prove-that-cos-4-x-sin-4-x-cos-4-x-sin-4-x-cos2A-sec-2A-2-

Question Number 134446 by physicstutes last updated on 03/Mar/21

prove that

\frac{\cos^{4} x + \sin^{4} x}{\cos^{4} x - \sin^{4} x} = \frac{\cos 2 A + \sec 2 A}{2}

Answered by Ar Brandon last updated on 03/Mar/21

\cos^{4} x = {(\frac{e^{ix} + e^{- ix}}{2})}^{4} = \frac{1}{16} (e^{4 ix} + {4 e}^{2 ix} + 6 + {4 e}^{- 2 ix} + e^{- 4 ix})

= \frac{1}{16} (2 \cos 4 x + 8 \cos 2 x + 6)

\sin^{4} x = {(\frac{e^{ix} - e^{- ix}}{2 i})}^{4} = \frac{1}{16} (e^{4 ix} - {4 e}^{2 ix} + 6 - {4 e}^{- 2 ix} + e^{- 4 ix})

= \frac{1}{16} (2 \cos 4 x - 8 \cos 2 x + 6)

\frac{\cos^{4} x + \sin^{4} x}{\cos^{4} x - \sin^{4} x} = \frac{4 \cos 4 x + 12}{16 \cos 2 x} = \frac{{8 \cos}^{2} 2 x + 8}{16 \cos 2 x}

= \frac{\cos 2 x + \sec 2 x}{2}