Trigonometric Identities/Double Angle Formula?

Question

What is sin(2x)^2 in terms of double angle? How do you get rid of the square? What about cos(2x)^2?

Answer 1

sin^2(2x) = 1 - cos^2(2x) = (1-cos2x)(1+cos2x)

cos^2(2x)= 1 - sin^2(2x) = (1-sin2x)(1+sin2x)

Answer 2

I kinda remember this from Math BII (this was a long time ago...) and if memory serves well...

sin^2(2x) (somehow I don't like how they write sin^2(2x)... it's the same exact meaning as sin(2x)^2... keep in mind)

sin(2x) is the same as 2sinxcosx
I'm guessing if sin^2(2x) would be (2sinxcosx)^2

For cos(2x)... it all equals either one of the following:
cos^2(x) - sin^2(x)
2cos^2(x) - 1
1 - 2sin^2(x)

cos^2(2x) might be
(cos^2(x) - sin^2(x))^2
(2cos^2(x) - 1)^2
(1-2sin^2(x))^2

That's the only legitimate thing I can think of... I don't think I ever learned a formula to directly solve squared double angles... if there are any...

Answer 3

be conscious that cos 2x = a million - 2sin^2 x sub to provide: sin^2 x + (3/2)(a million - 2sin^2 x) = 0 sin^2 x + 3/2 - 3sin^2 x = 0 3/2 = 2sin^2 x 3/4 = sin^2 x sin x = +/- sqrt(3) / 2 wide-spread answer: x = pi/3 + 2pi * n (the position n is an integer) x = 2pi/3 + 2pi * n x = 4pi/3 + 2pi * n x = 5pi/3 + 2pi * n recommendations between 0 and 2pi: x = pi/3 , 2pi/3 , 4pi/3, and 5pi/3