If the curve represented by the equation x ^ 2Sin α - y ^ 2cos α = 1 of X, y is an ellipse, then the center of the circle represented by equation (x + cos α) ^ 2 + (y + sin α) ^ 2 = 1 is located at?

If the curve represented by the equation x ^ 2Sin α - y ^ 2cos α = 1 of X, y is an ellipse, then the center of the circle represented by equation (x + cos α) ^ 2 + (y + sin α) ^ 2 = 1 is located at?

ellipse
So x and Y coefficients are positive
So sin α > 0, cos α < 0
Center of circle (- cos α, - sin α)
The abscissa is positive and the ordinate is negative
And (- cos α) 2 + (- sin α) 2 = 1
So the center of the circle is the part of the unit circle in the fourth quadrant

If α is an inner angle of a triangle and sin α + cos α = 1 / 5, then the equation x ^ 2Sin α - y ^ 2cos α = 1___ The answer is the ellipse with the focus on the y-axis. Why is the focus on the y-axis?

1/25=1+sin2a
sin2a-cosa
So 1 / Sina

Given that f (x) = [2cos 3 x + sin 2 (2 π - x) + sin (π / 2 + x) - 3] / 2 + 2cos 2 (π + x) + cos (- x), simplify as soon as possible

Finally, it is clear that the title is a fraction
f(x)=[2cos^3x+sin^2(2π-x)+sin(π/2+x)-3]/[2+2cos^2(π+x)+cos(-x)]
=(2cos^3x+sin^2x+cosx-3)/(2+2cos^2x+cosx)
=(2cos^3x+1-cos^2x+cosx-3)/(2+2cos^2x+cosx)
=(2cos^3x-cos^2x+cosx-2)/(2+2cos^2x+cosx)
=(cosx-1)(2x^2+cosx+2)/(2+2cos^2x+cosx)
=cosx-1

2 cos 3 θ + sin 2 (2 π - θ) + cos (- θ) simplification 2 cos 3 θ + sin 2 (2 π - θ) + cos (- θ) - 3 / 2 + 2cos 2 (π + θ) + cos (2 π - θ)

To use the induction formula cos (π + θ) = - cos θ, the original formula is simplified to 2cos 3 θ + cos 2 θ + 2cos θ - 1 / 2

Let f (θ) = 2cos3 θ - sin2 (θ + π) - 2cos (− θ - π) + 1 2 + 2cos2 (7 π + θ) + cos (− θ), find f (π) 3) Value of

f(θ)＝2cos3θ−sin2θ+2cosθ+1
2+2cos2θ+cosθ
=2cos3θ−(1−cos2θ)+2cosθ+1
2+2cos2θ+cosθ
=2cos3θ+cos2θ+2cosθ
2+2cos2θ+cosθ
=cosθ(2cos2θ+cosθ+2)
2cos2θ+cosθ+2＝cosθ，
∴f(π
3)=cosπ
3=1
2．

Let f (a) = [2cos3 power a + sin square (2 π - a) + sin (π / 2 + a) - 3] / [2 + 2cos square (π + a) + cos (- a), find the value of F (π / 3)]

f(A)=[2cos³A+sin²A+cosA-3]/[2+2cos²A+cosA]
sin(π/3)=√3/2
cos(π/3)=1/2
∴f(A)=-0.5

It is known that 0 ＜ α ＜ π, Tan α = - 2. (1) find the value of sin (α + π / 6); (2) find 2cos (π / 2 + α) - cos (π - α) sin (π / 2 - α) It is known that 0 ＜ α ＜ π, Tan α = - 2 (1) Find the value of sin (α + π / 6); (2) Calculate the value of 2cos (π / 2 + α) - cos (π - α) sin (π / 2 - α) - 3sin (π + α); （3）2sin²α-sinαcosα+cos²α

tanα=-2 -> sinα/cosα=-2 -> sinα=-2cosα
sinα^2+cosα^2=1 -> (-2cosα)^2+cosα^2=1 -> cosα^2=1/5
Because 0 < α < π, and Tan α = - 2
So alpha is the second quadrant angle
So sin α > 0, cos α < 0 > cos α = - √ 5 / 5, sin α = - 2cos α = 2 √ 5 / 5
(1)sin(α+π/6)=sinαcosπ/6+cosαsinπ/6=(2√15-√5)/10
(2)2cos(π/2+α)-cos(π-α)sin(π/2-α)-3sin(π+α)
=-2sinα+cosαcosα+3sinα
=sinα+cosα^2=(1+2√5)/5
(3)2sinα^2-sinαcosα+cos^α
=1+sinα^2-sinαcosα
=7/5

Known: Tan θ = 2, find the value of sin θ - cos θ / sin θ + 2cos θ

sinθ-cosθ/sinθ+2cosθ
=(tanθ-1)/(tanθ+2)
=(2-1)/(2+2)
=1/4

If Tan θ = 2, the value of (sin θ + 2cos θ) / (sin θ - cos θ) is

(sin θ + 2cos θ) / (sin θ - cos θ)
=(tanθ+2)/(tanθ-1)
=4

Evaluation: given sin alfa-2cos Alfa = 0, find sin (pat - alfa) cos (2 beats - alfa) sin [- Alfa + (3 beats / 2) / Tan (- Alfa -...) Evaluation: given sin alfa-2cos Alfa = 0, find the value of sin [- Alfa + (3 beats / 2) / Tan (- Alfa beat) sin (- pat AARF) cos (2 beats - alfa) sin

∵sinα-2cosα=0
∴sinα=2cosα ①
Then Tan α = sin α / cos α = 2
The original formula = sin (π - α) * cos (2 π - α) * sin (- α + 3 π / 2) / [Tan (- α - π) * sin (- π - α)]
=sinα*cosα*(-cosα)/(-tanα*sinα)
=(cos²α)/2
Sin? α + cos? α = 1? 2
From (1) to (2)
cos²α=1/5
Therefore, the original formula = (1 / 5) * (1 / 2) = 1 / 10