الحساب المثلثي (1_7)
للتذكير
| x | 0 | π | π | π | π | ||||
| 6 | 4 | 3 | 2 | ||||||
|---|---|---|---|---|---|---|---|---|---|
| sinx | 0 | 1 | √2 | √3 | 1 | ||||
| 2 | 2 | 2 | |||||||
| cosx | 1 | √3 | √2 | 1 | 0 | ||||
| 2 | 2 | 2 | |||||||
| tanx | 0 | √3 | 1 | √3 | × | ||||
| 3 |
العلاقات بين الخطوط المثلثية
| cos(-x) = cosx | sin(-x) = - sinx | |
| sin(x+2kπ) = sinx | cos(x+2kπ) = - cosx | tan(-x) = - tanx | tan(x+kπ) = tanx |
| sin(π-x) = sinx | cos(π-x)= - cosx | |
| sin(π+x) = - sinx | cos(π+x)= - cosx | |
| tan(π-x) =- tanx | tan(π+x)= tanx |
| sin( | π | - x) = cosx | |
| 2 | |||
| cos( | π | - x) = sinx | |
| 2 | |||
| sin( | π | + x) = cosx | |
| 2 | |||
| cos( | π | + x) = - sinx | |
| 2 |
| tan( | π | - x) = | 1 | |
| 2 | tanx | |||
| tan( | π | + x)=- | 1 | |
| 2 | tanx |
تمرين 1 tp
بسط ما يلي
A=cos(4π+x)+cos(3π-x)
B=sin(9π-x)+sin(x+8π)
C=tan(3π-x)+tan(4π+x)
تصحيح
لدينا A=cos(x+2.2π)+cos(π+2.1π)
A=cosx+cos(π-x) اذن A=0
B=sin(π+2.4π-x)+sin(x+2.4π)
=sin(π-x)+sinx=sinx+sinx
اذن B=2sinx.
لدينا C = tan(3π-x)+tan(4π+x)
=tan(π+2π-x)+tan(x+2.π)
=tan(π-x)+tan(π+x)
=tan(-x)+tan(x)
=-tanx + tanx=0
اذن C=0.