Question

Find the exact value of

Answer 1

\[\large \tan \frac{ 11\Pi }{ 12 } \]

Answer 2

\[\large \pi\] *

Answer 3

\[\tan(\frac{11\pi}{2})=\tan(6\pi-\frac{\pi}{2})=-\tan(\frac{\pi}{2})=-\infty\]

Answer 4

how did you come up with these values ?

Answer 5

because tan(2pi)=tan(6pi)
he gets tan(6pi) from 11Pi/2 (since 11=6*2-1)
then -tan(pi/2) is a vertical asymptote approaching negative infinity.

Answer 6

With trig functions, every 2pi you move, the answers will be the same, as a unit circle is an angle of 2pi. That is why 6pi=2pi for this problem

Answer 7

for example, cosine of 0 starts at 1, and if you follow the curve to 2pi, it is back at 1, then 4pi it is at 1 again. and 6pi 8pi ...... 2(n)pi

Answer 8

Oh okay thank you !

Answer 9

@BoredMathNerd so basically you just look for two numbers that would give you the sum ?

Answer 10

@darthjavier But the problem was 11pi/12, not 11pi/2.

Answer 11

You have to use the Half Angle Formula for Tangent.\[\large \tan\left(\frac{\theta}{2}\right)=\frac{1-\cos \theta}{\sin \theta}\]
We can rewrite our angle to match this identity.\[\huge \tan\left(\frac{11\pi}{12}\right)\quad \rightarrow \quad \tan\left(\frac{\frac{11\pi}{6}}{2}\right)\]

Answer 12

Where theta is 11pi/6. Which is one of our special angles... I think... I hope :3 lol

Answer 13

\[\huge =\frac{1-\cos\left(\frac{11\pi}{6}\right)}{\sin\left(\frac{11\pi}{6}\right)}\]

Answer 14

yes it is a special angle, its on the unit circle

Answer 15

is there another way without using those formulas because we're not allowed to use those yet :$

Answer 16

Not allowed to use formulas..? :o
Are you allowed to use a calculator?

Answer 17

we're not allwed to use the half angles because my teacher doesnt like them :/

Answer 18

LOL hmm that's weird..

Answer 19

i knooow ..

Answer 20

Ok I think I know another way we can do it then, it's not as easy as the half angle, but it should work.
\[\huge \tan\left(\frac{11\pi}{12}\right)=\tan\left(\frac{9\pi}{12}+\frac{2\pi}{12}\right)\]

Answer 21

\[\huge =\tan\left(\frac{3\pi}{4}+\frac{\pi}{6}\right)\]

Answer 22

From there, you can apply the Angle Sum Formula for Tangent.

Answer 23

\[\huge \tan(a+b)=\frac{\tan a+\tan b}{1-\tan a \cdot \tan b}\]

Answer 24

ohhh

Answer 25

\[\frac{ -1+\sqrt{3} }{ 1-(-1)(\sqrt{3}) }\]

Answer 26

Hmm I think pi/6 gives youuuuu... 1/sqrt3.

Answer 27

oh yeah it's cos over sin

Answer 28

\[\huge \frac{ -3+\sqrt{3} }{ 3 } \over \huge \frac{ 3+\sqrt{3} }{ 3 }\]

Answer 29

Prolly should simplify it down a bit, but yah it looks right! :D

Answer 30

\[\frac{ 3- 3 \sqrt 3 }{ 2 }\]

Answer 31

Yeah, \[\mathrm{LaTeX}\] everywhere haha