Question

Тригонометрия!!!!!!!!!

Answer 1

=(2cos^2(2a)-sin^2(2a)-cos^2(2a)+sqrt(3)sin4a)/(2sin^2(2a)-sin^2(2a)-cos^2(2a)+sqrt(3)sin4a))=

=(cos4a+sqrt(3)sin4a)/(sqrt(3)sin4a-cos4a)=sin(п/6+4a)/sin(4a-П/6)

Answer 2

sin(4x + π/6) = sin4xcosπ/6 + cos4xsinπ/6 = sin4x$sqrt{3}/2$ + cos4x/2.

cos4x = 2$Cos^{2}(2x)$ - 1.

sin(4x + π/6) = (2$Cos^{2}(2x)$  + $sqrt{3}$sin4x -1)/2.

sin(4x - π/6)  = sin4xcosπ/6 - cos4xsinπ/6 = sin4x$sqrt{3}/2$ - cos4x/2.

cos4x/2 = 1 - 2$Sin^{2}(2x)$.

sin(4x - π/6) = (2$Sin^{2}(2x)$ + $sqrt{3}$sin4x - 1)/2.

(2$Cos^{2}(2x)$  + $sqrt{3}$sin4x -1)/2)/(2$Sin^{2}(2x)$ + $sqrt{3}$sin4x - 1)/2) =

(2$Cos^{2}(2x)$  + $sqrt{3}$sin4x -1)/(2$Sin^{2}(2x)$ + $sqrt{3}$sin4x - 1).