Senior 1 math trigonometric function questions, senior 1 math trigonometry questions

The numerator and denominator are multiplied by cos10° at the same time

sin10°×sin50°×sin70°

sin10°×sin50°×sin70°× cos10°/cos10°

sin10°× cos10°×sin50°×sin70°/cos10°

1/2) ×sin20°× sin50°× sin70°/cos10°

1/2) ×sin20°× sin50°× cos20°/cos10°

1/4) ×sin40°× sin50°/cos10°

1/4) ×sin40°× cos40°/cos10°

1/8) ×sin80°/cos10°

1/8) ×sin80°/sin80°

lg（1/1-cosa）=n

lg（1-cosa）=-n ①

lg(1+cosa) = m clump dust is missing.

Collapse Zen: lg(1-cosa)+lg(1+cosa)=m-nlg [1-(cosa) 2]=m-n

lg [(sina)^2]=m-n

2lgsina=m-n

lgsina = (m-n) 2 choose d

If I'm not mistaken, I chose B for this question

The process is as follows:...

sinx cosx =, let sinx+cosx=t, according to the auxiliary angle formula, t=[-root2, root2], f(x)=[t 2-1] 2(t+1)=(t-1) 2, choose b

It's equal to 0, and you know that the period of the sinusoidal function is 2, and then you can bend the negative one into a stupid and boring positive one.

sin-1071')*sin99'+sin(-171')*sin(-261')=sin9°*sin99°+sin(π+9°）*sin99°

sin99°*[sin9+sin(+9°)ridge]

1.Simplification: f(x)=2sinwxcoswx+2 3cos 2wx- 3=sin2wx+ 3cos2wx=2sin(2wx+ 3) minimum positive period t=(2)*2= =2 2w, w=1

2.From 1, we can see that f(x)=2sin(2x+3), because f(a)=2 3=2sin(2a+3), sin(2a+3)=1 3

1-2sin^2(2a+π/3)=cos(4a+2π/3)=cos(-4a-2π/3)=7/9

cos(-4a+π/3)=cos(-4a-2π/3+π)=-cos（-4a-2π/3）=-7/9

sin(5π/6-4a)=sin(-4a+π/3+π/2)=cos(-4a+π/3)=-7/9

OK, that's pretty much it, I hope it helps.

When x belongs to the interval [0, 2], 2x+ 6 belongs to [ 6,7 6] and sin2x+ 6 belongs to [-1, 2,1].

And because a is greater than 0, it is known that -2a+2a+b=-5 a+2a+b=1 gives a=2 b=-5

2) From (1) f(x) = 4sin(2x+6)-1

Answer: 2 |tana|

Process: Original.

1-(sin^2)a)/(1-sina)-√1-(sin^2)a)/(1+sina)

The first numerator denominator is multiplied by (1-sina) and the second numerator denominator is multiplied by (1+sina).

(1-(sin 2)a) 2sina] 1-(sin 2)a) (This step is a general score).

2sina 1-(sin 2)a) (this step is the numerator and denominator divided by (1-(sin 2)a)).

Because 1-(sin 2)a = (cos 2)a

So the original formula = 2sina cos 2)a= 2sina cosa|

Be careful not to drop the absolute value symbol).

Because 2 0 sina 1

So the original formula = 2sina cosa|= 2 |tana|Be what you want.

Note that you still can't drop the absolute value symbol).

(1) y=sinx---y=sin(x- 10) (left plus right minus) --y=sin(

2)y=sinx---y=

When solving the problem, pay attention to the change so that the change is only for the x itself, and it is important to see whether it is to expand or pan first.

Question 1: Shift y=sin x to the right by 10 to get y=sin(x- 10), and then elongate the abscissa of each point to 2 times to get y=sin(1 2x- 10).(Here 10 is not multiplied by 1 2).

Question 2: Extend y=sinx abscissa to double to get y=sin(, and then translate 10 to the right to get y=sin[

1 translates 10 to the right and is: y=sin(x- 10), and after elongation twice is: y=sin(1 2x- 10);

2 elongation by a factor of two: y=sin(1 2x), after translating 10 to the right, y=sin[1 2(x- 10)].

The two questions distinguish the concepts and make them, beginners are easy to confuse, and if you are familiar with these, it is pediatrics.

Solution: f(x) is the number of odd letters, and the line refers to the symmetry with respect to the origin of the defined domain, so a+b=0

f(-x)=-f(x), i.e., -xcos(-x)+c=-(xcosx+c), the solution yields c=0

So a+b+c=0

This question is wrong, the other two angles of a right triangle will not add up to more than 90°, so it is wrong.

sin80=sin(180-100)=sin100

Because tan100 = sin100 cos100, the square of sin100 plus the square of cos100 is equal to 1, naturally find sin100!!