( 1)

f(x)=sin(2x)-cos(2x)

Find the derivative of the increasing interval of f(x), f'(x)=2*cos2x+2*sin2x.

When f' (x) >; =0, namely 2cos (2x)+2sin2x >; =0

sin(π/4)* cos2x+cos(π/4)* sin2x & gt； =0

sin(2x+π/4)>=0

2x+π/4∈[2nπ, 2nπ+π] where n is an integer.

X∈[nπ-π/8, nπ+3π/8] n is an integer, and f(x) is increasing.

(2)

0 = & ltx & lt=π/2

According to (1), we can know that [0,3 π/8] is monotonically increasing and [3π/8, π/2] is decreasing.

f(0)=- 1

F (3 π/8) = 2 0.5, where 0 stands for power number and 2 0.5 stands for 0.5 power of 2, which is the root number 2.

f(π/2)= 1

That is, the range of f(x) is [- 1, 2 0.5].

17, similar to the above question.

( 1)

f(a)= 8 cosa * Sina-8 cosa * cosa+6 = 4 sin(2a)-4 cos(2a)+2 = 4 *2^0.5 * sin(2a-π/4)+2

The maximum value f(a) is 4 * 2 0.5+2 (four times the root number two plus two).

The maximum value can be obtained when 2a-π/4=π/2+2nπ.

(2)

The internal angle range of an acute triangle is (0, π/2)

f(A)=6

That is, the root number of sin is two (2a-π/4) = 2 0.5/2.

A=π/4 45

Area S= 1/2 *b*c*sinA=3.

B * c = 6/Sina = 6 * 2 0.5 6th root number 2.

b+c=2+3*2^0.5

The solution is b, c=2 or 3 * 2 0.5.

Let's just say that B is bigger.

After passing through point B, make an AC vertical line and intersect with AC at point D (make a long-side vertical line), AB=2, A = 45, then AD = BD = 2 0.5.

CD=AC-AD=2*2^0.5

a=BC= 10^0.5