π/2-a）/sin(a-3π）×cos(8π-a）/sin(-a-4π）

sin(a-π）/cos(π-a）×sina/sin(a-π）cos(-a）/sin(-a）

sina/(-cosa)×sina/(-sina）×cosa/(-sina）

2。.f（x)=sin(nπ-x）cos(nx+x)/cos[(n+1)π-x]×tan(x-nπ）cot（nπ/2+x)

Divide n into odd and even numbers, simplify them separately, and then evaluate them.

You should take a good look at the textbook...

,x [ 4, 3 4],2x+ 6 [2 3, 5 3],sin(2x+ 6) 1, 3 2],a>0, the maximum value of the function is 2a +2a+b, and the minimum value is -3a+2a+b

So 2a +2a+b= 3-1,- 3a+2a+b=-3, the solution is a=1, b= is not a rational number, rounded off.

When a<0, the maximum value of the function is -3a+2a+b, and the minimum value is 2a+2a+b

So- 3a+2a+b = 3-1, 2a +2a+b =-3, the solution is a=-1, b=1

2.The original question is like this; y=4cos^4(x)+4sin4(x)-3

y=4cos^4(x)+4sin^4(x)-3

1+cos2x)^2+(1-cos2x)^2-3

1+2*cos2x+(cos2x)^2+1-2*cos2x+(cos2x)^2-3

2*(cos2x)^2-1

cos4x, in the form of y=asin(x+), the result is y=sin(4x+2), when y<0, cos4x<0,2k + 2<4x<2k +3 2,k z

kπ/2+π/8<4x

(1) Function f(x)=-2asin[2x+(6)]+2a+b∵x∈[π/4,3π/4].∴2π/3≤2x+(π/6)≤5π/3.

=>-1≤sin[2x+(π/6)]≤3/2.When a 0, there should be 2a+b- 3a f(x) 4a+bThere should be (2-3) a+b=-3, 4a+b=3-1

a=7-4 are irrational numbers and are not in place. When a 0, there should be 4a+b f(x) (2- 3)a+bThere should be 4a+b=-3, (2-3)a+b=3-1

a=-1,b=1.Meets the question. In summary, a=-1, b=1

b) Function f(x)=4cos x+4sin x-3=4(cos x+sin x)-3=4-3=1Please take another look at the question.

(1) Left = 2sina*cosa*sina cosa+cos2a*cosa sina sina+2sina*cosa

2sin^2a+(1-2sin^2a)*cosa/sina+2sina*cosa

2sin^2a+cosa/sina-2sina*cosa+2sina*cosa

2sin^2a+cosa/sina

There's something wrong with the question, and you're taking a good look at it.

2）=sin(-11/6)+f(11/6-1)-1=sin(-11/6)+f(5/6)-1

sin(-11/6)+f(5/6-1)-1-1=-sin(11/6)+f(-1/6)-2=-sin(11/6)-sin(1/6)-2=-2sin1cos(5/6)-2

sin2a×tana=2sinacosa×sina／cosa=2sina²

Original = 2sina 2sina cosa + cos2a cota cos2a 1+2sina cosa + (1 2sin a) cota

cos2a 1+2sina cosa+cota 2sinacosa 2sin a cota is there a mistake f(x)=sin(x 1) 1(x>0).

Original formula = sin(-11 6) + sin(5 6) 1 2sin( 1 2)cos(4 3) 1

The first question is divided into situations.

At 1,01, the increasing interval is k(pai)-(pai) 12, and the decreasing interval is k(pai)+5(pai) 12

Two: Method one: known by tana=3, so, |sina|=3 10 * change sign 10, |cosa|=1 10 * change number 10

The answer is 1 Method 2: It is known from the debate book banquet tana=3, and the silver sin2a=9cos2a, that is, the manuscript -1 3sin2a+3cos2a=0

The function f(x) is an even function f(x)=f(-x), and the sum difference product formula gives 2cos( +3)sin(2x)=0 because 0 so = 6

f(x)=2cos2x=1 cos2x=1/2x∈[-2x∈[-2π,2π]

x=±π6,±5π/6

For any m r, f(x) is maximized between [m,m+1] at least once, and once at the minimum.

The minimum positive period t<1 so 2 k<1,k>2 ,k n,k=7 minimum positive period t=2 7

The center ordinate of symmetry is 0

Let y=0, then 3cos(7x+4)=0

7x+π/4=nπ+π2（n∈z)

x=（nπ+π4）/7（n∈z)

So the center of symmetry ((n + 4) 7,0) (n z) I hope you can understand it, I think it is already more detailed.........

The first way is to replace the root number 3 with 2*sin (80°-20°), and finally use the double angle formula to talk about it.

In the second way, tan(2) is obtained by tan(-1 2), tan by sin =3 5, 2), and finally tan(-2).

Only the method is provided with smart fiber, and the old wide touch is not counted.

1、π=

So 2> -2>1

sin(π-2)=sin2

sinx is incrementing at (0, 2).

So sin2>sin1

sin3sin1>sin3

i.e. b>a>c

2. Let the radius of the circle corresponding to the fan be r

The center angle is the arc length of l

s=lr/2=4

Perimeter = ar + 2r = 10

where there is l=ar

then l+2r=10

Two-way standing. r(10-2r)=8

(r-1)(r-4)=0

r = 1 or 4 then l = 8 or 2

a = l r1 8 or 2

1. Because = so 1 is approximately equal to 3. In the same way 2 is approximately equal to 2 3 3 is approximately equal to So from the image of the trigonometric function, we can see that c