Let's first look at how sinx becomes a function of cos:

From the function image, we can know that changing cosx to the right by 1/4 period, that is, π/2 becomes the same function as sinx, and sinx=cos(x-π/2).

2sin(2x+B) is to translate sinx by b units to the left, then reduce the period and stretch it up and down. Or reduce the period to π first, then move B/2 to the left, and finally stretch it up and down.

In short, whether you translate or loop first, the result is the same, but the statement is different, that is, you have to find a way to turn a sine function into an equivalent cosine function. After the change, as long as it is equal.

In your words:

We see that the period of 2sin(2x+B) is π, and when 2 is put forward, it is 2sin[2(x+B/2)], which is equivalent to shortening the period of sinx to π, then shifting it to the left by B/2 units, and then stretching it up and down. Cos(2x+B) also changes in this way, and the difference between cos(2x+B) and sin(2x+B) is 1/4 cycles. Just shift cos(2x+B) to the right by 1/4 cycles, that is, x minus π/4 becomes

Cos[2(x-π/4)+B], which is cos(2x+B-π/2).

Hmm?

Is it in the book? I calculate how to subtract π/2. Am I wrong or am I wrong in the book?