If sin (π - a) = 1 / 2 + cosa, then sin 3 power (π + a) + cos 3 power (2 π - a)=

If sin (π - a) = 1 / 2 + cosa, then sin 3 power (π + a) + cos 3 power (2 π - a)=

sin(π-a)=1/2+cosa
sina=1/2+cosa
sina-cosa=1/2,(sina-cosa)^2=1-2sinacosa=1/4
sin^3(π+a)+cos^3(2π-a)=cos^3a-sin^3a=(cosa-sina)(1+cosasina)=-1/2(1+3/8)=-11/16

It is proved that sin 4-cos 4 = 2 sina-1

(sina)^4-(cosa)^4=[(sina)^2+(cosa)^2][(sina)^2-(cosa)^2]=-cos2a=-[1-2(sina)^2]=2(sina)^2-1.

It is proved that cos 4 power a-SiN 4 power a = cos 2A

cos^4a-sin^4a
=(cos^2a-sin^2a)(cos^2a+sin^2a)
=Cos ^ 2a-sin ^ 2A (angle doubling formula)
=cos2a

Proof: for any angle a, cos ^ 4a-sin ^ 4A = cos2a Is cos? A-SiN? A = cos2a? Is there a formula?

cos^4a-sin^4a
=(cos^2a-sin^2a)(cos^2a+sin^2a)
=Cos ^ 2a-sin ^ 2A (angle doubling formula)
=cos2a

Given the function f (x) = cosx, it is proved that 1 / 2 [F 2 (π / 4) + F 2 (π + x)] ≥ F 2 (π / 4) + F 2 (π + x)

1、1/2[f²（π/4－x)＋f²(π/4＋x)]
=[cos²(π/4－x)＋cos²(π/4＋x)]/2
=[((cos(π/2-2x) 1)/2 ((cos(π/2 2x) 1)/2]/2
=[sin2x 1 (-sin2x 1)]/4
=2/4=1/2
f²（π/4－x）*f²（π/4＋x）=cos²(π/4－x)*cos²(π/4＋x)
=((cos(π/2-2x) 1)/2*((cos(π/2 2x) 1)/2=(1 sin2x)(1-sin2x)/4
=(1-sin²2x)/4=cos²2x/4

Then the value of COS (x + 1 / 2) is the smallest

The minimum is 6.25
Then, let sin x = 1 x 2

The minimum and maximum values of the function f (x) = 1-2sin? X + 2cosx are

f(x)=1-2sin²x+2cosx
=2cos^2x+2cosx-1
=2(cos^2x+cosx+1/4)-3/2
=2(cosx+1/2)^2-3/2
When cosx = - 1 / 2, there is a minimum of - 3 / 2
When cosx = 1, there is a maximum value of 4

Given the function f (x) = 2Sin (x - π / 3) cosx + sinxcosx + √ 3sin ^ 2x (x belongs to R), find the monotone increasing interval of F (x)

f(x)=2sin(x-π/3)cosx+sinxcosx+√3sin^2x
=2（sinxcosπ/3-cosxsinπ/3)cosx+sinxcosx+√3sin^2x
=sinxcosx-√3cos^2x+sinxcosx+√3sin^2x
=2sinxcosx-√3(cos^2x-sin^2x)
=sin2x-√3cos2x
=2(1/2sin2x-√3/2cos2x)
=2sin(2x-π/3)
From 2K π - π / 2 ≤ 2x - π / 3 ≤ 2K π + π / 2, K ∈ Z
K π - π / 12 ≤ x ≤ K π + 5 π / 12, K ∈ Z
The increasing range of F (x) is [K π - π / 12, K π + 5 π / 12], K ∈ Z

It was proved that sin (2x + y) / sinx-2cos (x + y) = siny / SiNx

Certificate:
sin(2x+y)/sinx -2cos(x+y)
=[ sin[(x+y)+x] -2cos(x+y)sinx ]/sinx
=[sin(x+y)cosx+cos(x+y)sinx-2cos(x+y)sinx]/sinx
=[sin(x+y)cosx-cos(x+y)sinx]/sinx
=sin[(x+y)-x]/sinx
=siny/sinx
The equation holds

Simplification of sin (2x + y) / sinx-2cos (x + y)

sin(2x+y)/sinx-2cos(x+y) =sin(x+y+x)/sinx-2cos(x+y)=〔sin(x+y)cosx+cos(x+y)sinx〕/sinx-2cos(x+y)sinx/sinx=〔sin(x+y)cosx-cos(x+y)sinx〕/sinx=〔sin(x+y-x)〕/sinx=siny/sinx