Proving Trig Problem: sec (2x) - 1 = sin^2 x / 2 sec (2x)

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To prove the equation sec(2x) - 1 = sin^2 x / (2 sec(2x)), the discussion emphasizes starting with the left side and applying the double angle identity. The user suggests substituting terms for sec(2x) using cos(2x) and exploring identities for sin^2 x and cos^2 x. There is a focus on simplifying fractions to avoid confusion. The conversation highlights the importance of maintaining clarity while manipulating trigonometric identities. Overall, the goal is to demonstrate the equivalence of both sides of the equation through algebraic manipulation.
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I'm stuck :/ I have to prove the following:
sec (2x) - 1 = sin^2 x
____________
2 sec (2x)

Unfortunately, I don't know any terms that are equal to sec (2x). Any help would be awesome!
 
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sec(2x)=\frac{1}{cos(2x)}=\frac{1}{cos^2x-sin^2x}

Also you can change the denominator by substituting in for cos^2x or sin^2x.
 
I think I would start with the left side of the eq. and use the double angle identity to get it in terms of sin/cos.
 
I'll try that again. I kept getting way off track, but I probably just got confused with all the fractions haha. Thanks everyone!
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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