Tangens poluugla: uprosti 1 + tg kvadrat (π/4 − α/2)

Koristimo osnovni trigonometrijski identitet koji povezuje tangens i kosinus: $1 + \text{tg}^2 x = \frac{1}{\cos^2 x} .$ U našem slučaju, argument je $x = \frac{\pi}{4} - \frac{\alpha}{2} .$

1 + \text{tg}^2 \left(\frac{\pi}{4} - \frac{\alpha}{2}\right) = \frac{1}{\cos^2 \left(\frac{\pi}{4} - \frac{\alpha}{2}\right)}

Primenjujemo formulu za kvadrat kosinusa preko polovičnog ugla ili direktno transformišemo imenilac koristeći adicionu formulu za kosinus razlike: $\cos(x - y) = \cos x \cos y + \sin x \sin y .$

\cos \left(\frac{\pi}{4} - \frac{\alpha}{2}\right) = \cos \frac{\pi}{4} \cos \frac{\alpha}{2} + \sin \frac{\pi}{4} \sin \frac{\alpha}{2}

Zamenjujemo poznate vrednosti $\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$ i $\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} .$

\cos \left(\frac{\pi}{4} - \frac{\alpha}{2}\right) = \frac{\sqrt{2}}{2} \cos \frac{\alpha}{2} + \frac{\sqrt{2}}{2} \sin \frac{\alpha}{2} = \frac{\sqrt{2}}{2} \left( \cos \frac{\alpha}{2} + \sin \frac{\alpha}{2} \right)

Sada kvadriramo dobijeni izraz u imeniocu:

\cos^2 \left(\frac{\pi}{4} - \frac{\alpha}{2}\right) = \left[ \frac{\sqrt{2}}{2} \left( \cos \frac{\alpha}{2} + \sin \frac{\alpha}{2} \right) \right]^2 = \frac{2}{4} \left( \cos^2 \frac{\alpha}{2} + 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} + \sin^2 \frac{\alpha}{2} \right)

Koristimo identitete $\sin^2 x + \cos^2 x = 1$ i $2 \sin x \cos x = \sin 2x .$ U našem slučaju $2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} = \sin \alpha .$

\cos^2 \left(\frac{\pi}{4} - \frac{\alpha}{2}\right) = \frac{1}{2} (1 + \sin \alpha)

Vraćamo se na početni izraz i zamenjujemo izračunati imenilac:

1 + \text{tg}^2 \left(\frac{\pi}{4} - \frac{\alpha}{2}\right) = \frac{1}{\frac{1}{2}(1 + \sin \alpha)} = \frac{2}{1 + \sin \alpha}

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