Question

sin^2( 23пи/8 )как разложить?

Answer 1

$Sin ^{2} \frac{23 \pi }{8} =Sin ^{2} (2 \pi + \frac{7 \pi }{8})=Sin ^{2}( \pi - \frac{ \pi }{8})=Sin ^{2} \frac{ \pi }{8}$$= \frac{1-Cos \frac{ \pi }{4} }{2}= \frac{1- \frac{ \sqrt{2} }{2} }{2} = \frac{2- \sqrt{2} }{4}$

Answer 2

sin^2(23π/8)=sin^2(2π+π-π/8)=
sin^2(π-π/8)=sin^2(π/8)=
(1-cos2*π/8)/2=(1-cosπ/4)/2=
(1-√2/2)/2=(2-√2)/2*1/2=(2-√2)/4
sin^2(a)=(1-cos2a)/2
такой формула .есть