Esercitazione di Matematica 0 del 20/12/2004 Corso del prof ...
Esercitazione di Matematica 0 del 20/12/2004 Corso del prof ...
Esercitazione di Matematica 0 del 20/12/2004 Corso del prof ...
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<strong>Esercitazione</strong> <strong>di</strong> <strong>Matematica</strong> 0 <strong>del</strong> <strong>20</strong>/<strong>12</strong>/<strong>20</strong>04<br />
<strong>Corso</strong> <strong>del</strong> <strong>prof</strong>. Davide Vergni<br />
1) Utilizzando la circonferenza goniometrica verificare che:<br />
a) sin(π − α) = sin(α)<br />
( ) π<br />
b) sin<br />
2 − α = cos(α)<br />
c) sin(π + α) = − sin(α) d) sin(2π − α) = − sin(α)<br />
2) Semplificare le seguenti espressioni trigonometriche:<br />
(<br />
a) sin 4 (α) − sin 2 (α) − cos 4 (α) + cos 2 (α) b) cos α + π ) (<br />
cos α − π )<br />
+ 1 6<br />
6 4<br />
c) tan(π + α) sin(π − α) cos(π + α) − tan 2 (π − α) cos 2 (−α)<br />
d) (tan(α) + 1)(−1 − tan(π − α) + 2 cos 2 ( π<br />
2 − α )<br />
+ 2 cos 2 (α − 3π)<br />
e) 1 − sin ( π<br />
− 2 α)<br />
1 + cos(5π + α)<br />
+ tan(α) +<br />
tan(π + α)<br />
sin(−α) cos(−α)<br />
f) tan α + tan β −<br />
sin(α + β)<br />
cos(α) cos(β)<br />
3) Risolvere le seguenti equazioni trigonometriche:<br />
(<br />
a) sin 2x − π )<br />
= 1 b) 2 cos 2 (x)−cos(x)−1 = 0 c) cos(x) = sin 2 (x)−cos 2 (x)<br />
2 2<br />
( ) ( )<br />
π π<br />
d) sin(<br />
4 + x +sin(<br />
4 − x = 1 e) sin(x) = sin(2x) f) 2 cos(x)+2 sin(x) = √ 3+1<br />
g) sin(2x) = 1 h) sin(x) cos(x) + √ 3 cos 2 (x) = √ 3 i) tan 2 (x) = tan(x)<br />
4) Risolvere le seguenti <strong>di</strong>sequazioni trigonometriche:<br />
a) cos(x) > 1 2<br />
b) 2 cos 2 (x) − cos(x) < 0 c) sin(x) + cos(2x) < 1<br />
d) cos(x) − √ 3 sin(x) > 0 e) √ 3 sin(x) + cos(x) > 1 f) 2 cos(x) − 3<br />
sin(x)<br />
g) 4 cos 2 (x) − 3 < 0 h) | sin(x)| − cos(x) > 0 i) 4 cos2 (x) + 4 sin(x) − 1<br />
− cos(x) − sin(x) − 1 > 0<br />
≥ 0
[2d]<br />
[2a] 0 [2b] cos 2 (α) [2c] 0<br />
1<br />
cos 2 (α)<br />
[2e] sin(α) [2f] 0<br />
[3a] x = π 3 +2kπ , 2π<br />
3 +2kπ [3b] x = 2kπ , ±2π 3 +2kπ [3c] x = π+2kπ , ±π 3 +2kπ<br />
[3d] x = ± π 4 + 2kπ [3e] x = kπ , ±π 3 + 2kπ [3f] x = π 6 + 2kπ , π<br />
3 + 2kπ<br />
[3g] x = π 4 + kπ<br />
[3h] x = kπ , π<br />
6 + kπ [3i] x = kπ , π<br />
4 + kπ<br />
[4a] − π 3 +2kπ < x < π 3 +2kπ [4b] π 3 +2kπ < x < π 2 +2kπ ∪ 3π 2 +2kπ < x < 5π 3 +2kπ<br />
[4c] π 6 +2kπ < x < 5π 6<br />
+2kπ ∪ π+2kπ < x < 2π+2kπ [4d]<br />
−5π 6 +2kπ < x < π 6 +2kπ<br />
[4e] 2kπ < x < 2π 3 + 2kπ<br />
[4g] π 6 + kπ < x < 5π 6 + kπ<br />
[4f] π + 2kπ < x < 2π + 2kπ<br />
[4h] π<br />
4 + 2kπ < x < 7π 4 + 2kπ<br />
[4i] π + 2kπ < x < 7π 6 + 2kπ ∪ 3π 2<br />
+ 2kπ < x <<br />
11π<br />
6 + 2kπ