document.write( "Question 685106: Find the exact values of each trigonometric function at 0(theta)=-(7pi/12). For sin cos and tan. \n" ); document.write( "
Algebra.Com's Answer #424398 by lwsshak3(11628)\"\" \"About 
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Find the exact values of each trigonometric function at 0(theta)=-(7pi/12). For sin cos and tan.
\n" ); document.write( "-(7π/12) is in quadrant III where sin<0, cos<0, tan>0
\n" ); document.write( "-(7π/12)=[(-3π/12)+(-4π/12)]=[(-π/4)+(-π/3)]
\n" ); document.write( "sin-(7π/12)=sin[(-π/4)+(-π/3)]=[sin(-π/4)cos(-π/3)]+[cos(-π/4)sin(-π/3)]
\n" ); document.write( "=[(-√2/2)*(1/2)+√2/2*-√3/2]
\n" ); document.write( "=[-√2/4-√6/4]
\n" ); document.write( "=(-√2-√6)/4
\n" ); document.write( "sin-(7π/12)=-(√2+√6)/4
\n" ); document.write( "check with calculator:
\n" ); document.write( "sin-(7π/12)=-0.9659..
\n" ); document.write( "-(√2+√6)/4=-0.9659..
\n" ); document.write( "..
\n" ); document.write( "cos-(7π/12)=cos[(-π/4)+(-π/3)]=[cos(-π/4)cos(-π/3)]-[sin(-π/4)sin(-π/3)]
\n" ); document.write( "=[(√2/2)*(1/2)]-[-√2/2*-√3/2]
\n" ); document.write( "=[√2/4-√6/4]
\n" ); document.write( "=(√2-√6)/4
\n" ); document.write( "check with calculator:
\n" ); document.write( "cos-(7π/12)=-0.2588..
\n" ); document.write( "(√2-√6)/4=-0.2588..
\n" ); document.write( "..
\n" ); document.write( "tan-(7π/12)=tan[(-3π/12)+(-4π/12)]=tan[(-π/4)+(-π/3)]
\n" ); document.write( "=[tan(-π/4)+tan(-π/3)]/[1-tan(-π/4)*tan(-π/3)]=[-1+(-√3)]/[1-(-1*(-√3)]
\n" ); document.write( "=(-1-√3)/(1-√3)
\n" ); document.write( "check with calculator:
\n" ); document.write( "tan-(7π/12)=3.732..
\n" ); document.write( "(-1-√3)/(1-√3)=3.732.. \n" ); document.write( "
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