Using Sina and cosa to express sin3a and cos3a

Using Sina and cosa to express sin3a and cos3a

sin3a=sin2acosa+cos2asina
=2sinacosacosa+[1-2(sina)^2]sina
=-4(sina)^3+3sina.
cos3a=cos2acosa-sin2asina
=[2(cosa)^2-1]cosa-2sinacosasina
=4(cosa)^3-3cosa.

(cos3a / 2 + cosa / 2, sin3a / 2 + Sina / 2) = (2cosacosa / 2, - 2sinasina / 2) the specific process of simplification

cos3A/2+cosA/2=2 cos[(3A/2+A/2)/2] cos[(3A/2-A/2)/2] =2 cosA cos(A/2）sin3A/2+sinA/2=2 sin[(3A/2+A/2)/2] cos[(3A/2-A/2)/2] =2 sinA cos(A/2） cos3A/2-cosA/2=-2 sin[(3A/2+A/2)/2] sin[(3A/2-A/2)/2] =-2 s...

It is known that sin α + cos α = - 1 / 5 and 2 π / 3

Sina + cosa = - 1 / 5 because 2 π / 3

First simplify, then evaluate: (b ^ 2-A ^ 2) / (a ^ 2-AB) / (a + (2Ab + B ^ 2) / a) × (1 / A + 1 / b), where a = radical 2 + radical 3, B = radical 2-radical 3

(b^2-a^2)/(a^2-ab)÷(a+(2ab+b^2)/a)×(1/a+1/b)
=-（a+b)/a/[(a+b)^2/a]*(a+b)/(ab)
=-(a+b)/a*a/(a+b)^2*(a+b)/(ab)
=-ab
When a = radical 2 + radical 3, B = radical 2 - radical 3,
-ab=-(√2+√3）（√2-√3）
=-（√2^2-√3^2)
=-(2-3)
=1 .