Question

Find the exact value of tan 255º.

Answer 1

keep in mind that 255 = 135 + 120

so \(\bf tan(\alpha+\beta) = \cfrac{tan(\alpha)+tan(\beta)}{1-tan(\alpha)tan(\beta)}\\ \quad \\
tan(255^o)\implies tan(135^o+120^o)=\cfrac{tan(135^o)+tan(120^o)}{1-tan(135^o)tan(120^o)}\)


and you can get those angles from your Unit Circle

Answer 2

okay that looks familiar. but what would be my next step, canceling things out or dividing?

Answer 3

no.... just get the tangent values, they'd be fractions or just an a radical
and add them up and simplify the rational

since the tangents will give you a number, you'd end up with a number, either fractional or integer

Answer 4

would it look like this
\[3\pi/4+2\pi/3 \over 1- 3\pi/4 * 2\pi/3\]

Answer 5

scratch what i just said, it's wrong. i know tangent is y/x and i found out that tangent of 120 degrees is - square root of 3, but how would you get that?

Answer 6

hmmm how do we get tangent of 120 degrees is \(\bf -\sqrt{3}\) ?

Answer 7

say for example... what's the tangent now of 135 degrees?

Answer 8

would it be 1 or -1?

Answer 9

same/same = 1
-same/same = -1
same/-same = -1

Answer 10

so -1

Answer 11

i think i see how it's - square root of 3
the 2s cancel out and so does the one, so its - square root of 3?

Answer 12

so is -1... ok, now we know the tangents of 135 and 120 degrees, we know that thus \(\bf tan(135^o+120^o)=\cfrac{tan(135^o)+tan(120^o)}{1-tan(135^o)tan(120^o)}\\ \quad \\
\implies tan(135^o+120^o)=\cfrac{(-1)+(-\sqrt{3})}{1-(-1)(-\sqrt{3})}\)

Answer 13

okay i get that. but i was never good at simplifying. would the numerator be square root of 3?

Answer 14

no, it'll be just a simple addition

nothing to cancel out as far as I can tell

you may just be asked to leave it rational too, and that'd be ok

Answer 15

i think i got it. would it be \[2 + \sqrt{3}\]

Answer 16

\(\bf \cfrac{(-1)+(-\sqrt{3})}{1-(-1)(-\sqrt{3})}\implies \cfrac{-1-\sqrt{3}}{1-\sqrt{3}}\)

Answer 17

can't add the radical to the integer, unless you make it a real-number,