Question

Find the exact value of ....

Answer 1

\[\tan (\frac{ 17\pi }{ 12 })\]

Answer 2

how many 12's are in 17?

Answer 3

I know I have to use the equation \[\tan(a+b)=(\frac{ \tan(a)-\tan(b) }{ 1+\tan(a)\tan(b) })\]

Answer 4

1

Answer 5

right so 17=12(1)+5
\[\tan(\frac{17}{12} \pi) \\ \tan([1+\frac{5}{12}]\pi) \\ \tan(\pi+\frac{5\pi}{12}) =\tan(\frac{5\pi}{12} ) \text{ since tangent function has period } \pi \]
so now we are looking at tan(5pi/12) instead

you said you hated the radian stuff so you can concert 5pi/12 to degrees if you want

Answer 6

\[\frac{5\pi}{12} \cdot \frac{180^o}{\pi} = \frac{5 }{12} \cdot 180^o =\frac{5 \cdot 180}{12}^o =5(15)^o=75^o \\ \text{ and } 75^o=30^o+45^o\]

Answer 7

could I have done that with the 17pi/12 too or would that be too hard ?

Answer 8

\[\tan(30^o+45^o)=\frac{\tan(30^o)+\tan(45^o)}{1-\tan(30^o)\tan(45^o)}\]

Answer 9

you could have

Answer 10

\[\frac{17 \pi}{12} \cdot \frac{180^o}{\pi}=17(15)^o=255^o \\ \text{ and then instead use } \\ 255^o=225^o+30^o\]

Answer 11

or you could have chosen 120+135 for 255

Answer 12

there or other choices
you just want to choose the ones that are on the unit circle

Answer 13

do I change the degrees into the points ? Like from the unit circle

Answer 14

you mean change tan(30) and tan(45) using unit circle?
if so yes

Answer 15

yeah , that's what I mean . And sin tan is sin/cos it'd be
tan(30) = \[\frac{ \frac{ 1 }{ 2 } }{ \frac{ \sqrt{3} }{ 2 } }\]

Answer 16

thats right

Answer 17

which can be written as 1/sqrt(3) or sqrt(3)/3

Answer 18

SO I'd multiply the denominator and numerator by 2/rad(3) ?

Answer 19

to get that

Answer 20

yep

Answer 21

and you can rationalize the denominator if you prefer which is how I got also or sqrt(3)/3
but you can leave as 1/sqrt(3) for now if you prefer

Answer 22

and tan(45)=?

Answer 23

it'd be 1 right ?

Answer 24

yes tan(45) is 1
\[\tan(30^o+45^o)=\frac{\tan(30^o)+\tan(45^o)}{1-\tan(30^o)\tan(45^o)} \\ \text{
inserting values we just found: } \\ \tan(30^o+45^o)=\frac{\frac{1}{\sqrt{3}}+1}{1-\frac{1}{\sqrt{3}}(1)} \\ \tan(75^o)=\frac{\frac{1}{\sqrt{3}}+1}{1-\frac{1}{\sqrt{3}}}\]
or we could say:
\[\tan(\frac{17\pi}{12})=\frac{\frac{1}{\sqrt{3}}+1}{1-\frac{1}{\sqrt{3}}}\]

Answer 25

so your teacher probably doesn't want you to leave your answer in compound fraction from

Answer 26

that is tiny fractions inside a big fraction

Answer 27

you can chose to clear the "tiny fractions" by multiply the denominator and numerator of the big fraction by the denominators of the "tiny fractions"
so you only need to get rid of the compound action by multiply top and bottom of main fraction by sqrt(3)

Answer 28

\[\tan(\frac{17\pi}{12})=\frac{\sqrt{3}}{\sqrt{3}} \cdot \frac{\frac{1}{\sqrt{3}}+1}{1-\frac{1}{\sqrt{3}}} \\ \tan(\frac{17\pi}{12})=\frac{\sqrt{3} \cdot (\frac{1}{\sqrt{3}})+\sqrt{3}(1)}{\sqrt{3}(1)-\sqrt{3}(\frac{1}{\sqrt{3}})} \]

Answer 29

\[\sqrt{3}(1)=\sqrt{3} \\ \text{ and } \\ \sqrt{3} \cdot \frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{\sqrt{3}}=1\]

Answer 30

couldn't I multiply by 1+sqrt 3 ?

