Esercizio 20 ripasso goniometria e trigonometria

Ripasso goniometria e trigonometria

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Esercizio 20   (\bigstar\largewhitestar\largewhitestar\largewhitestar\largewhitestar). Esprimere le seguenti espressioni in funzione della funzione trigonometrica a fianco indicata e poi semplificare:

    \[\begin{aligned} &(1)\quad \frac{\displaystyle\frac{1}{\tan\alpha}-\tan\alpha}{\displaystyle\frac{1}{\tan\alpha}+\tan\alpha}-\cos^4\alpha,\qquad \sin\alpha;\\ &(2)\quad 3\tan^2\alpha+\sec^2\alpha-\sin^2\alpha(1+\cot^2\alpha),\qquad \cos\alpha;\\ &(3)\quad \frac{\sin^2\alpha}{\cot\alpha}+\frac{\cos^2\alpha}{\tan\alpha}+\frac{2\sin\alpha}{\sec\alpha}-\frac{1}{\cot\alpha},\qquad \tan\alpha. \end{aligned}\]

 

Svolgimento.  Risolviamo (1) ottenendo

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

Risolviamo (2) ottenendo

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

Risolviamo (3) ottenendo

    \[\begin{aligned} &\frac{\sin^2\alpha}{\cot\alpha}+\frac{\cos^2\alpha}{\tan\alpha}+\frac{2\sin\alpha}{\sec\alpha}-\frac{1}{\cot\alpha}=\\ &=\tan\alpha\cdot\frac{\tan^2\alpha}{1+\tan^2\alpha}+\frac{1}{\tan\alpha(1+\tan^2\alpha)}+2\sin\alpha\cos\alpha-\tan\alpha=\\ &=\frac{\tan^4\alpha+1-\tan^2\alpha-\tan^4\alpha}{\tan\alpha(1+\tan^2\alpha)}+\frac{2\tan\alpha}{1+\tan^2\alpha} =\\ &=\frac{1-\tan^2\alpha+2\tan^2\alpha}{\tan\alpha(1+\tan^2\alpha)}=\frac{1}{\tan\alpha}. \end{aligned}\]

 

Fonte: ignota.