In the trigonometric function line problem, if the terminal edge of any angle α intersects the unit circle at point P (x, y), why is it defined by trigonometric function that sin α = y, cos α = x

In the trigonometric function line problem, if the terminal edge of any angle α intersects the unit circle at point P (x, y), why is it defined by trigonometric function that sin α = y, cos α = x

Because the radius of unit circle is 1, sin α = Y / x = Y / 1 = y is defined by trigonometric function. Similarly, cos α = X

Knowing the trigonometric function of the two corners of a triangle and the length of an edge between the two corners, how to find the area of the triangle? Is there a formula?

There is only one formula s = 1 / 2absina a, where B is the angle of the triangle and a is the angle between the two sides

Find angle by known trigonometric function X1 and X2 are two of the equations x ^ 2-xsin (π / 5) + cos (4 π / 5) = 0. Find the value of arctanx1 + arctanx2

Then Tan (arctanx1 + arctanx2) = {[Tan (arctanx1)] + [Tan (arctanx2)] / [1-tan (arctanx1) * Tan (arctanx2)] = = (x1 + x2) / (1-x1 * x2) = sin (π / 5) / [1-cos (4 π / 5)]

Given Tana = - 3, and a is the angle of the second quadrant, find a to get other trigonometric function values

Because a is the second quadrant angle, sina is negative and cosa is positive. Tana = Sina / cosa = - 3
Sina = - 3 plant 10 / 10, cosa = plant 10 / 10. (factory representative root)

Given Tana = - radical 5, and a is the second quadrant angle, find the other five trigonometric function values of angle A

Sin α = root 5 / root 6, cos α = - 6 / 6, cot = minus 5 / 5, arcsin α = root 6 / root 5, arccos α = - negative root 6

If a is the angle of the second quadrant, Tana = - 1 / 2, then cos=

A is the angle of the second quadrant
tana=-1/2
Then cos = - 1 / root sign (1 + Tan ^ 2 α) = - 1 / root sign (1 + (- 1 / 2) ^ 2) = - 1 / root sign (5 / 4) = - 2 root sign 5 / 5

A problem about symmetric center of trigonometric function The symmetry center of the fourth power of y = SiNx + the fourth power of cosx is______

The original analytic formula is equivalent to
(the 2nd power of SiNx + the 2nd power of cosx) ^ 2 - 2sinx to the 2nd power of cosx
The second power of cosx reduced to 1-2 SiNx
It is obtained from the simplification of the formula of double angle
Square of 1-sin2x
The formula of double angle is reduced to
3\4+cos4x \4
According to the properties of trigonometric function, its symmetry center is (4x = k π + π - 2)
(K π \ \ 4 + π \ \ 8,3 \ \ 4) k is a real integer
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② If cos (α is known to be the third quadrant angle, f (α) = sin (π - α) cos (2 π - α) Tan (- α - π) / Tan (- α) sin (- π - α), find the value of F (α) by simplifying f (α) ② calculating the value of F (α) if cos (α - 3 / 2 π) = 1 / 5, and ③ if α = - 1860 °

f(α)=sin(π-α)cos(2π-α)tan(-α-π)/tan（-α）sin（-π-α）
=sinα * cosα * (-tanα) /[(-tanα)*sinα]
=cosα
cos(α-3/2 π)=-cosα=1/5
f(α)=cosα=-1/5
f(-1860°)=cos(-1860°)=cos(1860°）=cos(10*180°+60）=cos60°=1/2

Let tan2 θ = - 2 radical sign 2,2 θ∈ (π / 2, π) find (2cos ^ 2 θ - 2-sin θ - 1) / (sin θ + cos θ)

In this paper, the author [2cos (θ / 2) - sin [1] / (sinθ + cos θ) tan2 θ = 2tan θ / (1-tan 2 θ) = 2 √ 2tan θ = ((2) (1-tan 2) θ) = 2 √ 2tan θ = ((2) (1-tan 2) (1-tan 2) ((2) Tan (2) ta + Tan θ (2) (1-tan (2) ((2) n (3) / (2 √ 2) θ ∈ (π / 2, π) (2, 2, π) (2, π) (π (π / 2, π) (π) (π / 2, π) (π) ta

If sin α + 2cos α = root 10 / 2, tan2 α =?

sinα+2cosα=√10/2
The square of both sides can be obtained as follows:
∴sin²α+4cos²α+4sinαcosα=5/2
∴【sin²α+4sinαcosα+4cos²α】/(sin²α+cos²α）=5/2
The left numerator and denominator is divided by cos 2 α
【tan²α+4tanα+4】/(tan²α+1）=5/2
The results are as follows: 1
3tan²α-8tanα-3=0
The solution is as follows:
Tan α = - 1 / 3 or 3
tan2α=2tanα/【1-tan²α】
When Tan α = - 1 / 3, tan2 α = - 3 / 4
When Tan α = 3, tan2 α = - 3 / 4
To sum up, tan2 α = - 3 / 4