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savac

  • 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

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

  2. PeterPan
    • 3 years ago
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    Cosine is adj/hyp, right?

  3. savac
    • 3 years ago
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    yes

  4. PeterPan
    • 3 years ago
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    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. savac
    • 3 years ago
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    i understand

  6. PeterPan
    • 3 years ago
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    Now, what's the length of this missing side... |dw:1360758005338:dw| (Hint: Pythagorean theorem)

  7. savac
    • 3 years ago
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    \[\sqrt{106}\] right?

  8. PeterPan
    • 3 years ago
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    Remember, 9 is your hypotenuse When you have a^2 + b^2 = c^2 9 is your c :)

  9. mathsmind
    • 3 years ago
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    \[\cos ^2(\theta)+\sin^2(\theta)=1\]

  10. mathsmind
    • 3 years ago
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    \[\sin^2(\theta) = 1-\cos^2(\theta)\]

  11. savac
    • 3 years ago
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    \[\sqrt{56}\]

  12. mathsmind
    • 3 years ago
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    \[\sin^2(\theta)=(\frac{- 5 }{ 9 })^2 -1\]

  13. mathsmind
    • 3 years ago
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    i prefer the method provided by @PeterPan, because it shows you the inside out

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

  15. PeterPan
    • 3 years ago
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    @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. mathsmind
    • 3 years ago
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    yes that is what i said

  17. mathsmind
    • 3 years ago
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    i said i prefer the other method to know the inside out

  18. savac
    • 3 years ago
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    we are allowed to draw on the test

  19. PeterPan
    • 3 years ago
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    @savac REMEMBER THIS |dw:1360758381571:dw|

  20. mathsmind
    • 3 years ago
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    although the identity was proven using pyth.

  21. savac
    • 3 years ago
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    whats that?

  22. PeterPan
    • 3 years ago
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    ACTS!!!! |dw:1360758433801:dw|

  23. savac
    • 3 years ago
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    ohhhh thats smart

  24. mathsmind
    • 3 years ago
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    savac PeterPan is doing a great job

  25. PeterPan
    • 3 years ago
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    @mathsmind Thanks for the kudos! Much appreciated :)

  26. PeterPan
    • 3 years ago
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    So back to this... |dw:1360758603756:dw| Can you now determine the rest of the trig functions of this theta?

  27. savac
    • 3 years ago
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    give me a mint

  28. savac
    • 3 years ago
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    min

  29. PeterPan
    • 3 years ago
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    :D I was about to say something clever, but you beat me to it :)

  30. PeterPan
    • 3 years ago
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    REMEMBER In this particular quadrant, only sine (and cosecant) are positive

  31. savac
    • 3 years ago
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    lol sorry running low on sleep

  32. savac
    • 3 years ago
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    ok i got sin=\[\sqrt{56}\]/9 tan=-\[\sqrt{56}\]/5 csc=9/\[\sqrt{56}\] sec=-9/5 cot=-5/\[\sqrt{56}\]

  33. PeterPan
    • 3 years ago
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    What about cos?

  34. savac
    • 3 years ago
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    cos=-5/9

  35. PeterPan
    • 3 years ago
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    And there you have it :)

  36. savac
    • 3 years ago
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    that was rright?

  37. PeterPan
    • 3 years ago
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    Well yeah, unless you want to make it \[\huge \sqrt{56}=2\sqrt{14}\]

  38. savac
    • 3 years ago
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    wow i surprised myself. lol thank you very much

  39. PeterPan
    • 3 years ago
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    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. savac
    • 3 years ago
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    thats what i thought it was

  41. hartnn
    • 3 years ago
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    actually since theta is in 2nd quadrant, that is its measure is between 90 to 180, the triangle will be , |dw:1360722028091:dw|

  42. PeterPan
    • 3 years ago
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    Looks more complicated.

  43. hartnn
    • 3 years ago
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    but the way its solved and answers are correct.

  44. hartnn
    • 3 years ago
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    sin and cosec will be positive, all other ratios will be negative.

  45. PeterPan
    • 3 years ago
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    Yay :)

  46. hartnn
    • 3 years ago
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    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. PeterPan
    • 3 years ago
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    Identities are harder to swallow, and they taste kinda awful too :(

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