cos−1x−cos−12y=αcos−1(2xy+(−x2)(1−4y2))=αcos−1(2xy+4−y2−4x2+x2y2)=α⇒4−y2−4x2+x2y2=4cos2α+x2y2−4xycosα⇒4x2+y2−4xycosα=4sin2α
If cos−1x−cos−12y=α, then 4x2−4xycosα+y2 is equal to
Held on 30 Apr 2005 · Verified 6 Jul 2026.
2sin2α
4
4sin2α
−4sin2α
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