题目
利用和差公式求函数值 sin (-7π/12). cos (-61π/12) tan 35π/12
sin (-7π/12). cos (-61π/12) tan 35π/12 .、过程.谢谢
sin (-7π/12)
cos (-61π/12)
tan 35π/12 ....
sin (-7π/12). cos (-61π/12) tan 35π/12 .、过程.谢谢
sin (-7π/12)
cos (-61π/12)
tan 35π/12 ....
提问时间:2020-07-31
答案
sin(-7/12)=sin(5π/12-π)=-sin(π-5π/12)=-sin5π/12=sin(3π/12+2π/12).
=-sin(π/4+π/6)=-(sinπ/4cosπ/6+cosπ/4sinπ/6)=-[(√2/2*√3/2+√2/2*(1/2)]
= -(√6+√2)/4.
cos(-61π/12)=cos61π/12=cos[12*5π+π)/12=cos(5π+π/12)=cos(4π+(π+π/12].
=-cos(π+π/12)=cosπ/12=cos(4π/12-3π/12)=cos(π/3-π/4).
=cos(π/3)*cos(π/4)+sin(π/3)sin(π/4.
=(1/2)*√2/2+√3/2√2/2.
=(√2+√6)/4.
tan35π/12=tan(36π-π)/12=tan(3π-π/12)=-tanπ/12.
=-tan(π/3-π/4)=-(tanπ/3-tanπ/4)/(1+tanπ/3*tanπ/4).
=(1-√3)/(1+√3*1).
=(1-√3)^2/(1+√3)(1-√3).
=(1-2√3+3)/(-2).
=-(2-√3).
=(√3-2).
/
=-sin(π/4+π/6)=-(sinπ/4cosπ/6+cosπ/4sinπ/6)=-[(√2/2*√3/2+√2/2*(1/2)]
= -(√6+√2)/4.
cos(-61π/12)=cos61π/12=cos[12*5π+π)/12=cos(5π+π/12)=cos(4π+(π+π/12].
=-cos(π+π/12)=cosπ/12=cos(4π/12-3π/12)=cos(π/3-π/4).
=cos(π/3)*cos(π/4)+sin(π/3)sin(π/4.
=(1/2)*√2/2+√3/2√2/2.
=(√2+√6)/4.
tan35π/12=tan(36π-π)/12=tan(3π-π/12)=-tanπ/12.
=-tan(π/3-π/4)=-(tanπ/3-tanπ/4)/(1+tanπ/3*tanπ/4).
=(1-√3)/(1+√3*1).
=(1-√3)^2/(1+√3)(1-√3).
=(1-2√3+3)/(-2).
=-(2-√3).
=(√3-2).
/
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