Question

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

要怎么求？求详细过程！！！

Answer 1

①sin(－7π/12)＝－sin(7π/12)＝－sin﹙π/3＋π/4﹚
＝－枯镇﹛sinπ/3·cosπ/4＋cosπ/3·sinπ/4﹜没答粗＝,,,,,,,,,,
②cos（-61π/12）＝cos﹙61π/12)＝cos(5π＋π/12)
＝cos﹙π＋π/12﹚＝－cos﹙π/12﹚＝－cos﹙π/3－π/4﹚
＝－﹙cc＋ss﹚举租＝－,,,,,,
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﹜
＝..........

Answer 2

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)/拍薯4

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

tan(35π/12)
= tan(3π-π/激皮12)
= tan(π-π/12)
= -tan(π/12)
= -tan(π/3-π/4)
= -(tanπ/3-tanπ/4) / (1+tanπ/3tanπ/4)
= -(√3-1) / (1+√3)
= -(√3-1)^2 /{ (√3+1)(√3-1)}
= -(4-2√3)/(3-1)
= √3 - 2

Answer 3

sin（-7π/12）=-sin（7π/12)=-sin（π/3+π/4））
cos（-61π/12）=cos（-7π/12-π/2-4π）=cos（7π/12+π/散铅侍颂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)/冲谈好sin(5π/12)=cos(7π/12)/sin(7π/12)=cos（π/3+π/4）/sin（π/3+π/4））

Answer 4

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