Example 1 Known 0 ≤ α≤ 90. (1) Prove: sin2α+cos2α =1;
(2) Prove: sinα+cosα≥ 1, and discuss under what circumstances the equal sign holds;
(3) Given sinα+cosα= 1, find the value of sinα+cosα.
Proof (1) as shown in figure 6- 1, when 0
When α = 0, sinα=0, cos α =1; When α = 90, SIN α = 1 and COS α = 0. So it still exists in both cases.
sin2α+cos2α= 1。
(2) As shown in Figure 6- 1, when 0
When α = 0, sin α+cos α = 0+1=1; When α = 90, sinα+cosα= 1+0= 1. So when 0 ≤ α≤ 90, there is always.
sinα+cosα≥ 1，
The equal sign holds if and only if α = 0 or 90.
(3) Since Sina +cosα= 1 is known. According to (2), α = 0 or α = 90, so there is always.
sin3α+cos3α= 1。
Example 2 Verification: For 0 ≤ α≤ 90,
Prove that 1 is shown in Figure 6- 1, and let BC = A, AC = B, AB = C. It is composed of acute trigonometric functions.
When α = 0 or α = 90, it is easy to verify that the above equation still holds.
Syndrome method 2
The first method is based on the definition of acute trigonometric function; Proof 2 uses the formula sin2α+cos2α= 1.
It is proved that a trigonometric identity is established and the formula at the left (right) end of the equal sign can be transformed. If you get the formula of the right (left) end of the equal sign, prove the original identity. Generally, complicated formulas can be converted, and the left and right formulas of equal sign can be converted. If you get the same formula, prove the original identity.
Tangent and cotangent of 1.2
Prove when 0 (1)
When α = 0, tgα=0, sinα=0, cosα= 1. So there is still tgα=
(2)α must satisfy the inequality:
0 & ltα& lt； 90 .
As shown in Figure 6-2,
So tgα ctgα = 1.
Example 2: The acute angle α is known, and tgα is the root of equation x2-2x-3=0. Find it.
The two roots of the solution 1:x2-2x-3 = 0 are 3 and-1. Here it can only be tgα=3.
As shown in Figure 6-3, since tgα=3, BC = 3 and AC = 1 can be set, therefore,
Solution 2 tgα=3, divide the original molecule and denominator by cos2α, and get
Proof 1 As shown in Figure 6-2, assuming BC = A, AC = B, AB = C, then
So the original formula holds.
Prove the left end of the second equation
Comment on α ≠ 0,90 here.
How to understand the concept of acute trigonometric function
A: The definition of acute triangle function given in the current junior high school geometry textbook is based on the basic fact that when the acute angle is fixed in a right triangle, the ratio of its opposite side, adjacent side and hypotenuse is a fixed value.
In this regard, let's take a look at the figure 1. The right triangles AB 1C 1, AB2C2, AB3C3, … in the figure all have equal acute angles A, that is, the acute angle A takes a fixed value. As shown in the figure, the equal acute angles in many right-angled triangles are superimposed together, and one right-angled side falls on the same straight line, then the hypotenuse must fall on the other.
B 1C 1‖B2C2‖B3C3‖…,
∫△ab 1c 1∽△ab2c 2∽△ab3c 3 ∽…,
Therefore, in these right triangles, the ratio of the opposite side to the hypotenuse of ∠A is a constant value.
Similarly, from the knowledge of "similarity", we can know that in these right-angled triangles, the ratio of the opposite side to the adjacent side of ∠A and the ratio of the adjacent side to the hypotenuse of ∠A are all constant values.
Thus, in △ABC, ∠C is a right angle, and we call the ratio of the opposite side to the hypotenuse of acute angle A ∠A sine, the ratio of the adjacent side to the hypotenuse of sinA acute angle A ∠A cosine, the ratio of the opposite side to the adjacent side of cosA acute angle A ∠A tangent, and the ratio of the adjacent side to the opposite side of tgA acute angle A ∠.
To deeply understand the definition of acute trigonometric function, we should pay attention to the following points:
The acute trigonometric function of (1) angle A has nothing to do with the size of the triangle, that is, the side length.
As long as the angle a is determined, the four ratios are determined accordingly; When the angle α changes, the four ratios change accordingly. This reflects the characteristics of the function, and the acute trigonometric function is also a function, in which the angle A is the independent variable, and for each certain angle A, the above four ratios have unique definite values corresponding to it. Therefore, the acute trigonometric function is a function with the angle as the independent variable and the ratio as the function value.
(2) To accurately understand the definition of acute trigonometric functions, we should memorize how each acute trigonometric function is defined and the ratio of which side of the angle to which side; When applying the definition, we should pay attention to distinguish which side is the opposite side of the angle, which side is the adjacent side of the angle and which side is the hypotenuse.
[Example] Find the values of sind and tge in Figure 2.
(3) "Sina" is a complete symbol.
Integer symbols cannot be regarded as the product of sin and a, and the "sin" leaving the angle A is meaningless, as are the other three cosA, tgA, ctgA, etc. So you can't separate "sin" from "a" when writing.
The definition of acute trigonometric function combines shape and number, and is observed from the mutual relationship of things. For a right triangle, we don't look at its angle and its edge in isolation, but grasp the relationship between them, thus opening our minds and laying the foundation for further research. It is not difficult to see from the derivation of the definition that the acute trigonometric function is the result of the perfect combination of number (ratio) and shape (angle A), and students should understand and master the ideas of this kind of research problems well in their study.
calculate
answer the question
3. In Rt△ABC, ∠ c = 90. If Sina is a root of the equation 5x2-14x+8=0, find Sina, tgA.
Q is an angle of a triangle. If the equation10x2-(10cosq) x-3cosq+4 = 0 has two equal real roots, find tgq.
answer
3. Solution: ∵sinA is a root of the equation 5x2- 14x+8=0.
Then 5sin2A- 14sinA+8=0.
4. Solution: ∫ 100 cos2q-40(4-3 cosq)= 0.
That is 5cos2q+6cosq-8=0.