Esercizio 22 ripasso goniometria e trigonometria

Ripasso goniometria e trigonometria

Home » Esercizio 22 ripasso goniometria e trigonometria
Generic selectors
Exact matches only
Search in title
Search in content
Post Type Selectors
post
page

 

Esercizio 22   (\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. 

Semplifichiamo (1):

    \[\begin{aligned} &\frac{\sin 3\alpha}{\sin\alpha}-\frac{\cos 3\alpha}{\cos\alpha}=\frac{\sin(\alpha+2\alpha)}{\sin\alpha}-\frac{\cos(\alpha+2\alpha)}{\cos\alpha}=\\ &=\frac{\sin\alpha\cos 2\alpha+\sin 2\alpha\cos\alpha}{\sin\alpha}-\frac{\cos\alpha\cos 2\alpha-\sin\alpha\sin 2\alpha}{\cos\alpha}=\\ &=\cos 2\alpha+2\cos^2\alpha-\cos 2\alpha+2\sin^2\alpha=2 \end{aligned}\]

Semplifichiamo (2):

    \[\begin{aligned} &\left(\frac{\cos\alpha-\sin\alpha}{\cos\alpha+\sin\alpha}\right)^2-\frac{1-\sin2\alpha}{1+\sin 2\alpha}=\\ &=\frac{\cos^2\alpha-2\sin\alpha\cos\alpha+\sin^2\alpha}{\cos^2\alpha+2\sin\alpha\cos\alpha+\sin^2\alpha} -\frac{1-2\sin\alpha\cos\alpha}{1+2\cos\alpha\sin\alpha}=0. \end{aligned}\]

Semplifichiamo (3):

    \[\begin{aligned} \frac{\cos^2\alpha(\sin 8\alpha-\sin 4\alpha)}{\sin 4\alpha\cos 6\alpha}&=\\ &=\frac{\cos^2\alpha(\sin(2\cdot4\alpha)-\sin 4\alpha)}{\sin 4\alpha\cos(2\alpha+4\alpha)}=\\ &=\frac{\cos^2\alpha(2\sin4\alpha\cos4\alpha-\sin4\alpha)}{\sin4\alpha(\cos2\alpha\cos4\alpha-\sin2\alpha\sin4\alpha)} =\\ &=\frac{\cos^2\alpha(2\cos4\alpha-1)}{\cos2\alpha\cos4\alpha-\sin2\alpha\sin4\alpha}=\\ &=\frac{\cos^2\alpha(2\cos^2 2\alpha-2\sin^2 2\alpha-1)}{\cos^3 2\alpha-\cos2\alpha\sin^2 2\alpha-2\sin^2 2\alpha\cos2\alpha}=\\ &=\frac{\cos^2\alpha(4\cos^2 2\alpha-3)}{\cos 2\alpha(4\cos^2 2\alpha-3)}=\\ &=\frac{\cos^2\alpha}{\cos 2\alpha}. \end{aligned}\]

Semplifichiamo (4):

    \[\begin{aligned} &\tan\frac{\alpha}{2}+\cot\frac{\alpha}{2}= \pm\sqrt{\frac{1-\cos\alpha}{1+\cos\alpha}}\pm\sqrt{\frac{1+\cos\alpha}{1-\cos\alpha}}=\\ &=\frac{1+\cos\alpha+1-\cos\alpha}{\pm\sqrt{1-\cos^2\alpha}}=\frac{2}{\sin\alpha}. \end{aligned}\]

Fonte: ignota.