Answer 31

and then use foil

Answer 32

wait you mean 1+1/sqrt(3)?

Answer 33

I rationalized the denominator to just have the sqrt 3

Answer 34

we can rationalize the denominator after what I have...
or if you prefer to rationalize first then get rid of the compound fraction action you may do that
but you will want to multiply top and bottom by the conjugate of what we have on the bottom which is going to be 1+1/sqrt(3)

Answer 35

What I have right now is \[\frac{ 1+\sqrt{3} }{ 1-\sqrt{3} }\]

Answer 36

that is close to what you should have your bottom should be sqrt(3)-1

Answer 37

why is -1 ?

Answer 38

why is it *

Answer 39

tan(30)=1/sqrt(3) and tan(45)=1
\[\tan(30^o+45^o)=\frac{\tan(30^o)+\tan(45^o)}{1-\tan(30^o)\tan(45^o)} \\ \text{ inserting values we just found: } \\ \tan(30^o+45^o)=\frac{\frac{1}{\sqrt{3}}+1}{1-\frac{1}{\sqrt{3}}(1)} \\ \tan(75^o)=\frac{\frac{1}{\sqrt{3}}+1}{1-\frac{1}{\sqrt{3}}} \]
\[\tan(\frac{17\pi}{12})=\frac{\sqrt{3}}{\sqrt{3}} \cdot \frac{\frac{1}{\sqrt{3}}+1}{1-\frac{1}{\sqrt{3}}} \\ \tan(\frac{17\pi}{12})=\frac{\sqrt{3} \cdot (\frac{1}{\sqrt{3}})+\sqrt{3}(1)}{\sqrt{3}(1)-\sqrt{3}(\frac{1}{\sqrt{3}})} \\ \tan(\frac{17\pi}{12})=\frac{1+\sqrt{3}}{\sqrt{3}-1}\]

Answer 40

into rationalize the denominator we multiply top and bottom by sqrt(3)+1

Answer 41

oooh okay , I rationalized the 1/sqrt 3 before plugging it all in

Answer 42

ok and you winded up with sqrt(3)/3 when you did that right?

Answer 43

\[\frac{1}{\sqrt{3}}=\frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}=\frac{1 \cdot \sqrt{3}}{ \sqrt{3} \cdot \sqrt{3}}=\frac{\sqrt{3}}{3}\]

Answer 44

this does not equal just sqrt(3)

Answer 45

lol oops , I just canceled them out

Answer 46

\[\tan(\frac{17\pi}{12})=\frac{\sqrt{3}}{\sqrt{3}} \cdot \frac{\frac{1}{\sqrt{3}}+1}{1-\frac{1}{\sqrt{3}}} \\ \tan(\frac{17\pi}{12})=\frac{\sqrt{3} \cdot (\frac{1}{\sqrt{3}})+\sqrt{3}(1)}{\sqrt{3}(1)-\sqrt{3}(\frac{1}{\sqrt{3}})} \\ \tan(\frac{17\pi}{12})=\frac{1+\sqrt{3}}{\sqrt{3}-1} \]
so basically one last step
you multiply top and bottom by sqrt(3)+1

Answer 47

\[\sqrt{3}+\sqrt{3} = 2\sqrt{3}\] ??

Answer 48

so it would be \[2+\sqrt{3}\]

Answer 49

for the final answer

Answer 50

\[\frac{1+\sqrt{3}}{\sqrt{3}-1} \cdot \frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{(1+\sqrt{3})(\sqrt{3}+1)}{(\sqrt{3})^2-1}\]
\[\frac{1+\sqrt{3}+\sqrt{3}+3}{3-1}=\frac{4+2 \sqrt{3}}{2}= \frac{4}{2}+\frac{2}{2} \sqrt{3} \\ =2+1 \sqrt{3}=2+\sqrt{3}\]
yes you are right

Answer 51

thank you ! I get this a lot more than the last problems lmao

Answer 52

cool stuff!:)

Answer 53

what about sec(pi/8) ?