[tex](\sin 2 x+\cos 2 x)^{2}=1+\sin 4 x[/tex]
Solution:
Given trigonometric function:
[tex](\sin 2 x+\cos 2 x)^{2}[/tex]
Using perfect square formula:
[tex](a+b)^{2}=a^{2}+2 a b+b^{2}[/tex]
[tex]a=\sin 2 x, b=\cos 2 x[/tex]
[tex](\sin 2 x+\cos 2 x)^{2}=\sin ^{2}(2 x)+2 \sin 2 x \cos 2 x+\cos ^{2}(2 x)[/tex]
Arranging square terms together.
[tex]=\sin ^{2}(2 x)+\cos ^{2}(2 x)+2 \sin 2 x \cos 2 x[/tex]
Using the trigonometric identity: [tex]\sin ^{2}x+\cos ^{2}x=1[/tex]
[tex]=1+2 \cos 2 x \sin 2 x[/tex]
Using the trigonometric identity: [tex]2 \cos \sin x=\sin 2 x[/tex]
[tex]=1+\sin (2 \cdot 2 x)[/tex]
[tex]=1+\sin 4 x[/tex]
Hence [tex](\sin 2 x+\cos 2 x)^{2}=1+\sin 4 x[/tex].
