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Anonymous users2024-02-08Right triangle definition.
It has six basic functions (elementary basic representations): a table of trigonometric numerical functions (r, y, and x. In the planar Cartesian coordinate system xoy, a ray op is drawn from the point o, let the rotation angle be , let op=r, and the coordinates of the p point are (x,y) with the sine function sin =y r sine (sin):
Opposite side ratio of angle hypotenuse cosine function cos = x r cosine (cos): ratio of the adjacent edge of the angle hypotenuse tangent function tan = y x tangent (tan): the opposite side of the angle is relative to the side cotangent function cot = x y cotent (cot):
The adjacent edge of the angle is compared to the opposite edge sec = r x secant(sec): the hypotenuse of the angle is the adjacent edge cosecant function csc = r y cosecant (csc): the hypotenuse of the angle is compared to the opposite edge and two functions that are not commonly used and tend to be obsolete
Positive vector function versin =1-cos cos covector function covers =1-sin sin , cos , tan definition domain: sin defines the domain infinity, the value range [-1,1] cos defines the domain infinity, and the value range [-1,1] tan defines the domain (- 2+k , 2+k), k belongs to the integer and the value range is infinite.
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Anonymous users2024-02-07Sine sin a is the opposite side of the angle compared to the hypotenuse.
Cosine cos a is the angular adjacent edge than the hypotenuse.
tan a is the opposite side next to the next side.
sin30, 45, 60 degrees correspond to 1 2, 2 points of the root number 2, 2 points of the root number 3, respectively
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Anonymous users2024-02-06Sine sina is next to the branches.
The horns are thick on the side of the ant. Compare. Hypotenuse.
Cosine cosa
It is. Corner adjacent edges. Compare. Hypotenuse.
tan a is the opposite side next to the next side.
SIN 3045, 60 degrees.
The corresponding ones are 1 2
2 points of the root of the oak No. 2, 2 points of the root No. 3
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Anonymous users2024-02-05Solution: Let the vertical feet be on respectively.
qa=2qb=11
Because mon = 60 degrees, oca = 30 degrees.
bq=1 2cq so cq=22 ac=24 in the rt triangle oca.
Because OCA = 30 degrees. So OA=1 2QC
Because OA2+AC2=OC2
Let OA be X, then the original formula is.
x^2+24^2=4x^2
x^2=192
In the RT triangle OQA.
oq^2=oa^2+aq^2
So oq=14
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Anonymous users2024-02-04Didn't your high school teacher teach you such a simple question?
Or is it your homework?
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Anonymous users2024-02-03y^2/4=(sina)^2 * cosa)^2/2 * cosa)^2/2
From the mean inequality: (xyz) (1 3)=<(x+y+z) 3:
y^2/4)^(1/3)=<[(sina)^2+(cosa)^2/2+(cosa)^2/2]/3=1/3
Therefore, the maximum value is: root number (4 27).
At this point, it should be satisfied: (sina) 2=(cosa) 2 2, omitted.
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Anonymous users2024-02-02y=4sinxcosx=2sin2x, while y=(root number 3)sin2x-cos2x
2sin(2x- 6)=2sin2(x- 12), so just shift the image of the function y=4sinxcosx to the right by 12.
Using sina + sinb = 2 sin((a+b) 2)cos((a-b) 2
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