It is known that α is an acute angle, satisfying Sin & sup2; α + cos & sup2; α = 1, sin α = (90 & ordm; - α) (1) Find the value of Sin & sup2; 36 & ordm; + Sin & sup2; 54 & ordm (2) Find the value of Sin & sup2; 1 & ordm; + Sin & sup2; 2 & ordm; + Sin & sup2; 3 & ordm; + ··· + Sin & sup2; 88 & ordm; + Sin & sup2; 89 & ordm

It is known that α is an acute angle, satisfying Sin & sup2; α + cos & sup2; α = 1, sin α = (90 & ordm; - α) (1) Find the value of Sin & sup2; 36 & ordm; + Sin & sup2; 54 & ordm (2) Find the value of Sin & sup2; 1 & ordm; + Sin & sup2; 2 & ordm; + Sin & sup2; 3 & ordm; + ··· + Sin & sup2; 88 & ordm; + Sin & sup2; 89 & ordm

Because sin α = cos (90 & ordm; - α),
So sin54 & ordm; = cos (90 & ordm; - 54 & ordm;) = cos36 & ordm;,
So Sin & sup2; 36 & ordm; + Sin & sup2; 54 & ordm;
=sin²36º+cos²36º ( sin²α＋cos²α＝1 )
=1.
In the same way:
sin²1º+sin²89º=1
sin²2º+sin²88º=1
.
sin²44º+sin²46º=1
sin²1º+sin²2º+sin²3º+····+sin²88º+sin²89º
=44*1+sin²45º.
also
sin²45º+sin²45º=1
2sin²45º=1
sin²45º=1/2=0.5.
So Sin & sup2; 1 & ordm; + Sin & sup2; 2 & ordm; + Sin & sup2; 3 & ordm; + ··· + Sin & sup2; 88 & ordm; + Sin & sup2; 89 & ordm;
=44.5.

2cos20*cos40*cos80=?
RT

Original formula = 2sin20 degree cos20 degree cos40 degree cos80 degree / sin20 degree
=Sin40 degree cos40 degree cos80 degree / sin20 degree
=Sin80 degree cos80 degree / 2sin20 degree
=Sin160 degrees / 4sin20 degrees
=1/4

How to simplify f (x) = cos2x + 2sinx?

f(x)=cos2x+2sinx
＝1-2sin^2x+2sinx
=-2(sin^2-sinx)+1
=-2(sin^2-sinx+1/4-1/4)+1
=-(sinx-1/2)^2+3/2
And then there's the result you want