Let f (x) = a × B, a = (2Sin (π / 4), cos2x), B = (sin (π / 4 + x), - √ 3 1. Find f (x) analytic expression 2. Find f (x) period and monotone increasing interval

Let f (x) = a × B, a = (2Sin (π / 4), cos2x), B = (sin (π / 4 + x), - √ 3 1. Find f (x) analytic expression 2. Find f (x) period and monotone increasing interval

0. 0 problem error can only tell you the solution
f（x）＝a×b
a=（sinA ,cosB） b=（cosA ,sinB）
From above we can get f (x) = sinacosb + cosasinb = sin (a + b)
Or when
a=（cosA ,sinB） b=（cosA ,sinB）
From the above we can get f (x) = cosacosa + sinbsinb = cos (a-b) 0.0 sin2a = 2sinacosa cos2a = 1-2sinasina = 2cosacosa-1 = cosacosa Sina

The maximum value of F (x) = 2Sin (2x + π / 6) + sin (2x - π / 6) + cos2x + A is 1, (1) find a, (2) find the set of X that holds f (x) > = 0, (3) explain how the image of F (x) is transformed by y = SiNx
The maximum value of F (x) = 2Sin (2x + π / 6) + sin (2x - π / 6) + cos2x + A is 1, (1) find a, (2) find the set of X that makes f (x) > = 0, (3) show how the image of F (x) is transformed by y = SiNx, (2) find the set of X that makes f (x) > = 0, (3) show how the image of F (x) is transformed by y = SiNx

f(x)=2(sin2xcosπ/6+cos2xsinπ/6)+sin2xcosπ/6cos-cos2xsinπ/6+cos2x+a
=3sin(2x+π/6)+a
The maximum value is 1, then a = - 2

How to calculate 2Sin (x - π / 4) sin (x + π / 4)?
How to get (SiNx cosx) (SiNx + cosx)? Is it to separate sin (x - π / 4) and sin (x + π / 4 and multiply them by 2? Is there no simple method

Solution
2sin(x-π/4)sin(x+π/4)
=2sin(x-π/4)sin[π/2+(x-π/4)]
=2sin(x-π/4)cos(x-π/4)
=sin[2(x-π/4)]
=sin(2x-π/2)
=sin[-(π/2-2x)]
=-sin(π/2-2x)
=-cos2x
=-(cos²x-sin²x)
=sin²x-cos²x
=(sinx-cosx)(sinx+cosx)