Trigonométrie (1)
Exercice 1 tp
Calculer
| cos | 7π | et | tan | 7π |
| 12 | 12 |
Correction
| cos | 7π | = cos( | 3π | + | 4π | ) |
| 12 | 12 | 12 |
| = cos | π | cos | π | - sin | π | sin | π |
| 4 | 3 | 4 | 3 |
| = | √(2) | - | √(3) |
| 4 | 4 |
donc
| cos | 7π | = | √(2)-√(3) |
| 12 | 4 |
| tan | 7π | = tan( | 3π | + | 4π | ) |
| 12 | 12 | 12 |
| = | tan | π | + tan | π |
| 4 | 3 | |||
| 1-tan | π | ×tan | π | |
| 4 | 3 | |||
ou encore
| tan | 7π | = | 1+√(3) | = | (1+√(3))² |
| 12 | 1-√(3) | -2 |
Donc
| tan | 7π | =-2-√(3) |
| 12 |
Exercice 2 tp
Calculer
| cos | 5π | et | tan | 5π |
| 12 | 12 |
Exercice 3 tp
Soit x∈IR. Simplifier
| A = sin( | π | -x)+sin( | π | +x) |
| 4 | 4 |
| B = cos( | π | -x)+cos( | π | +x) |
| 4 | 4 |
Exercice 4 tp
Calculer
| cos² | π | et sin² | π |
| 8 | 8 |
Correction
Notons que
| { | cos²(x) = | 1+cos(2x) |
| 2 | ||
| sin²(x) = | 1-cos(2x) | |
| 2 |
| cos²( | π | ) = | 1+cos(π/4) |
| 8 | 2 | ||
| cos²( | π | ) = | 2+√(2) |
| 8 | 4 |
de la même façon on obtient
| sin²( | π | ) = | 1-cos(π/4) |
| 8 | 2 | ||
| sin²( | π | ) = | 2-√(2) |
| 8 | 4 |
Exercice 5 tp
Calculer
| cos | 11π | tan | 11π | |
| 12 | 12 |
| cos² | π | sin² | π | |
| 12 | 12 |