Given the function f (x) = a {2Sin (x / 2) + SiNx} + B, find out (1) when a = 1, the monotone decreasing interval of F (x) please 3Q Given the function f (x) = a {2Sin (x / 2) + SiNx} + B, find (1) when a = 1, find the monotone decreasing interval of F (x) (2) when a < 0, the value range of F (x) on [0, π] is [2,3], and find the value of a, B

(1) when a = 1, f (x) = √ 2Sin (x-π / 4) + B + 1 when a = 1, then the monotone decreasing interval is [2K π + 3 π / 4,2k π + 7 π / 4]; (2) ∵ f (x) = a {2Sin (X / 2) + SiNx} + B = a × √ 2Sin (x-π / 4) + B + a = a × √ 2Sin (x-π / 4) + B + a ? x ∈ [0, π],, x-π / 4 ∈ [- π / 4,3 π / 4,3 π / 4]; (4) 9 (4) 9 (4) sin (x - π

How to simplify the function f (x) = 2Sin square X / 2 + SiNx + B and find its monotone decreasing interval

Revise the answer on the first floor
Using cosine multiple angle formula
After simplification, it is = SiNx cosx + B-1
=√ 2 * sin (x-45 °) + B-1 (coefficient is not √ 2 / 2)
The monotone decreasing interval is 2K * Pai + Pai / 2 < x-pai / 4 < 2K * Pai + 3pai / 2
That is 2K * Pai + 3pai / 4 < x < 2K * Pai + 7pai / 4
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Function SiNx / 2cos / 2 + cos ^ 2x / 2-2 Simplify to asin (Wx + φ) + B and write the period of FX

=1/2 *2sinx/2 *cosx/2 + cos^2x/2-2
=1/2 sinx + 1/2(cosx+1)-2
=1/2(sinx+cosx)-3/2
=√2/2sin（x+π/4）-3/2
The period is 2 π

(mathematics is not good, please) given the function y = (SiNx + COS) ^ 2 + 2cos ^ 2, find its decreasing interval, maximum and minimum

The function y = (SiNx + COS) ^ 2 + 2cos ^ 2 = sinx2 + 2sinxcosx + cosx2 + 2cosx2 = sinx2 + 2sinxcosx + 3cosx = [1-cos2x] / 2 + sin2x + [cos2x + 1] / 2 * 3 = co2x + sin2x + 2 = radical 2cos [x - π / 4] + 2 is obtained by combining the number forms: π / 2 + 2K π=

It is known that f (x) = 2Sin (x - π / 3) cos (x - π / 3) + 2 radical 3cos ^ 2 (x - π / 3) - radical 3. ① find the maximum value of F (x) and the corresponding value of X ② Write a symmetric axis equation

The reduction formula of F (x) = 2Sin (2x-2 π / 3) + 2 √ 3cos 2 (x - π / 3) -√ 3
=Sin (2x-2 π / 3) + √ 3cos (2x-2 π / 3) extracts the sine formula of two angular sum
=2sin(2x-π/3)
When 2x - π / 3 = (π / 2) + 2K π, i.e. x = (5 π / 12) + K π, the maximum value of F (x) is 2
When - π = (- 2x) π = (- 2x) + 2
Symmetry axis equation: 2x - π / 3 = (π / 2) + K π
Let k = 0 to get the equation of symmetry axis: x = 5 π / 12

The known function f (x) = log2[ 2sin（2x-π 3）]． (1) Find the definition domain of function; (2) Find the value range of X satisfying f (x) = 0

(1) In this paper, the definition domain of the function is (K π + π 6, K π + 23 π), K ∈ Z, K ∈ Z ′ K π + π 6 ＜ x ＜ K π + 23 π, K ∈ z.therefore, the definition domain of the function is (K π + π 6, K π + 23 π), K ∈ Z. (2) ∵ f (x) = 0, sin (2x - π 3) = 22 Κ 2x 2x π 3 = 3 = 22  2x2x π 3 = 3 = 3 = 22  2x π 3 = 3 = 3 = 22  2x 2x π 3 = 3 = 3 = 22  2x2k π + π 4 or

Let f (x) = log2 [radical 2Sin (- π X / 4 + π / 3) + 1] (1) Find the domain of definition (2) Finding monotone increasing interval (3) If the image of function y = g (x) and function f (x) are symmetric about x = 2 / 3, the analytic formula of G (x) is obtained

(1) : the definition domain only needs to satisfy 2Sin (- π X / 4 + π / 3) + 1 > 0, and then we can solve the range of X! The solution x is - 2 + 8K

Given cos (x + P / 4) = 4 / 5, find the value of sin2x-2sin ^ 2x / 1-tanx

(sin2x-2sin^2x)/(1-tanx)=[2sinx*(cosx-sinx)]/[1-sinx/cosx]=[2sinx*(cosx-sinx)]/[(cosx-sinx)/cosx]=2sinxcosx cos(pi/4+x)=3/5,√2/2cosx-√2/2sinx=3/5,cosx-sinx=(3*√2)/5 (cosx-sinx)^2=18/25=1-2sinxcosx,...

It is known that cos (45 + x) = 3 / 5 and 105

What grade is the classmate? Remind you, expand the known conditions, and then square can find sin2x, after you think about it! Come on!

（sin（180°+2x））/（1+cos2x）*（cos2x0/（cos（90°+x））=

（sin（180°+2x））/（1+cos2x）*（cos2x0/（cos（90°+2x））＝[-sin2x/(1+cos2x)][cos2x/(-sinx)]=(sin2xcos2x)/(sinx(1+cos2x))=(sin2xcos2x)/[2snxcos²x]=(sin2xcos2x)/sin2xcosx=cos2x/cosx