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.