The values of sin9 ° cos15 ° - sin24 ° / sin9 ° sin15 ° + cos24 ° are___

The values of sin9 ° cos15 ° - sin24 ° / sin9 ° sin15 ° + cos24 ° are___

sin9°cos15°-sin24°/sin9°sin15°+cos24°=sin9°cos15°-sin(9°+15°)/sin9°sin15°+cos(9°+15°)=sin9°cos15°-sin9°cos15°-cos9°sin15°/sin9°sin15°+cos9°cos15°-sin9°sin15°=-cos9°sin15°/c...

(sin15cos9 ° - cos66 °) / (sin15 ° sin9 ° + sin66 °) can be simplified as an important process

Cos66 = sin24 sin24 = sin15 * cos9 + sin9 * cos15 sin15 * cos9-cos66 = - sin9 * cos15 sin66 = cos24 cos24 = cos15 * cos9-sin9 sin15 * sin9 + sin66 = cos15 * cos9-cos66) / (sin15 * sin9 + sn66) = - sin9cos15 / (cos15cos9) = - tan9

It is known that the minimum positive period of the function f (x) = 2cos2 (Wx / 2) + cos (Wx + π / 3) - 1 is π （1） Find the value of W （2） In the acute triangle ABC, a, B, C are the opposite sides of angles a, B and C respectively. Find the value range of F (a) （3） Under the condition of (2), if f (a) = - 3 / 2, C = 2, and the area of triangle ABC is 2 √ 3, find the value of A W>0

If it is, then: (1) f (x) = cos ω x + cos (ω x + π / 3 / 3) = (3 / 2) cos ω x + cos (ω x + π / 3) = (3 / 2) cos ω X - (√ 3 / 2) sin ω x = √ 3cos (ω x + π / 6), from the minimum positive period of F (x) is π, we can get ω = 2. (2) from (1) we know that f (x) = √ 3cos (2x + π / 6), \ff f (1) know that f (x) = √ 3cos (2x + π / 6), \f \ (1) know that f (x) (x) (x) (...)

It is known that the function f (x) = - 2 x power + B / 2 x + 1 power + A is odd function (1) The definition domain is monotone decreasing (2) if for any t belongs to R, the inequality f (square of t-2t) + F (2t square-k)

If f (0) = 0, and f (- x) = - f (x) = - f (x) = f (x) = f (x) = 0, and f (- x) = - f (x) f (x) = f (x) according to f (0) = 0, {2 ^ 0 + B} / {2 ^ (0 + 1) + a}, {- 1 + B} / {2 + 2 + a}, {1 + B} / {2 + 2 + a}, \\\\\\\\\\\\\\\\\\\

It is known that a, B are two adjacent intersection points of the line y = 0 and the function f (x) = 2cos2 (Wx) / 2 + cos (Wx + π / 3) - 1, and ab = π / 2, (1) find the value of W (2) in the acute angle triangle ABC, a, B, C are the opposite sides of angles a, B and C respectively. If f (a) = - 3 / 2, C = 3, the surface product of triangle ABC is 3 times the root sign 3, and then find the value of A

1.f（x）=cos(wx)+1+1/2coswx-√3/2sinwx-1
=3/2coswx-√3/2sinwx
=√3cos(wx+π/6)
∴T=2π/w
∴T/2=π/w=π/2
∴w=2
2. So f (x) = 1 + √ 3cos (2x + π / 6)
So f (a) = 1 + √ 3cos (2a + π / 6) = - 3 / 2
A=
By cosine theorem
a^2=b^2+c^2-2bccosA
S△ABC=b*csinA /2
The idea is like this. I have something to do now, and I will continue at night

It is known that the distance between the adjacent symmetry axes of the function f (x) = sin 4 power Wx + cos 4 power Wx is π / 2. Find the value of positive number W 2. Find the maximum value of the function g (x) = 2F (x) + sin square (x + π / 6) and the value of X when the maximum value is obtained

1. Because [(sinwx) ^ 2 + (coswx) ^ 2] ^ 2 = (sinwx) ^ 4 + (coswx) ^ 4 + 2 (sinwxcoswx) ^ 2
=1
That is, f (x) = (sinwx) ^ 4 + (coswx) ^ 4 = 1-2 (sinwxcoswx) ^ 2 = 1 - [(sin2wx) ^ 2] / 2
=1-(1-cos4wx)/4
=(cos4wx)/4+3/4
T / 2 = π / 2, t = 2 π / (4W) = π
Then w = 1 / 2
2. From (1), then G (x) = (cos2x) / 2 + 3 / 2 + [sin (x + π / 6) ^ 2]
=(cos2x)/2+3/2+[1-cos(2x+π/3)]/2
=2+[(cos2x)-cos(2x+π/3)]/2
=2+cos(2x-π/3)/2
Then G (x) max = 5 / 2, and COS (2x - π / 3) = 1
That is, 2x - π / 3 = 2K π, that is, x = k π + π / 6 K belongs to integers
Similarly, when G (x) min = 3 / 2, cos (2x - π / 3) = - 1
That is, 2x - π / 3 = 2K π + π, that is, x = k π + 2 π / 3 K belongs to integers

If the minimum positive period of the function y = 2Sin ω xcos ω x (ω > 0) is π, then f (x) = 2Sin (ω x + π) 2) A monotone increasing interval of is () A. [−π 2，π 2] B. [π 2，π] C. [π，3π 2] D. [0，π 2]

Because the minimum positive period of the function y = 2Sin ω xcos ω x (ω > 0) is π, that is, the minimum positive period of the function y = 12sin2 ω x is π, so 2 π 2 ω = π, so 2 ω = 1, and the function f (x) = 2Sin (x + π 2), because 2K π − π 2 ≤ x + π 2 ≤ 2K π + π 2 K ∈ Z, 2K π - π

Given the function f (x) = Cos4 power x-2sinxcosx-sin4 power X, when x belongs to [0,2 / π], find the minimum value of F (x) and the set of X when the minimum value is taken

f(x)=(cos²x+sin²x)(cos²x-sin²x)-sin2x
=cos2x-sin2x
=-√2sin(2x-π/4)
So sin (2x - π / 4) = 1 is the minimum
2x-π/4=2kπ+π/2
So the minimum value = - √ 2
x∈{x|x=kπ+3π/4}

F (x) = Cos4 power-2 roots 3sinxcosx-sin4 power 1. Find period 2. X belongs to [0, the school of half, f (x) min and the set with the minimum value 3. Increasing interval

f(x)=(cosx)^4-(sinx)^4-2sinxcosx
=[(sinx)^2+(cosx)^2][(cosx)^2-(sinx)^2]-2sinxcosx
=1*[(cosx)^2-(sinx)^2]-2sinxcosx
=cos2x-sin2x
=cos(2x+π/4)
So the minimum positive period is 2 π / 2 = π~

The minimum positive period of the function y = cos 4 power 3x-sin 4 power 3x is () A ∏/4 B ∏/3 C ∏ D 2∏

y=(cos3x)^4-(sin3x)^4
=[(cos3x)^2+(sin3x)^2]*[(cos3x)^2-(sin3x)^2]
=1*cos6x
=cos6x
The minimum positive period is 2pi / 6 = pi / 3
Choose B