Trigonometric function: cos (α - β / 2) = 1 / 9, sin (α / 2 - β) = 2 / 3, 0

Trigonometric function: cos (α - β / 2) = 1 / 9, sin (α / 2 - β) = 2 / 3, 0

Because 0

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?

Because the curve represented by X? Sin α - y? Cos α = 1 is an ellipse
Therefore, sin α > 0, cos α < 0
Alpha is in the second quadrant
The center of the circle represented by (x + cos α) 2 + (y + sin α) 2 = 1 is
（-cosα,-sinα）
Because - cos α > 0, - sin α < 0
The center of the circle is in the fourth quadrant

Let θ be an inner angle of △ ABC, and sin θ + cos θ = 7 Then x2sin θ - y2cos θ = 1 represents () A. Ellipse with focus on X-axis B. Ellipse with focus on y-axis C. Hyperbola with focus on X-axis D. Hyperbola with focus on y-axis

Because θ∈ (0, π), and sin θ + cos θ = 7
13, squared, 2Sin θ cos θ = - 120
169＜0，
So, theta ∈ (π)
2, π), and sin θ > 0, and cos θ < 0, and | sin θ | > cos θ|,
Thus, X 2Sin θ - y 2cos θ = 1 represents the ellipse with focus on the Y axis
Therefore, B

If α is an inner angle of the triangle and sin α - cos α = 1 / 5, then α is? A. Arcsin4 / 5 B. π - arcsin4 / 5 C. π + arcsin (- 4 / 5) d. uncertain

Option a
The reasons are as follows.
The square of sin α - cos α = 1 / 5 is used to calculate 2Sin α * cos α = 24 / 25,
So sin α * cos α = 12 / 25,
Because α is an inner angle of the triangle and sin α * cos α = 12 / 25,
So alpha is an acute angle,
Then square sin α + cos α to obtain 1 + 2Sin α * cos α = 49 / 25
So sin α + cos α = 7 / 5 and simultaneous sin α - cos α = 1 / 5
Sin α = 4 / 5, so choose a

Simplify f (x) = (sin α + cos α) 2 + 2cos α, First reduce the order, and then use the unified formula

F (x) = 1 + 2Sin α cos α + 2cos α = 1 + sin2 α + 2cos α can only be simplified. Is there a problem with the problem

F (&) = 2cos3 & + sin 2 (2 π - &) + cos (2 π -- -) - 3 / 2 + 2cos2 (π + &) + cos (- &), find the value of F (π / 3) This is a higher mathematical problem. It is the induction formula of sine and cosine

The formula: F (x) = 2cos3x + sin (4 π - 2x) + cos (2 π - x) - 3 / 2 + 2cos2 (π + x) + cos (- x) to sort out the original formula: F (x) = 2cos3x + sin (4 π - 2x) + cos (2 π - x) - 3 / 2 + 2cos (2 π + 2x) + cos (- x) = 2cos3x + sin (- 2x) + cos (- x) - 1.5 + 2cos2x + cos (- x) = 2cos3x-sin2x + cosx-1 + cosx-1-1-1-1.5 + 2cos2x + cos (- x) = 2cos3x-sin2x + cosx-1 + cosx-1 + cosx-1 + cosx-1 + cosx-1 + cosx 5 + 2cos2x + cosx = 2cos3x-sin2x + 2cosx-1.5 + 2cos2x substituting x = π / 3, The results show that f (x) = 2cos π - sin (2 π / 3) + 2cos (π / 3) - 1.5 + 2cos (2 π / 3) = 0 - (2 / 2 root 3) + 1-1.5-1 = - [(radical 3) + 3] / 2

The second power of [cos α - 1] and the second power of sin α = 2-2cos α

〔cosα-1〕²+sin²α
=cos²α-2cosα+1+sin²α
=cos²α+sin²α-2cosα+1
=1-2cosα+1
=2-2cosα

Let Tan α = 1 2, then sin α + 2cos α sinα−cosα=______ .

∵tanα＝1
2，
∴sinα+2cosα
sinα−cosα
=tanα+2
tanα−1
=1
2+2
One
2−1
=-5．
So the answer is: - 5

Sin θ + 2cos θ is known Sin θ − cos θ = 3 （1）tanθ；  （2）sinθ•cosθ．

(1) From the known: sin α + 2cos θ = 3 (sin α - cos α), sin θ = 5
2 cos θ, so tan θ = 5
2．
(2) From (1) sin θ = 5
And cosin 2, cos 2 = 1
Twenty-nine
∴sinαcosα=5
2cos2α=5
2×4
29=10
29… (10 points)

Given Tan θ = 1, find sin θ cos θ - Sin ^ 2 θ - 2cos ^ 2 θ, and 1 / sin ^ 2 θ - sin θ cos θ - cos ^ 2 θ

Tan θ = 1 -- > θ = k * PI + pi / 4,2 θ = 2K * PI + pi / 21. Sin θ * cos θ = sin (2 θ) / 2 = 1 / 2; sin ^ 2 θ = (1 - Cos2 θ) / 2 = 1 / 2; Cos ^ 2 θ = (1 + Cos2 θ) / 2 = 1 / 2, so sin θ cos θ - Sin ^ 2 θ - 2cos ^ 2 θ = 1 / 2 / (1 / 2 - 2 * 1 / 2) = -