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Oblicz liczbę \(a= \frac{41}{5+39\sin2 \alpha } \), jeśli \(\tg \alpha = \frac{3}{2} \)

\( \frac{\sin \alpha }{\cos \alpha } = \frac{3}{2} \\
\sin \alpha =1,5\cos \alpha \\
3,25\cos^2 \alpha=1\\
|\cos \alpha |= \frac{2}{ \sqrt{13} }\\
|\sin \alpha |= \frac{3}{ \sqrt{13} } \\
\sin 2 \alpha =2\sin \alpha \cos \alpha \\
\)
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Albo:
\(a= \dfrac{41}{5+39\sin2 \alpha }= \dfrac{41\cdot1}{5\cdot1+39\sin2 \alpha }=\dfrac{(41\cos^2\alpha+41\sin^2\alpha):\cos^2\alpha}{(5\cos^2\alpha+5\sin^2\alpha+78\sin\alpha\cos\alpha):\cos^2\alpha}=\\ \qquad=\dfrac{41+41\tg^2\alpha}{5+5\tg^2\alpha+78\tg\alpha}=\dfrac{41+41\cdot\left({3\over2}\right)^2}{5+5\cdot\left({3\over2}\right)^2+78\cdot{3\over2}}=\ldots\)

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