document.write( "Question 626806: if cos(t)= -1/5 and pi < t < 3pi/2, then tan( pi/3 + t) = ???? \n" ); document.write( "
Algebra.Com's Answer #394508 by jsmallt9(3758)\"\" \"About 
You can put this solution on YOUR website!
To find \"tan%28pi%2F3%2Bt%29\" we will be using the tan(A+B) formula:
\n" ); document.write( "\"tan%28A%2BB%29+=+%28tan%28A%29%2Btan%28B%29%29%2F%281-tan%28A%29tan%28B%29%29\"
\n" ); document.write( "So we will need \"tan%28pi%2F3%29\" and tan(t). Since \"pi%2F3\" is a special angle we will be able to find its tangent without a calculator. We will use the given information about t and cos(t) to find tan(t).

\n" ); document.write( "Since \"pi+%3C+t+%3C+3pi%2F2\" t terminates in the 3rd quadrant. In the 3rd quadrant tan is positive. (Knowing that tan will be positive means we can ignore signs for the rest of finding tan(t).)

\n" ); document.write( "Since tan is opposite/adjacent, we will need those numbers to find it. cos is adjacent/hypotenuse. So we can use a 1 for the adjacent side. But how do we find the opposite side? Answer: The Pythagorean Theorem! Let's call the opposite side \"x\". Then
\n" ); document.write( "\"1%5E2+%2B+x%5E2+=+5%5E2\"
\n" ); document.write( "Solving for x...
\n" ); document.write( "\"1+%2B+x%5E2+=+25\"
\n" ); document.write( "\"x%5E2+=+24\"
\n" ); document.write( "\"x+=+sqrt%2824%29\"
\n" ); document.write( "(Remember, we don't care about signs so we can forget the negative square root of 24 which, in this case, is actually the number you want to use!)
\n" ); document.write( "Simplifying:
\n" ); document.write( "\"x+=+sqrt%284%2A6%29\"
\n" ); document.write( "\"x+=+sqrt%284%29%2Asqrt%286%29\"
\n" ); document.write( "\"x+=+2sqrt%286%29\"

\n" ); document.write( "So \"tan%28t%29+=+2sqrt%286%29%2F1+=+2sqrt%286%29\"

\n" ); document.write( "Since \"pi%2F3\" is a special angle we should know that its tan is \"sqrt%283%29\". Now that we have tan(t) and \"tan%28pi%2F3%29\" we can use tan(A+B):
\n" ); document.write( "
\n" ); document.write( "Substituting in the values we have found we get:
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\n" ); document.write( "First we multiply in the denominator:
\n" ); document.write( "\"tan%28pi%2F3+%2B+t%29+=+%28sqrt%283%29%2B2sqrt%286%29%29%2F%281-2sqrt%2818%29%29\"
\n" ); document.write( "The square root in the denominator will simplify:
\n" ); document.write( "\"tan%28pi%2F3+%2B+t%29+=+%28sqrt%283%29%2B2sqrt%286%29%29%2F%281-2sqrt%289%2A2%29%29\"
\n" ); document.write( "
\n" ); document.write( "\"tan%28pi%2F3+%2B+t%29+=+%28sqrt%283%29%2B2sqrt%286%29%29%2F%281-2%2A3%2Asqrt%282%29%29\"
\n" ); document.write( "\"tan%28pi%2F3+%2B+t%29+=+%28sqrt%283%29%2B2sqrt%286%29%29%2F%281-6sqrt%282%29%29\"
\n" ); document.write( "This may be an acceptable answer. But it does have a square root in the denominator so you may want/need to rationalize it. To rationalize a two-term denominator like this we take advantage of the \"%28a%2Bb%29%28a-b%29+=+a%5E2-b%5E2\" pattern. To rationalize \"1-6sqrt%282%29\" we multiply it by \"1%2B6sqrt%282%29\":
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\n" ); document.write( "we from the pattern know how the denominators multiply out. For the numerators we must use FOIL:
\n" ); document.write( "
\n" ); document.write( "Simplifying:
\n" ); document.write( "
\n" ); document.write( "\"tan%28pi%2F3+%2B+t%29+=+%28sqrt%283%29%2B12sqrt%286%29%2B12sqrt%2812%29%29%2F%28-71%29\"
\n" ); document.write( "\"tan%28pi%2F3+%2B+t%29+=+%28sqrt%283%29%2B12sqrt%286%29%2B12sqrt%284%2A3%29%29%2F%28-71%29\"
\n" ); document.write( "
\n" ); document.write( "\"tan%28pi%2F3+%2B+t%29+=+%28sqrt%283%29%2B12sqrt%286%29%2B12%2A2%2Asqrt%283%29%29%2F%28-71%29\"
\n" ); document.write( "\"tan%28pi%2F3+%2B+t%29+=+%28sqrt%283%29%2B12sqrt%286%29%2B24sqrt%283%29%29%2F%28-71%29\"
\n" ); document.write( "\"tan%28pi%2F3+%2B+t%29+=+-%2812sqrt%286%29%2B25sqrt%283%29%29%2F71\" \n" ); document.write( "
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