There is a reward for answering well in math trigonometry

sin +sin 2 (sin +sin) 2=1 2......Let cos +cos = t then (cos +cos) 2=t 2......Get (sin +sin ) 2+(cos +cos ) 2=1 2+t 2 get sin 2+sin 2+2sin *sin +cos 2+cos 2+2cos *cos =1 2+t 2 sort out 2+2cos( -=1 2+t 2 t 2=3 2+2cos( -because -1 cos( -1 so 0 t 2 7 2 so -2 of 14 t 2 of 14 so -2 of 14 cos +cos 2 of 2 14 cos +cos 2 of 14

1.。Because of sin

a +cos

a =2/3

So the equation can be preceded by a root number 2, which becomes a sin (45...) that is twice the root number+a), and then easy to get: sin(45+a) = root number 2 3, less than 1 2, that is, 30 degrees, easy to get a is an obtuse angle, so the first question is b

2..Breaking down this cubic equation, it is easy to get the original question = 117 128

3.。According to the first condition and the second condition, the radian angle of this sector and its radius length can be set, and the relationship between the two equations can be listed, and in solving the system of equations, it can be obtained that the radian of the central angle of this circle is 2

4.。Let the central angle of this sector be a and the radius is r, so (a2)*

2 r = arc length, so, the arc length is ar, so, 2r + ar = 20 cm, so r = 20 (a+2), so, the area of the sector s = r 2

Multiply a 2 and express the radius r with a formula, and then use the inequality relationship to obtain that s is less than or equal to 25, and when a = 2, s is up to 25cm 2

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When =0°, x=r and y=0

Therefore, sin 0°=0 r=0

cos 0°=r/r=1

tan 0°=0/r=0

In fact, you just need to draw a unit circle in the center of the axis.

r is actually the radius of the circle, x is the abscissa of a point on the circle, and y is the ordinate of this point on the circle. is the size of the angle between the point and the axis origin and the x-axis.

Then sin 0°=y r=0 r=0

cos 0°=x/r=r/r=1

tan 0°=y/x=0/r=0

Do you know now? Triangular culverts are actually not difficult, learn hard.

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cosec=1/sin

cot=1/tan

sec=1/cos

This is the definition of these three functions, just substitute the original learning things into it, when =0°, x=r, y=0, this paragraph is about how to represent trigonometric functions in the coordinate system taught in high school, you can understand it by reading the book.

lz should be remembered.

tan=1/cot

sin=1/cosec

cos=1/sec

Convert the problem to an elementary function such as sin.

In the unit circle, sina = r y cosa=x r tana = y x

And sina = 1 csca = y r cosa = 1 seca = r x tana = 1 cota = x y

where sin is sin, cos is cosine, tan is tangent, sec is secant, csc is cosecant, and cot is cotangent.

Borrow a book and look at it from the beginning, don't rush it. In fact, the book says it very clearly. That's how I came over.

The first question is wrong.

f(t)=10-(√3)cos(πt/12)-sin(πt/12)=10-2

10-2[cos(π/6)·cos(πt/12)+sin(π/6)·sin(πt/12)]

10-2cos[(6)-(t 12)]=10-2cos[(t 12)-(6)] The second question and the first question are answered.

The first question is the answer to the wrong conclusion, the basis of the wrong answer, and the second question is self-understanding.

You have to give a range of angles A and B!!

t=2π/w

So w=4f(x)=tan(4x+4).

4x+π/4≠kπ+π/2

i.e. x≠k 4 + 16, and k is an integer.

f(a/2)=tan(2a+π/4)

tan2a+tan 4) (1-tan2a tan 4)=(tan 2a+1) (1-tan 2a)=3 to get tan2a=1 2