eq \phi $, so $ S $ is purely imaginary. However, the original question may expect a real value. Re-examining:
Related Articles
- ight)}{2i\sin\left( rac{ heta - \phi}{2}
- ight)} = -2 \cot\left( rac{ heta - \phi}{2}
- eq b $ implies $ heta
- Let $ S = rac{(a + b)^2 + (a - b)^2}{a^2 - b^2} = rac{2a^2 + 2b^2}{a^2 - b^2} $. Let $ a = 1 $, $ b = i $:
- S = rac{2(1 + (-1))}{1 - (-1)} = 0, ext{ but } a^2 - b^2 = 1 - (-1) = 2, ext{ numerator } 2(1 + (-1)) = 0 \Rightarrow S = 0.
- Another example: $ a = 1 $, $ b = e^{i2\pi/3} $:
Trending Articles
Scott Pilgrim 20th Anniversary B: Why This Glitch Became a Legendary Feature?
Scott Pilgrim 20th Anniversary B: The Untold Story Behind the Final Boss?
Scott Pilgrim 20th Anniversary B: The Secret Move That Broke the Game?
The Most Overpowered Ironclad Build in Slay the Spire 2?
Dominate Slay the Spire 2 with OP Ironclad Build!
Scott Pilgrim 20th Anniversary B: Hidden Content That Only 5% of Players Found?
Why Does the Diamondback Snake Avatar Look Like It's Decaying in This New Game?
The Ultimate Ironclad Build for Slay the Spire 2 is Finally Here!
Ironclad Builds Slay the Spire 2: This One's Broken!
The Blockbuster Party Game That Chews Gum and Wins Too