To prove that √(sinx*sin3x) = cosx, we can square both sides and simplify:

√(sinxsin3x) = cosx
Square both sides:
(sin x sin 3x) = cos^2(x)
sin x sin 3x = 1 - sin^2(x) (Using the Pythagorean identity cos^2(x) = 1 - sin^2(x))
sin x sin 3x = cos^2(x) (As cos^2(x) = 1 - sin^2(x))

Now, we know that sin 3x = 3sinx - 4sin^3(x) from the triple angle formula of sin 3x.

Substitute sin 3x = 3sinx - 4sin^3(x) into the equation sin x * sin 3x = cos^2(x):

sin x * (3sin x - 4sin^3(x)) = cos^2(x)
3sin^2(x) - 4sin^4(x) = cos^2(x) (Distribute and simplify)

Using the Pythagorean identity sin^2(x) + cos^2(x) = 1, we know that cos^2(x) = 1 - sin^2(x):

3sin^2(x) - 4sin^4(x) = 1 - sin^2(x)
3sin^2(x) - 4sin^4(x) + sin^2(x) = 1
3sin^2(x) + sin^2(x) - 4sin^4(x) = 1
4sin^2(x) - 4sin^4(x) = 1
4sin^2(x)(1 - sin^2(x)) = 1
4sin^2(x)cos^2(x) = 1

Therefore, we have shown that √(sinx*sin3x) = cosx.