You can put this solution on YOUR website! Find the exact values of each trigonometric function at 0(theta)=-(7pi/12). For sin cos and tan.
-(7π/12) is in quadrant III where sin<0, cos<0, tan>0
-(7π/12)=[(-3π/12)+(-4π/12)]=[(-π/4)+(-π/3)]
sin-(7π/12)=sin[(-π/4)+(-π/3)]=[sin(-π/4)cos(-π/3)]+[cos(-π/4)sin(-π/3)]
=[(-√2/2)*(1/2)+√2/2*-√3/2]
=[-√2/4-√6/4]
=(-√2-√6)/4
sin-(7π/12)=-(√2+√6)/4
check with calculator:
sin-(7π/12)=-0.9659..
-(√2+√6)/4=-0.9659..
..
cos-(7π/12)=cos[(-π/4)+(-π/3)]=[cos(-π/4)cos(-π/3)]-[sin(-π/4)sin(-π/3)]
=[(√2/2)*(1/2)]-[-√2/2*-√3/2]
=[√2/4-√6/4]
=(√2-√6)/4
check with calculator:
cos-(7π/12)=-0.2588..
(√2-√6)/4=-0.2588..
..
tan-(7π/12)=tan[(-3π/12)+(-4π/12)]=tan[(-π/4)+(-π/3)]
=[tan(-π/4)+tan(-π/3)]/[1-tan(-π/4)*tan(-π/3)]=[-1+(-√3)]/[1-(-1*(-√3)]
=(-1-√3)/(1-√3)
check with calculator:
tan-(7π/12)=3.732..
(-1-√3)/(1-√3)=3.732..
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