Question 626806
To find {{{tan(pi/3+t)}}} we will be using the tan(A+B) formula:
{{{tan(A+B) = (tan(A)+tan(B))/(1-tan(A)tan(B))}}}
So we will need {{{tan(pi/3)}}} and tan(t). Since {{{pi/3}}} 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).<br>
Since {{{pi < t < 3pi/2}}} 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).)<br>
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
{{{1^2 + x^2 = 5^2}}}
Solving for x...
{{{1 + x^2 = 25}}}
{{{x^2 = 24}}}
{{{x = sqrt(24)}}}
(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!)
Simplifying:
{{{x = sqrt(4*6)}}}
{{{x = sqrt(4)*sqrt(6)}}}
{{{x = 2sqrt(6)}}}<br>
So {{{tan(t) = 2sqrt(6)/1 = 2sqrt(6)}}}<br>
Since {{{pi/3}}} is a special angle we should know that its tan is {{{sqrt(3)}}}. Now that we have tan(t) and {{{tan(pi/3)}}} we can use tan(A+B):
{{{tan(pi/3 + t) = (tan(pi/3)+tan(t))/(1-tan(pi/3)*tan(t))}}}
Substituting in the values we have found we get:
{{{tan(pi/3 + t) = (sqrt(3)+2sqrt(6))/(1-sqrt(3)*2sqrt(6))}}}
First we multiply in the denominator:
{{{tan(pi/3 + t) = (sqrt(3)+2sqrt(6))/(1-2sqrt(18))}}}
The square root in the denominator will simplify:
{{{tan(pi/3 + t) = (sqrt(3)+2sqrt(6))/(1-2sqrt(9*2))}}}
{{{tan(pi/3 + t) = (sqrt(3)+2sqrt(6))/(1-2sqrt(9)*sqrt(2))}}}
{{{tan(pi/3 + t) = (sqrt(3)+2sqrt(6))/(1-2*3*sqrt(2))}}}
{{{tan(pi/3 + t) = (sqrt(3)+2sqrt(6))/(1-6sqrt(2))}}}
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 {{{(a+b)(a-b) = a^2-b^2}}} pattern. To rationalize {{{1-6sqrt(2)}}} we multiply it by {{{1+6sqrt(2)}}}:
{{{tan(pi/3 + t) = ((sqrt(3)+2sqrt(6))/(1-6sqrt(2)))((1+6sqrt(2))/(1+6sqrt(2)))}}}
we from the pattern know how the denominators multiply out. For the numerators we must use FOIL:
{{{tan(pi/3 + t) = (sqrt(3)*1+sqrt(3)*6sqrt(2)+2sqrt(6)*1+2sqrt(6)*6sqrt(2))/((1)^2- (6sqrt(2))^2)}}}
Simplifying:
{{{tan(pi/3 + t) = (sqrt(3)+6sqrt(6)+2sqrt(6)+12sqrt(12))/(1-36*2)}}}
{{{tan(pi/3 + t) = (sqrt(3)+12sqrt(6)+12sqrt(12))/(-71)}}}
{{{tan(pi/3 + t) = (sqrt(3)+12sqrt(6)+12sqrt(4*3))/(-71)}}}
{{{tan(pi/3 + t) = (sqrt(3)+12sqrt(6)+12sqrt(4)*sqrt(3))/(-71)}}}
{{{tan(pi/3 + t) = (sqrt(3)+12sqrt(6)+12*2*sqrt(3))/(-71)}}}
{{{tan(pi/3 + t) = (sqrt(3)+12sqrt(6)+24sqrt(3))/(-71)}}}
{{{tan(pi/3 + t) = -(12sqrt(6)+25sqrt(3))/71}}}