document.write( "Question 277603: Use an addition or subtraction formula to find the exact value of the expression. \r
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document.write( "Tan(-5pie/12)\r
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document.write( "The formula is tan(s-t)=tan s + tan t/ 1+tans tant
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document.write( "or
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document.write( "tan(s+t)= tan s-tan t/ 1+tans tant \n" );
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Algebra.Com's Answer #202101 by nabla(475) ![]() You can put this solution on YOUR website! Try: \n" ); document.write( "To make this more natural, remember that the primitive tangent function has period of pi. So we have Tan(-5pi/12) congruent to Tan(7pi/12) mod pi.\r \n" ); document.write( "\n" ); document.write( "Now: What functions do we easily know the value to for trigonometrics? pi/3 is certainly one. So, pi/3=4pi/12---> We need to know 3pi/12=pi/4 ! This is also an easy find.\r \n" ); document.write( "\n" ); document.write( "So, we take Tan(7pi/12)=Tan(3pi/12 + 4pi/12) Now we use your formulae:\r \n" ); document.write( "\n" ); document.write( "Tan(3pi/12 + 4pi/12)=[Tan(pi/4)-Tan(pi/3)]/(1+Tan(pi/4)Tan(pi/3))\r \n" ); document.write( "\n" ); document.write( "Now it remains to remember what Tan(pi/4) and Tan(pi/3) are. We use the definition of Tangent. \n" ); document.write( "Tan(x)=Sin(x)/Cos(x). \n" ); document.write( "So Tan(pi/4)=1. \n" ); document.write( "Tan(pi/3)=(sqrt(3)/2)/(1/2)=sqrt(3)\r \n" ); document.write( "\n" ); document.write( "Finally, substitute into our expression:\r \n" ); document.write( "\n" ); document.write( "(1-sqrt(3))/(1+sqrt(3))\r \n" ); document.write( "\n" ); document.write( "Now, simplify:\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "This could be done quicker, but I hope you can see the full details of what is going on through this procedure. \n" ); document.write( " |