Question 596062
using the half angle formulas how do i find the exact value of the sin,cos,tan for (7/12)pi, i just dont understand trig at all.
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(7π/6)=2*(7π/12
or
(7π/6)/2=(7π/12)
using half-angle formulas
..
sin(7π/12)=sin(7π/6)/2)=±√[(1-cos(7π/6)/2]
select + root because 7π/12 is in quadrant II where sin>0
=√[(1-cos(7π/6)/2]
=√[(1+√3/2/2]
=√[(2+√3)/4]
..
cos(7π/12)=cos(7π/6)/2)=±√[(1+cos(7π/6)/2]
select - root because 7&#960;/12 is in quadrant II where cos<0
=-&#8730;[(1+cos(7&#960;/6)/2]
=-&#8730;[(1-&#8730;3/2)/2]
=-&#8730;[(2-&#8730;3)/4]
..
tan(7&#960;/12)=tan(7&#960;/6)/2
=sin (7&#960;/6)/(1+cos((7&#960;/6)
=(-1/2)/(1-&#8730;3/2)
=-1/(2-&#8730;3)
multiply top and bottom by (2+&#8730;3)
=-(2+&#8730;3)/4-3
=-(2+&#8730;3)