Then there is sin (π/4+α) cos (π/4+α) =1/6.

Sin(π/2+2α)= 1/3, which means cos2a= 1/3.

Then (tan2a) 2 = (sin2a) 2/(cos2a) 2 = (1-(cos2a) 2)/(cos2a) 2 = 8.

As cos2a & gt0, so 2a∈(-π/2+2kπ, π/2+2kπ), that is, a∈(-π/4+kπ, π/4+kπ).

Combining α∈(π/2, π), we can know that a∈(3π/4, π) is 2a∈(3π/2, 2π).

Tan2a=-2√2

tan4a=(2tan2a)/( 1-(tan2a)^2)=(4√2)/7

2. √ 3 (tana+tanb) = tanatanb-1to1-tanatanb =-√ 3 (tana+tanb).

Replace tanB+tanC+(√3)tanBtanC=√3 with (√ 3/3) (tanb+tanc) =1-tanb tanc.

tan(A+B)=(tanA+tanB)/( 1-tanA * tanB)=(tanA+tanB)/-√3(tanA+tanB)=-(√3/3)

Then A+B= 150 degrees.

tan(B+C)=(tanC+tanB)/( 1-tanC * tanB)=√3

Then B+C=60 degrees.

Combined with the properties of triangle, A+B+C= 180 degrees is obtained.

Solve three equations to get A= 120 degrees, B=30 degrees, and C=30 degrees.

3. Do this: let (dsinx-f cosx) 2.

Extend this formula to get d 2 sin 2 x-2 dfsinxcosx+f 2 cos 2 x.

Cos 2x is converted into 1-sinx 2 ... and substituted into the above formula to get (d 2-f 2) sin 2x-2 dfsinxcosx+f 2.

The equations d 2-f 2 = 2a, 2df = 2 √ 3 are obtained.

The solution is d = √ (3a) and f = √ a.

Therefore, f(x) can be expressed as (√ (3a) sinx-√ acosx) 2-a+b.

F (x) = 4a (√ 3/2sinx-(1/2) cosx) 2-a+b = 4a (sin (x+30 degrees)) 2-a+b can be obtained by using the auxiliary angle formula.

Therefore, the minimum value of f(x) in [0 degree, 90 degree] is f(0 degree or 90 degree) =b, and the maximum value is f(60 degree) = 3a+b.

Combining the range, we can see that B =-5 and A = 3.

Ask the first question

b-2c=(sinb-2cosb，4cosb+8sinb)

From the vertical lines of A and b-2c, we can see that 4cosasinb-8cosacosb+4cinacosb+8cinasinb = 0, and the product of quantities is 0.

Get 4sin(a+b)=8cos(a+b).

That is, sin(a+b)=2cos(a+b)

tan(a+b)= sin(a+b)/cos(a+b)= 2cos(a+b)/(cos(a+b))= 2

the second question

b+c=(sinb+cosb，4cosb-4sinb)

| b+c | =√(sinb+cosb)^2+(4cosb-4sinb)^2=√( 17(sin^2 x +cos^2 x)-30 sinbcosb)=√ 17- 15 sin2b

Because the range of 15sin2b is [- 15, 15].

So when 15sin2b is-15 |b+c|=4√2.

Third question

Calculate 4cosα * 4cosb-Sina * sinb =16cosacobb-sinasinb ...1.

Sinasinb= 16cosacosb is obtained from tanαtanβ= 16.

So 1 is 0.

So a//b

5 ... Let ω x+ψ = t.

Then f(t)=√3 sint-cost=-2cos(t+60 degrees).

Substitute ωx+ψ=t into the above formula to get f(x)=-2cos(ωx+ψ+60 degrees).

If f(x) is an even function, then ψ+60 degrees =k∏, combined with 0 < ψ < π, ψ= 120 degrees.

From the dyadic function y=f(x), we know that the distance between two adjacent symmetry axes of an image is π/2, and we know that T=π=2π/ω, ω=2.

So f(x)=2cos2x.

The first question f(π/8)=2cosπ/4=√2.

The decreasing interval of the second question f(x) is [kπ, π/2+kπ].

So the decreasing interval of g(x) [π/6+kπ, 2π/3+kπ]

The monotone interval of g(x) is actually adding π/6 to the monotone interval of f(x).

Some people who don't know QQ say that my QQ is 5 120 19720.