Esercizio 23 ripasso goniometria e trigonometria

Ripasso goniometria e trigonometria

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Esercizio 23   (\bigstar\largewhitestar\largewhitestar\largewhitestar\largewhitestar). Semplificare le seguenti espressioni:

    \[\begin{aligned} &(1)\quad \frac{\sin 3\alpha}{\sin\alpha}-\frac{\cos 3\alpha}{\cos\alpha}=\\ &(2)\quad \left(\frac{\cos\alpha-\sin\alpha}{\cos\alpha+\sin\alpha}\right)^2-\frac{1-\sin 2\alpha}{1+\sin 2\alpha};\\ &(3)\quad \frac{\cos^2\alpha(\sin 8\alpha-\sin 4\alpha)}{\sin 4\alpha\cos 6\alpha};\\ &(4)\quad \tan\frac{\alpha}{2}+\cot\frac{\alpha}{2}. \end{aligned}\]

 

Svolgimento. Risolviamo la (1) applicando le formule parametriche:

    \[\begin{aligned} &\frac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}+1= \frac{\displaystyle\frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2}}{\displaystyle\frac{2t}{1+t^2}-\frac{1-t^2}{1+t^2}}+1=\\ &=\frac{2t+1-t^2+2t-1+t^2}{t^2+2t-1}=\frac{4t}{t^2+2t-1}. \end{aligned}\]

Risolviamo la (2) applicando le formule parametriche:

    \[\begin{aligned} &\frac{2\sin\alpha-\sin 2\alpha}{2\sin\alpha+\sin\alpha}=\frac{2\sin\alpha(1-\cos\alpha)}{2\sin\alpha(1+\cos\alpha)}=\\ &=\frac{\displaystyle 1-\frac{1-t^2}{1+t^2}}{\displaystyle 1+\frac{1-t^2}{1+t^2}}=\frac{1+t^2-1+t^2}{1+t^2+1-t^2}=t^2. \end{aligned}\]

fonte: ignota.