## anonymous one year ago Find the exact value of ....

1. anonymous

$\tan (\frac{ 17\pi }{ 12 })$

2. freckles

how many 12's are in 17?

3. anonymous

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

4. anonymous

1

5. freckles

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

6. freckles

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

7. anonymous

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

8. freckles

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

9. freckles

you could have

10. freckles

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

11. freckles

or you could have chosen 120+135 for 255

12. freckles

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

13. anonymous

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

14. freckles

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

15. anonymous

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

16. freckles

thats right

17. freckles

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

18. anonymous

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

19. anonymous

to get that

20. freckles

yep

21. freckles

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

22. freckles

and tan(45)=?

23. anonymous

it'd be 1 right ?

24. freckles

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: $\tan(\frac{17\pi}{12})=\frac{\frac{1}{\sqrt{3}}+1}{1-\frac{1}{\sqrt{3}}}$

25. freckles

26. freckles

that is tiny fractions inside a big fraction

27. freckles

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)

28. freckles

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

29. freckles

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

30. anonymous

couldn't I multiply by 1+sqrt 3 ?

31. anonymous

and then use foil

32. freckles

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

33. anonymous

I rationalized the denominator to just have the sqrt 3

34. freckles

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)

35. anonymous

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

36. freckles

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

37. anonymous

why is -1 ?

38. anonymous

why is it *

39. freckles

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

40. freckles

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

41. anonymous

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

42. freckles

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

43. freckles

$\frac{1}{\sqrt{3}}=\frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}=\frac{1 \cdot \sqrt{3}}{ \sqrt{3} \cdot \sqrt{3}}=\frac{\sqrt{3}}{3}$

44. freckles

this does not equal just sqrt(3)

45. anonymous

lol oops , I just canceled them out

46. freckles

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

47. anonymous

$\sqrt{3}+\sqrt{3} = 2\sqrt{3}$ ??

48. anonymous

so it would be $2+\sqrt{3}$

49. anonymous

50. freckles

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

51. anonymous

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

52. freckles

cool stuff!:)

53. anonymous