Given the function f (x) = 2cos x (sin x-cos x) + 1, find the minimum positive period, minimum value and maximum value of function f (x) Given the function f (x) = 2cos x (sin x-cos x) + 1, find the minimum positive period, minimum value and maximum value of function f (x) My little brother is in the final exam What about the minimum and maximum?

Given the function f (x) = 2cos x (sin x-cos x) + 1, find the minimum positive period, minimum value and maximum value of function f (x) Given the function f (x) = 2cos x (sin x-cos x) + 1, find the minimum positive period, minimum value and maximum value of function f (x) My little brother is in the final exam What about the minimum and maximum?

Given the function f (x) = 2cos x (sin x-cos x) + 1, find the minimum positive period, minimum value and maximum value of function f (x) = 2sinxcos x-2cos ² X + 1 = sin2x - (1 + cos2x) + 1 = sin2x cos2x = (√ 2) sin (2x - π / 4), so the minimum positive period T = 2 π / 2 = π, minf (x) = - √ 2, MAXF (x) = √ 2

Given 1 + SiNx / cosx = - 1 / 2, what is the value of CoS / sinx-1?

★ Hello, I'm glad to be able to answer your questions ★ \ x0d ------------------------------------------------------------------------ \ x0d ∵ 1 + SiNx = [sin (x / 2) + cos (x / 2)] ^ 2, \ x0d cosx = [cos (x / 2) + sin (x / 2)]] [cos (x / 2) - sin (x / 2)], two

If cosx * cosy + SiNx * siny = 1 / 3, then cos (2x-2y) =?

The original condition is equivalent to cos (X-Y) = 1 / 3
According to the angle doubling formula
cos 2（x-y）=2cos^2 (x-y)-1=2/9-1=-7/9

Proof (SiNx) '= cosx;

△y=sin(x+△x)-sinx=2cos(x+△x/2)sin
△y/△x=[cos(x+△x/2)sin△x/2]/(△x/2)
△x→lim[sin△x/2]/(△x/2)=1
△x→lim△y/△x= cosx
(sinx)' = cosx

Prove that 2sinxcosx / (SiNx + cosx-1) (sinx-cosx + 1) = (I + cosx) / SiNx

The denominator on the left of the equation = (SiNx + cosx-1) (SiNx cosx + 1) = sin2x - (cosx-1) 2 = sin2x cos2x + 2cosx-1 = 2cosx-2cos2x and the numerator are about 2cosx, the left is SiNx / (1-cosx) and the numerator denominator is multiplied by (1 + cosx), and SiNx (1 + cosx) / (1-cos2x) is obtained, that is, SiNx (1 + cosx) / sin2x is about SiNx, which is equal to

Verify that: (sin2x) divided by (SiNx + cosx-1) (SiNx cosx + 1) is equal to Cotx / 2

On the left side of the equation, the left side of the equation is the left side of the equation = 2sinxccosx / [SiNx + (cosx-1) [SiNx (cosx-1)]] = 2sinxccosx / (sin ^ 2x2x-a-cosx-1 (1-cosx-1). The left side of the left side of the equation = 2sinxccosx / [SiNx [SiNx + [SiNx + (1-cosx-1-cosx-1-cosx (1-cosx-1). The left side of the equation is the left side of the equation = 2sinxccosx / [SiNx / X / 1] = 2scostx / [sin ^ 2x2x2x (cosx-1) ^ 1) ^ 2 ^ 1 ^ 2 ^ 2 ^ 2 ^ 2] = 2 = 2] = 2sinxccosx = 2sx = 2 = 2sinxccosx / 2 = 2 = 2sinxcoscx / COSIX / (2x / (2x / 2x / 2x / 2 / 2cos / 2(X / 2) = cot (x / 2)
The process is as detailed as possible. If you are skilled in the angle change formula, many steps can be omitted and the answer can be obtained directly