Answer:
1
Step-by-step explanation:
sin^2+cos^2 A=1 is an indentity and is true for all A, including A= 3x and hence sin^2+cos^2 3x=1
However, let us try is using values of sin3x and cos3x, but before this as sin^2x + cos^2x =1, squaring sin^4 x + cos^4 x =2sin^2 x cos^2 x=1 or sin^4 x +cos^4 x= 1 - 2 sin^2 x and sin^6 x + cos^6 x=1 -3 sin^2 x cos^2 x (sin^2 x + cos^2 x) =1 -3 sin^2 x cos^2 x -as a^3 =b^3= (a+b)^3 -3ab(a+b)
Now coming to proof as
sin3x= 3 sin x- 4 sin^3 x and cos x= 4 cos^3 x - 3 sin x
Therefore
sin^2 3x +cos^2 3x=(3 sin x- 4 sin^3 x)^2 +(4cos^3 x - 3 cos x)^2
=
9sin^2 x = 16 sin^6 x -24 sin^4 x + 16 cos^6 x + 9 cos^2 x- 24 cos^4 x
=
9(sin^2 x +cos^2 x) + 16(sin^6 x + cos^6 x) - 24 (sin^4 x + cos^4 x)
=
9*1 +16(1-3 sin^2 x cos^2 x) - 24 (1-2 sin^2 x cos^2 x)
=
9+16-48sin^2 x cos^2 x -24+48 sin^2 x cos^2 x
=1
