## anonymous one year ago If cosine of x equals 1 over 2, what is sin(x) and tan(x)? Explain your steps in complete sentences.

1. freckles

make a right triangle then use cos(x)=1/2 then use Pythagorean theorem to figure out opposite side of x

2. freckles

then you have everything you need to find sin(x) and tan(x)

3. anonymous

can u guide me @freckles

4. freckles

which part do you need help on?

5. freckles

did you draw the right triangle yet?

6. anonymous

i guess all.

7. anonymous

yes

8. freckles

ok then use cos(x)=1/2 remember cosine, think adjacent/hyp

9. anonymous

ok

10. freckles

then use the Pythagorean theorem to find the opp side can you do that?

11. anonymous

no

12. freckles

well we know the Pythagorean theorem says: $opp^2+adj^2=hyp^2 \\ \text{ and we know the } adj \text{ and the hyp } \\ \text{ just solve for opp}$

13. anonymous

adj is 1 and hyp is 2?

14. freckles

yes this was given by cos(x)=1/2 you could also use any thing that reduces down to 1/2 for example cos(x)=10/20 and call the adj=10 and hyp=20 but it doesn't matter in the end we will get the same reduced fractions so yes you can use adj=1 and hyp=2

15. anonymous

ok

16. freckles

have you solved for opp yet?

17. freckles

if you have you can find sin(x) and tan(x)

18. anonymous

still working on it

19. anonymous

opp is root 5?

20. freckles

$oop^2+1^2=2^2 \\ oop^2+1=4$ 4-1 isn't 5 should be 3

21. freckles

$opp^2=4-1 \\ opp^2=3 \\ opp= \pm \sqrt{3}$

22. anonymous

23. freckles

so you have two answers for both sin(x) and tan(x) because of the plus or minus thing

24. freckles

25. anonymous

ok

26. anonymous

sin(x) = 3/2 tan(x) = 3/1 = 3

27. anonymous

@freckles

28. freckles

what happen to the sqrt( ) part ?

29. freckles

also you should have two answers for each

30. freckles

$\sin(x)=\frac{ \pm \sqrt{3}}{2} \text{ this is the two answers for } \sin(x)$ now you try tan(x)=?

31. anonymous

ok

32. anonymous

±root 3?

33. freckles

yep sounds great

34. anonymous

how did you get the root sighn tho

35. anonymous

sign

36. anonymous

jk never mind

37. freckles

$opp^2+adj^2=hyp^2 \\ opp^2+1^2=2^2 \\ opp^2=4-1 \\ opp^2=3 \\ \text{ \to solve an equation like this you take square root of both sides }$

38. freckles

but it will also mean you wind up with two possibilities since sqrt(opp^2)=|opp|=opp if opp>0 but =-opp if opp<0

39. anonymous

ok. Thank You so so so so much!!!!!!!!!!!!!!!!!!!