利用和(差)角公式求下列各三角函数的值 (1) sin(-7π/12) (2)cos(-61π/12) (3)tan(35π/12

问题描述:

利用和(差)角公式求下列各三角函数的值 (1) sin(-7π/12) (2)cos(-61π/12) (3)tan(35π/12

1、sin(-7π/12)=-sin(7π/12)=-sin[π/3+π/4]
2、cos(-61π/12)=cos(61π/12)=cos(6π+11π/12)=cos(11π/12)=cos[3π/4+π/6]
3、tan(35π/12)=tan(3π-π/12)=-tan(π/12)=-tan[π/3-π/4]

sin(-7π/12)=-sin(7π/12)=-sin(π/3+π/4))
cos(-61π/12)=cos(-7π/12-π/2-4π)=cos(7π/12+π/2)=-sin(7π/12)
tan(35π/12)=tan(11π/12)=tan((5π/12+π/2)=sin(5π/12+π/2)/cos(5π/12+π/2)
=-cos(5π/12)/sin(5π/12)=cos(7π/12)/sin(7π/12)=cos(π/3+π/4)/sin(π/3+π/4))

sin(-7π/12)= -sin(7π/12)= -sin﹙π/3+π/4﹚= -(sinπ/3cosπ/4+cosπ/3sinπ/4)= -(√3/2 * √2/2 + 1/2 * √2/2)= -(√6+√2)/4cos(-61π/12)= cos(61π/12)= cos(5π+π/12)= cos(π+π/12)= -cos(π/12)= -...

①sin(-7π/12)=-sin(7π/12)=-sin﹙π/3+π/4﹚
=-﹛sinπ/3·cosπ/4+cosπ/3·sinπ/4﹜=,,,,,,,,,,
②cos(-61π/12)=cos﹙61π/12)=cos(5π+π/12)
=cos﹙π+π/12﹚=-cos﹙π/12﹚=-cos﹙π/3-π/4﹚
=-﹙cc+ss﹚=-,,,,,,
③tan﹙35π/12﹙=tan[﹙36π-π)/12]
=tan﹙3π-π/12﹚=tan﹙-π/12﹚=-tan﹙π/12﹚
=-tan﹙π/3-π/4﹚=-﹛tanπ/3-tanπ/4﹜/﹛1+tanπ/3·tanπ/4﹜
=..........