Let vector a = (cosx / 2, SiNx / 2), vector b = (sin3x / 2, cos3x / 2), and the function f (x) = radical 2 * vector a + vector B. the minimum positive period, maximum and minimum value of function f (x) are obtained

Let vector a = (cosx / 2, SiNx / 2), vector b = (sin3x / 2, cos3x / 2), and the function f (x) = radical 2 * vector a + vector B. the minimum positive period, maximum and minimum value of function f (x) are obtained

If (x) = (x) = (2) vector a ＋ vector b ｜ = √ 2 √ (sin3x / 2 + cosx / 2) ^ 2 + (cos3x / 2 + SiNx / 2) ^ 2 + (cos3x / 2 + SiNx / 2) ^ 2 = √ 2 (2 + 2 (sin3x / 2 cosx / 2 + cos3x / 2 SiNx / 2 SiNx / 2) = 2 √ (2 + 2sin2x) = 2 √ (1 + sin2x) = 2 √ (sin ^ 2x + cos ^ 2x + 2cosxsinx = 2 | SiNx + cosx | 2 | SiNx + cosx | 2 | SiNx + cosx | 2 | 2 | 2 | | = 2 √ 2 | sin (x + π / 4) |

It is known that f (x) = 3 ^ x, u and V belong to R Proof: for any u, V, there is f (U). F (V) = f (U + V)

F (x) = 3 ^ x, u and V belong to R
So for any u, V, f (U). F (V) = 3 ^ u * 3 ^ v = 3 ^ (U + V) = f (U + V)

Given that a (1,2) B (2,1) O is the origin of coordinates, then the equation of the straight line where the angle AOB bisector is located is as follows:

Y=x

Mathematics problems in senior one (simple) Let m = {x | x ^ 2 ≥ x}, n = {x | 1 / x > 2} m ∪ n =?

M ∪ n = {x > 1 and X

If Tan (α + 8 π / 7) = a, then [sin (15 π / 7 + α) + 3cos (α - 13 π / 7)] / [sin (20 π / 7 - α) - cos (α + 22 π / 7)]=

tan（α+8π/7）=tan(α+π/7)=a
sin（15π/7+α）=sin（π/7+α）
cos（α-13π/7)=cos（13π/7-α)=cos(π+6π/7-α)=-cos(6π/7-α)=-cos(π-π/7-α)=cos(π/7+α)
sin（20π/7-α)=sin（6π/7-α)=sin（π-π/7-α)=sin(π/7+α)
cos（α+22π/7)=cos(α+8π/7)=-cos(π/7+α)
therefore
[ sin（15π/7+α）+3cos（α-13π/7）]/[sin（20π/7-α）-cos（α+22π/7）]=
[sin（π/7+α）+3cos(π/7+α)]/[sin(π/7+α)+cos(π/7+α)]
=1+2cos(π/7+α)/[sin(π/7+α)+cos(π/7+α)]
Because [sin (π / 7 + α) + cos (π / 7 + α)] / 2cos (π / 7 + α) = (1 / 2) * Tan (α + π / 7) + 1 / 2 = (a + 1) / 2
So 2cos (π / 7 + α) / [sin (π / 7 + α) + cos (π / 7 + α)] = 2 / (a + 1)
Therefore, the original formula = 1 + 2 / (a + 1)

Tan (x + 8) is known 7 π) = t, use t to denote sin (15) 7π+x)+3cos(x−13 7π) sin(20 7π−x)−cos(x+22 7π)．

∵tan（x+87π）=tan（x+π+π7）=tan（x+π7）=t，∴sin(157π+x)+3cos(x−137π)sin(207π−x)−cos(x+227π)=sin(x+π7+2π)+3cos(x+π7−2π)sin(3π−π7−x)−cos(x+π7+3π)=sin(x+π7)+3cos(x+π7)sin(π7+x...

Let Tan (α + 8 / 7 π) = a, prove sin (22 π / 7 + α) + 3cos (α - 20 π / 7) / sin (20 π / 7 - α) - cos (α + 22 π / 7) = a + 3 / - A

Tan (α + 8 π / 7) = Tan (π + α + α + π / 7) = Tan (α + π / 7), namely: Tan (α + π / 7) = asin (22 π / 7 + α) + 3cos (α - 20 π / 7) / sin (20 π / 7 - α) - cos (α + 22 π / 7 / 7) = (sin (3 π + π / 7 + 7 + α) + 3cos (3 π - 2 π / 7 - α)) / (sin (3 π - 2 π / 7 - α)) / (sin (3 π - 2 π / 7 - α) - cos (3 π + 7 - α) - cos (3 π + 3 π + 7 - α) - cos (3 π + 3 π + 7 - α) - cos (3 π + 3 π π / 7

If cos (11 π - 3) = P, then Tan (- 3)=__ Thank you very much!

cos(11π-3)=cos(π-3)=-cos3=p
So cos3 = - P
Because π / 2 ＜ 3 ＜ π
So cos3 < 0, SIN3 > 0
So p > 0, SIN3 = √ (1-cos ^ 23) = √ (1-p ^ 2)
So tan (- 3) = - tan3 = - SIN3 / cos3 = [√ (1-p ^ 2)] / P

It is known that sin (π / 2-B) * cos (a + b) - sin (π + b) * sin (a + b) = 3 / 5, where a ∈ (3 π / 2,2 π) finds Tan (π / 4-A / 2)

solution
sin(π/2-b)*cos(a+b)-sin(π+b)*sin(a+b)=3/5
Namely
cosbcos(a+b)+sinbsin(a+b)
=cos[b-(a+b)]
=cos(-a)
=cosa
∴cosa=3/5
∵a∈(3π/2,2π)
∴sina

It is known that f (α) = sin (π / 2 - α) cos (2 π - α) Tan (- α + 3 π) / Tan (π + α) sin (π / 2 + α) (1) Simplify f (α) (2) If α is the angle of the third quadrant and COS (α - 3 π / 2) = 1 / 5, find the value of F (α) (3) If α = - 1860 °, find the value of F (α)

1F (α) = sin (π / 2 - α) cos (2 π - α) Tan (- α + 3 π) / Tan (π + α) sin (π / 2 + α) = cos α * cos α (- Tan α) / (Tan α cos α) = - cos α 2 ∵ cos (α - 3 π / 2) = 1 / 5 ? sin α = 1 / 5, sin α = - 1 / 5 ∵ α is the angle of the third quadrant