F (x) = asinx bcosx (a is not equal to 0) for any real number, f (π / 4 + x) = f (π / 4-x) holds, let Tan α = 2A / B, (α ∈ (0, π)) find Pay attention to distinguish between a and α mentioned in the following topic. What is required is α, which is represented by inverse triangle,

F (x) = asinx bcosx (a is not equal to 0) for any real number, f (π / 4 + x) = f (π / 4-x) holds, let Tan α = 2A / B, (α ∈ (0, π)) find Pay attention to distinguish between a and α mentioned in the following topic. What is required is α, which is represented by inverse triangle,

f(π/4+x)=f(π/4-x),
Let t = π / 4-x, then: x = t + π / 4,
So f (π / 2 + T) = f (T),
That is, f (x) = f (π / 2 + x)
F (x) = asinx bcosx,
So asinx bcosx = asin (π / 2 + x) - bcos (π / 2 + x) = acosx + bsinx,
(a-b)(sinx+cosx)=0,
Because x is any real number, SiNx + cosx is not constant = 0,
So a = B
So tan α = 2A / b = 2,
α∈（0,π）
Then: α = arctan2

What is trigonometric inequality
1.|a-b|≤|a-c|+|b-c|
2. When | x + 1|

|a-b|=|(a-c)+(c-b)|≤|a-c|+|b-c|
If and only if a-c and C-B have the same sign, the equal sign is taken
|x+1|