Question 596365
When you hear or read "exact value" in Trig, you should immediately think: "I'll need special angles for this."<br>
To solve problems like this one, you start by expressing the given angle in terms of one ore more special angles. Some possibilities for this one are:<ul><li>tan(2*60)
For this we could use the tan(2x) formula</li><li>tan(60+60)
For this we could use the tan(A+B) formula</li><li>tan(90+30)
For this we could use the tan(A+B) formula. But we would end up with a tan(90) and tan(90) is undefined. So this will not work.</li><li>tan(30+30+60)
There is no tan(A+B+C) formula so the 4th one is not easy. But, if we want to work that hard, we could use tan(A+B) twice. (I'll show you at the end.)<\li><li>tan(4*30)
There is no tan(4x) formula. But we could use tan(2x) twice on that one.</li><li>So either of the first two look good. I'll do them both:
Using {{{tan(2x) = (2tan(x))/(1-tan^2(x))}}} on tan(2(60)), making the "x" a 60 we get:
{{{tan(120) =tan(2(60)) = (2tan(60))/(1-tan^2(60)) = (2sqrt(3))/(1-(sqrt(3))^2) = (2sqrt(3))/(1-3) = 2aqrt(3)/(-2) = -sqrt(3)}}}<br>
Using tan(A+B) on tan(60+60), making both A and B 60's we get:
{{{tan(120) = tan(60+60) = (tan(60)+tan(60))/(1-tan(60)*tan(60)) = (sqrt(3) + sqrt(3))/(1-sqrt(3)*sqrt(3)) = (2sqrt(3))/(1-3) = (2sqrt(3))/(-2) = -sqrt(3)}}}<br>
I'm only going to start the last two so you can see ways to handle I-don't-have-a-formula-for-that situations.
Using tan(A+B) twice on tan(30+30+60). First I will designate the 30+30 as the "A" and 60 as the "B":
{{{tan(120) =tan((30+30)+60) = (tan(30+30) + tan(60))/(1-tan(30+30)*tan(60))}}}
Using tan(A+B) again on tan(30+30), making the A and B both 30's:
{{{tan(120) =tan((30+30)+60) = (tan(30+30) + tan(60))/(1-tan(30+30)*tan(60)) = (((tan(30)+tan(30))/(1-tan(30)tan(30))) + tan(60))/(1-((tan(30)+tan(30))/(1-tan(30)tan(30)))*tan(60))}}}
If you replace tan(60) with {{{sqrt(3)}}} and tan(30) with {{{sqrt(3)/3}}} this will work out to the right answer! (Try it if you don't believe me!)<br>
Using tan(2x) twice on tan(4(30)). The first time the "x" will be "2*30"
{{{tan(120) = tan(4x) = tan(2(2*30))) = (2tan(2*30)/(1-tan^2(2*30))}}}
Using tan(2x) again on tan(2*30), making the "x" a 30:
{{{tan(120) = tan(4x) = tan(2(2*30)) = (2tan(2*30))/(1-tan^2(2*30)) = (2((2tan(30))/(1-tan^2(30))))/(1-((2tan(30))/(1-tan^2(30)))^2)}}}