## anonymous one year ago Use the half-angle formula to evaluate tan(17pi/12)

1. freckles

well $\tan(\frac{17\pi}{12}) \\ y=\tan(x) \text{ has period } \pi \\ \tan(\frac{17\pi}{12}-\pi)=\tan(\frac{17\pi}{12}-\frac{12\pi}{12})=\tan(\frac{5\pi}{12}) \\ \text{ now } 2 \cdot \frac{5 \pi}{12}=\frac{5\pi}{6} \\ \text{ so we have that we want to use the half-angle formula on } \\ \tan( \frac{1}{2} \cdot \frac{5 \pi}{6})$

2. freckles

do you know the half angle identity for tan?

3. anonymous

no, only for sin

4. freckles

so you can't use the half angle identity for tan?

5. freckles

you can only use the one for sin?

6. freckles

the weird thing is you have tan here not sin

7. freckles

or do you mean you can use it but you just don't know it?

8. freckles

9. freckles

notice the three different ones they wrote for tan(x/2)

10. freckles

use one of them

11. anonymous

im confused because I don't really know how to do this lol

12. freckles

well first let's just copy of of the half-angle identities over from that page $\tan(\frac{1}{2}x)=\frac{\sin(x)}{1+\cos(x)} \\ \text{ and we want to evaluate } \\ \tan(\frac{1}{2} \frac{5\pi}{6})=?$ do you see what to replace x with ?

13. freckles

compare tan(1/2*5pi/6) to tan(1/2*x)

14. freckles

x has to be 5pi/6 right?

15. freckles

so replace the x's in the identity with 5pi/6

16. anonymous

ok one sec

17. anonymous

tan(1/2*5pi/6)= sin(5pi/6) / 1+cos(5pi/6) ??

18. freckles

ok and ( ) around the bottom

19. freckles

so sin(5pi/6)=? and cos(5pi/6)=?

20. anonymous

1/2 and -sqrt 3/2 ??

21. freckles

$\tan(\frac{1}{2} \cdot \frac{5\pi}{6})=\frac{\frac{1}{2}}{1+\frac{-\sqrt{3}}{2}}$ multiply top and bottom by 2

22. freckles

$\frac{\frac{2}{2}}{2+\frac{2(-\sqrt{3})}{2}}$

23. freckles

can you simplify from there?

24. freckles

you know 2/2=1

25. anonymous

Yes one second

26. anonymous

2+ sqrt3

27. anonymous

28. freckles

let me check one sec

29. freckles

$\frac{\frac{2}{2}}{2+\frac{2(-\sqrt{3})}{2}} \\ \frac{1}{2-\sqrt{3} } \\ \frac{1}{2 -\sqrt{3}} \cdot \frac{2 +\sqrt{3}}{2+\sqrt{3}} \\ \frac{2+\sqrt{3}}{4-3} \\ \frac{2 +\sqrt{3}}{1} \\ 2+\sqrt{3}$

30. freckles

yep looks fabulous

31. anonymous

thanks so much!!!

32. freckles

np