Question 685106
Find the exact values of each trigonometric function at 0(theta)=-(7pi/12). For sin cos and tan.
-(7&#960;/12) is in quadrant III where sin<0, cos<0, tan>0
-(7&#960;/12)=[(-3&#960;/12)+(-4&#960;/12)]=[(-&#960;/4)+(-&#960;/3)]
sin-(7&#960;/12)=sin[(-&#960;/4)+(-&#960;/3)]=[sin(-&#960;/4)cos(-&#960;/3)]+[cos(-&#960;/4)sin(-&#960;/3)]
=[(-&#8730;2/2)*(1/2)+&#8730;2/2*-&#8730;3/2]
=[-&#8730;2/4-&#8730;6/4]
=(-&#8730;2-&#8730;6)/4
sin-(7&#960;/12)=-(&#8730;2+&#8730;6)/4
check with calculator:
sin-(7&#960;/12)=-0.9659..
-(&#8730;2+&#8730;6)/4=-0.9659..
..
cos-(7&#960;/12)=cos[(-&#960;/4)+(-&#960;/3)]=[cos(-&#960;/4)cos(-&#960;/3)]-[sin(-&#960;/4)sin(-&#960;/3)]
=[(&#8730;2/2)*(1/2)]-[-&#8730;2/2*-&#8730;3/2]
=[&#8730;2/4-&#8730;6/4]
=(&#8730;2-&#8730;6)/4
check with calculator:
cos-(7&#960;/12)=-0.2588..
(&#8730;2-&#8730;6)/4=-0.2588..
.. 
tan-(7&#960;/12)=tan[(-3&#960;/12)+(-4&#960;/12)]=tan[(-&#960;/4)+(-&#960;/3)]
=[tan(-&#960;/4)+tan(-&#960;/3)]/[1-tan(-&#960;/4)*tan(-&#960;/3)]=[-1+(-&#8730;3)]/[1-(-1*(-&#8730;3)]
=(-1-&#8730;3)/(1-&#8730;3)
check with calculator:
tan-(7&#960;/12)=3.732..
(-1-&#8730;3)/(1-&#8730;3)=3.732..