## anonymous 3 years ago Use the given value and the trigonometric identities to find the remaining trigonometric functions of the angle. cos θ = − 5/9, sin θ > 0

1. anonymous

Okay, let's make that drawing again, though following @hartnn 's advice, I'll do it properly this time ;) DRAW!!! |dw:1360757734426:dw|

2. anonymous

3. anonymous

yes

4. anonymous

So, we have cosθ = -(5/9) We place the 5 and the 9, accordingly... |dw:1360757883210:dw| (Stop me when you don't understand)

5. anonymous

i understand

6. anonymous

Now, what's the length of this missing side... |dw:1360758005338:dw| (Hint: Pythagorean theorem)

7. anonymous

$\sqrt{106}$ right?

8. anonymous

Remember, 9 is your hypotenuse When you have a^2 + b^2 = c^2 9 is your c :)

9. anonymous

$\cos ^2(\theta)+\sin^2(\theta)=1$

10. anonymous

$\sin^2(\theta) = 1-\cos^2(\theta)$

11. anonymous

$\sqrt{56}$

12. anonymous

$\sin^2(\theta)=(\frac{- 5 }{ 9 })^2 -1$

13. anonymous

i prefer the method provided by @PeterPan, because it shows you the inside out

14. anonymous

So, we can put sqrt(56) on here...|dw:1360758287438:dw| And you can easily get the other trig functions. Be careful though...

15. anonymous

@mathsmind You do need to do this to illustrate the pythagorean identities you posted. Though if you're not allowed to draw, better know the pythagorean (and other) identities by heart!!! :D

16. anonymous

yes that is what i said

17. anonymous

i said i prefer the other method to know the inside out

18. anonymous

we are allowed to draw on the test

19. anonymous

@savac REMEMBER THIS |dw:1360758381571:dw|

20. anonymous

although the identity was proven using pyth.

21. anonymous

whats that?

22. anonymous

ACTS!!!! |dw:1360758433801:dw|

23. anonymous

ohhhh thats smart

24. anonymous

savac PeterPan is doing a great job

25. anonymous

@mathsmind Thanks for the kudos! Much appreciated :)

26. anonymous

So back to this... |dw:1360758603756:dw| Can you now determine the rest of the trig functions of this theta?

27. anonymous

give me a mint

28. anonymous

min

29. anonymous

:D I was about to say something clever, but you beat me to it :)

30. anonymous

REMEMBER In this particular quadrant, only sine (and cosecant) are positive

31. anonymous

lol sorry running low on sleep

32. anonymous

ok i got sin=$\sqrt{56}$/9 tan=-$\sqrt{56}$/5 csc=9/$\sqrt{56}$ sec=-9/5 cot=-5/$\sqrt{56}$

33. anonymous

34. anonymous

cos=-5/9

35. anonymous

And there you have it :)

36. anonymous

that was rright?

37. anonymous

Well yeah, unless you want to make it $\huge \sqrt{56}=2\sqrt{14}$

38. anonymous

wow i surprised myself. lol thank you very much

39. anonymous

Just place that right triangle in the correct quadrant, and you can't go wrong :) You'll notice that the horizontal bit was to the left of the x-axis, so strictly speaking, that's a -5, not a 5 :D

40. anonymous

thats what i thought it was

41. hartnn

actually since theta is in 2nd quadrant, that is its measure is between 90 to 180, the triangle will be , |dw:1360722028091:dw|

42. anonymous

Looks more complicated.

43. hartnn

but the way its solved and answers are correct.

44. hartnn

sin and cosec will be positive, all other ratios will be negative.

45. anonymous

Yay :)

46. hartnn

if you want to solve using diagram, then what i have drawn must be used . here, its solved using identities, which is altogether different method and doesn't need drawing triangle.

47. anonymous

Identities are harder to swallow, and they taste kinda awful too :(