Question

The exact value of tan5pi/12 is

Answer 1

Answer:

Step-by-step explanation:

This is the sum identity for tangent(x + y), where x and y are angles measured in radians.  The formula for tan(x + y) is  $\frac{tan(x)+tan(y)}{1-tan(x)tan(y)}$

The first order of business is to find which 2 angles, when expressed in terms of the denominator 12, add up to equal $\frac{5\pi}{12}$

We will use the first quadrant angles, and only the "important" ones:

$\frac{\pi}{4},\frac{\pi}{3},\frac{\pi}{6},\frac{\pi}{2},\pi$

When expressed in terms of the denominator of 12, these angles have the equivalent angles, in order from above:

$\frac{3\pi}{12},\frac{4\pi}{12},\frac{2\pi}{12},\frac{6\pi}{12},\frac{12\pi}{12}$

We need to find the 2 whose numerators add up to a 5.  That would be:

$\frac{3\pi}{12} +\frac{2\pi}{12}=\frac{5\pi}{12}$

Remember that

$\frac{3\pi}{12}=\frac{\pi}{4}$ and $\frac{2\pi}{12}=\frac{\pi}{6}$  so

angle x is $\frac{\pi}{4}$  and angle y is $\frac{\pi}{6}$, making our tangent sum:

$tan(\frac{\pi}{4}+\frac{\pi}{6})$

Filling that into our formula for the sum of tan(x + y):

$tan(\frac{\pi}{4}+\frac{\pi}{6})=\frac{tan(\frac{\pi}{4})+tan(\frac{\pi}{6}) }{1-tan(\frac{\pi}{4})tan(\frac{\pi}{6}) }$

It just so happens that

$tan(\frac{\pi}{4})=1$  and

$tan(\frac{\pi}{6})=\frac{\sqrt{3} }{3}$ so our formula then becomes

$tan(\frac{\pi}{4}+ \frac{\pi}{6})=\frac{1+\frac{\sqrt{3} }{3} }{1-(1)(\frac{\sqrt{3} }{3}) }$ which simplifies to

$\frac{\frac{3+\sqrt{3} }{3} }{\frac{3-\sqrt{3} }{3} }$ and then bring up the lower fraction and flip it to multiply giving you:

$tan(\frac{\pi}{4} +\frac{\pi}{6} )=\frac{3+\sqrt{3} }{3-\sqrt{3} }$

I have the feeling that you need to rationalize that denominator, and if you do that, the final answer will be:

$2+\sqrt{3}$

I know I kind of left you hanging at the very end with rationalizing, but there was so much already that went into this problem in such depth, that I didn't want to risk possibly confusing you even more than I may have already done so.  Try and follow the best that you can.