First, the definition of acute angle trigonometric function
The trigonometric function of acute angle is a function with acute angle as the independent variable and this value as the function value. As shown in the figure: We call sine, cosine, tangent and cotangent of acute angle ∠ acute angle function.
The sine (sin), cosine (cos) and tangent (tan), cotangent (cot), secant (sec) and cotangent (csc) of acute angle A are all called acute trigonometric functions of angle A, and junior middle school mathematics mainly examines sine (sin), cosine (cos) and tangent (tan).
Sine is equal to the hypotenuse of the opposite side; Sina = account
Cosine (cos) is equal to the ratio of adjacent side to hypotenuse; cosA=b/c
Tangent (tan) is equal to the opposite side of the adjacent side; tanA=a/b
Cotangent is equal to the comparison of adjacent edges; cotA=b/a
Second, the acute angle formula of trigonometric function
Regarding the junior high school formula of trigonometric function, the most commonly used in the exam is the special value of special trigonometric degree. For example:
sin30 = 1/2
sin45 =√2/2
sin60 =√3/2
cos30 =√3/2
cos45 =√2/2
cos60 = 1/2
tan30 =√3/3
tan45 = 1
tan60 =√3[ 1]
cot30 =√3
cot45 = 1
cot60 =√3/3
Secondly, there is the formula of the sum of two angles, which is easy to use in the formula of trigonometric function in junior high school mathematics examination. Two-angle sum formula
sin(A+B)=sinAcosB+cosAsinB
sin(A-B)=sinAcosB-sinBcosA
cos(A+B)=cosAcosB-sinAsinB
cos(A-B)=cosAcosB+sinAsinB
tan(A+B)=(tanA+tanB)/( 1-tanA tanB)
tan(A-B)=(tanA-tanB)/( 1+tanA tanB)
ctg(A+B)=(ctgActgB- 1)/(ctg B+ctgA)
ctg(A-B)=(ctgActgB+ 1)/(ctg b-ctgA)
In addition to the promotion of trigonometric functions, there are also half-angle formulas and differential product formulas used in multiple-choice questions in junior high schools. So students still have to master it.
half-angle formula
sin(A/2)=√(( 1-cosA)/2)sin(A/2)=-√(( 1-cosA)/2)
cos(A/2)=√(( 1+cosA)/2)cos(A/2)=-√(( 1+cosA)/2)
tan(A/2)=√(( 1-cosA)/(( 1+cosA))
tan(A/2)=-√(( 1-cosA)/(( 1+cosA))
ctg(A/2)=√(( 1+cosA)/(( 1-cosA))
ctg(A/2)=-√(( 1+cosA)/(( 1-cosA))
Sum difference product
2 sinA COSB = SIN(A+B)+SIN(A-B)2 COSA sinB = SIN(A+B)-SIN(A-B)2 COSA COSB = COS(A+B)-SIN(A-B)-2 sinA sinB = COS(A+B)-COS(A-B)sinA+sinB = 2 SIN((A+B)/ 2 COS((A-B)/2 COSA+COSB = 2 COS((A+B)/2)SIN((A-B)/2)TANA+TANB = SIN(A+B)/COSA COSB TANA-TANB =
Fourth, the comprehensive application of acute angle trigonometric function
It is known that the image of linear function y=-2x+ 10 and the image of inverse proportional function y = k/x (k > 0) intersect at two points A and B (A is on the right side of B).
(1) When a (4 4,2), find the analytical expression of the inverse proportional function and the coordinates of point B;
(2) Under the condition of (1), whether there is a point p on the other branch of the inverse proportional function image, and if there is a right triangle with △PAB as the right, find out the coordinates of all qualified points p; If it does not exist, please explain why.
(3) When A(a, -2a+ 10) and B(b, -2b+ 10), the straight line OA intersects with another branch of the inverse proportional function image at another point C, and connects BC and Y axis at point D. If BC/BD=5/2, find the △ABC.
Test center:
Inverse proportional function synthesis problem; Solving the resolution function by undetermined coefficient method; Intersection of inverse proportional function and linear function; Similar triangles's judgment and nature.
Answer:
Solution: (1) Substitute A (4 4,2) into y=k/x to get K = 4× 2 = 8.
The analytical formula of inverse proportional function is y = 8/x.
Solve equation y = 2x+ 10.
Y = 8/x，x = 1y = 8。
Or x = 4 y = 2,
∴ The coordinate of point B is (1, 8);
(2)① If BAP = 90,
Let point A be AH⊥OE in H, and let the intersection of AP and X axis be M, as shown in figure 1.
For y=-2x+ 10,
When y=0, -2x+ 10=0, x=5,
Point e (5 5,0), OE = 5.
∵A(4,2),∴OH=4,AH=2，
∴HE=5-4= 1.
∵AH⊥OE,∴∠AHM=∠AHE=90。
∫∠BAP = 90，
∴∠AME+∠AEM=90，∠AME+∠MAH=90，
∴∠MAH=∠AEM，
∴△AHM∽△EHA，
∴AH/EH=MH/AH，
∴2/ 1=MH/2，
∴MH=4，
∴M(0,0)，
The analytical formula of straight AP can be set to y=mx.
Then there is 4m=2, and the solution is m= 1/2.
∴ The analytical formula of linear AP is y= 1/2x,
Solve equation y = 1/2x,
Y = 8/x，x = 4 y = 2。
Or x = 4 y=？ 2,
The coordinates of point P are (-4, -2).
② If ∠ ABP = 90,
Similarly, the coordinates of point P are (-16,-1/2).
To sum up, the coordinates of qualified point P are (-4, -2), (-16,-1/2);
(3) The crossing point B is the BS⊥y axis in S, and the crossing point C is the CT⊥y axis in T, connecting OB, as shown in Figure 2.
Then there are bs∑CT, ∴△CTD∽△BSD,
∴CD/BD=CT/BS.
∫BC/BD = 5/2，
∴CT/BS=CD/BD=3/2.
∫A(A，-2a+ 10)，B(b，-2b+ 10)，
∴C(-a,2a- 10),CT=a,BS=b，
∴a/b=3/2
B = 2/3a。
∵A(a, -2a+ 10) and B(b, -2b+ 10) are both on the image of inverse proportional function y=k/x,
∴a(-2a+ 10)=b(-2b+ 10)，
∴a(-2a+ 10)=2/3
a(-2×2/3a+ 10)。
∵a≠0，
∴-2a+ 10=2/3
(-2×2/3a+ 10)，
Solution: A = 3.
∴A(3,4),B(2,6),C(-3,-4).
Let the analytical formula of BC line be y=px+q,
Then 2p+q = 6.
3p+q=？ 4,
Solution: p = 2q = 2,
The analytical formula of line BC is y = 2x+2.
When x=0 and y=2, then points d (0 0,2), OD=2,
∴S△COB=S△ODC+S△ODB= 1/2
ODCT+ 1/2 odbs = 1/2×2×3+ 1/2×2×2 = 5。
OA = OC，
∴S△AOB=S△COB，
∴ s △ ABC = 2s △ Cob = 10。 The above is a summary of the knowledge points of acute trigonometric function in junior high school mathematics. Bian Xiao suggested that students continue to browse the special summaries of junior high school mathematics knowledge points. For students who want to get high-quality math learning resources and skills by attending junior high school math cram school, so as to improve their grades, Only New Curriculum recommends the following courses:
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