Then sin163 = sin (180-17) = sin17,

sin 223 = sin( 180+43)=-sin 43，

sin 253 = sin(270- 17)=-cos 17，

sin 3 13 = sin(270+43)=-cos43

So sin163 sin223+sin253 sin313 =-sin17 sin43+cos17 cos43.

=cos(43 - 17 )=cos26

2. Because α and β are acute angles, 0 < α, β < π/2, then 0 < α+β < π, so both sinα and sin(α+β) are greater than 0.

For sin? α+cos？ α= 1, which means that sinα=3/5 and sin(α+β)=27/65.

Therefore, cos β = cos (α+β-α) = cos (α+β) cos α+sin (α+β) sin β =17/325.

3. According to the analysis in 2, 0 < α, β < π/2, then 0 < α+β < π.

Because sinx has no monotonicity at (0, π) and cosx has monotonicity at (0, π), we can determine α-β by finding cos(α-β).

Because 0 < α, β < π/2, press sin? α+cos？ α= 1 available, cosα=2 elements 5/5, sinβ=3 elements10/0.

Then cos(α-β)=cosαcosβ+sinαsinβ= root number 2/2.

Therefore, α-β=π/4.

4. Let sinα-sinβ=- 1/2 be the equation (1), and cosα-cosβ= 1/2 be the equation (2).

( 1)? +(2)? Available, sin? α+cos？ α+sin? β+cos？ β-2(cosαcosβ+sinαsinβ)= 1/2

That is, 2-2 (cos α cos β+sin α sin β) =1/2.

Therefore, cosαcosβ+sinαsinβ=3/4, that is, cos(α-β)=3/4.

Because α, β∈(0, π/2), -π/2 < α-β < π/2.

And sinα-sinβ =- 1/2 < 0, then sinα < sinβ. According to the monotonic increase of y=sinx on x∈(0, π/2), α < β.

Therefore, -π/2 < α-β < 0.

∴sin(α-β)=- radical sign [1-cos? (α-β)] =-radical sign 7/4

5.sin2A=2sinAcosA=2/3

Then (Sina +cosA)? = sin? A+cos？ A+2sinAcosA= 1+2 2/3=7/3

So Sina +COSA = root number 2 1/3.

6. It is known that cos2θ= two thirds of the radical sign, then the value of sin quartic θ+cos quartic θ is _ _ _

Because cos2θ= root 2/3, so sin? 2θ= 1-cos？ 2θ=7/9

sin^4θ+cos^4θ=(sin？ θ+cos？ θ)? -2 symptons? θcos？ θ= 1- 1/2(2sinθcosθ)？ = 1- 1/2sin？ 2θ= 1- 1/2 7/9= 1 1/ 18