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anonymous

  • one year ago

5(cos^2)60 + 4(sec^2)30 - (tan^2)45 = x Find x

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  1. freckles
    • one year ago
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    well cos(60) and sec(30) and tan(45) can all be evaluated using the unit circle

  2. freckles
    • one year ago
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    http://buildingaunitcircle.weebly.com/uploads/6/2/2/3/622393/9409299.gif

  3. anonymous
    • one year ago
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    Use the trig table........

  4. freckles
    • one year ago
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    do you need help using that chart above ?

  5. anonymous
    • one year ago
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    .... I need to the values like cos = 1/2 and all.......

  6. freckles
    • one year ago
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    cos(60)=1/2 that is correct so squaring both sides gives cos^2(60)=1/4

  7. freckles
    • one year ago
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    sec(30) means you need to find cos(30) and flip whatever number that is

  8. freckles
    • one year ago
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    tan(45) means you need to find sin(45) and cos(45) but both sin(45) and cos(45) are the same so tan(45)=1

  9. freckles
    • one year ago
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    \[5 \cos^2(60)+4\sec^2(30)-\tan^2(45) \\ 5(\frac{1}{4})+4(\frac{1}{\cos(30)})^2-1\]

  10. freckles
    • one year ago
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    so you still need to find cos(30)

  11. anonymous
    • one year ago
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    4/3 = sec^2(30)

  12. freckles
    • one year ago
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    \[\cos(30)=\frac{\sqrt{3}}{2} \\ \cos^2(30)=\frac{3}{4} \\ \frac{1}{\cos^2(30)}=\frac{4}{3} \\ \sec^2(30)=\frac{4}{3}\] that is right so far you have this now: \[5(\frac{1}{4})+4(\frac{4}{3})-1\]

  13. freckles
    • one year ago
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    that can be simplify just be doing a few multiplications and additions(subtractions)

  14. anonymous
    • one year ago
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    \[\frac{ 5 }{ 4 } +\frac{ 16 }{ 3 } - 1\]

  15. freckles
    • one year ago
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    right so far

  16. anonymous
    • one year ago
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    shall I take LCM?

  17. freckles
    • one year ago
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    yes which the lcm(4,3,1)=12

  18. freckles
    • one year ago
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    \[\frac{5}{4}+\frac{16}{3}-\frac{1}{1} \\ \frac{5(3)}{4(3)}+\frac{16(4)}{3(4)}-\frac{1(12)}{1(12)} \\ \frac{5(3)}{12}+\frac{16(4)}{12}-\frac{12}{12} \\ \frac{5(3)+16(4)-12}{12}\]

  19. anonymous
    • one year ago
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    67/12

  20. freckles
    • one year ago
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    looks awesome!

  21. freckles
    • one year ago
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    and totally correct :)

  22. anonymous
    • one year ago
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    Thanks! Please help me a bit more?

  23. freckles
    • one year ago
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    Sure. I can take a stab at it.

  24. triciaal
    • one year ago
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    @freckles would using identities just take longer?

  25. freckles
    • one year ago
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    @triciaal if you want to try something with identities you can

  26. anonymous
    • one year ago
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    \[\frac{ cotA-1 }{ 2 -\sec^2(A) } = \frac{ \cot A }{ 1 + tanA }\]

  27. freckles
    • one year ago
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    some people don't need to put in terms of sin and cos but I always do because I remember more identities with sin and cos

  28. anonymous
    • one year ago
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    Help me?

  29. freckles
    • one year ago
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    \[\frac{\frac{\cos(a)}{\sin(a)}-1}{2-\frac{1}{\cos^2(a)}} =\frac{\frac{\cos(a)}{\sin(a)}}{1+\frac{\sin(a)}{\cos(a)}}\] so this is what I would do as a first step second step would be to remove the compound fraction action going on

  30. anonymous
    • one year ago
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    I'm suppose to use LHS and get RHS

  31. freckles
    • one year ago
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    ok then get rid of the compound fraction action on the the left hand side

  32. anonymous
    • one year ago
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    \[\frac{ \cos(a) - \sin(a) }{ \sin(a) }\]

  33. anonymous
    • one year ago
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    and for the denominator 2sin^(a)/cos^2(a)

  34. freckles
    • one year ago
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    \[\frac{\frac{\cos(a)}{\sin(a)}-1}{2-\frac{1}{\cos^2(a)}} \cdot \frac{\sin(a) \cos^2(a)}{\sin(a)\cos^2(a)} \\ \frac{\cos^3(a)-\cos^2(a)\sin(a)}{2 \cos^2(a)\sin(a)-\sin(a)} \\ \text{ so doing a bit of factoring } \\ \frac{\cos^2(a)(\cos(a)-\sin(a))}{\sin(a)(2 \cos^2(a)-1)} \\ \\ \text{ now recall on the other side we wanted } \cot(a) \\ \text{ on \top } \\ \text{ well we notice } \frac{\cos(a)}{\sin(a)}=\cot(a) \\ \cot(a) \frac{\cos(a)(\cos(a)-\sin(a))}{2 \cos^2(a)-1} \\ \] now somehow we have to show that \[\frac{\cos(a)(\cos(a)-\sin(a))}{2 \cos^2(a)-1}=\frac{1}{1+\tan(a)}\] any ideas @BlackDranzer

  35. freckles
    • one year ago
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    if not read this if you want spoilers: \[\frac{\cos(a)(\cos(a)-\sin(a))}{2 \cos^2(a)-1} \cdot \frac{\cos(a)+\sin(a)}{\cos(a)+\sin(a)} \\ \frac{\cos(a)(\cos^2(a)-\sin^2(a))}{(2 \cos^2(a)-1)(\cos(a)+\sin(a))} \\ \text{ well recall double angle identity for } cosine \\ \frac{ \cos(a)}{\cos(a)+\sin(a)}\] now divide both top and bottom by cos(a)

  36. freckles
    • one year ago
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    that is I used \[\cos(2a)=\cos^2(a)-\sin^2(a) \\ \text{ and also } \cos(2a)=2\cos^2(a)-1\]

  37. freckles
    • one year ago
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    yep and we can bring down that cot(a) factor we had from earlier (though I normally write my fractions like this 1/(1+tan(a)) )

  38. freckles
    • one year ago
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    so the key in our proof was the use of double angle identity for cosine

  39. anonymous
    • one year ago
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    Nice.. Thanks... I need help in quadratic equations too...

  40. freckles
    • one year ago
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    ok let's see it

  41. freckles
    • one year ago
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    hey @triciaal did you come up with a way using identities for the first one ?

  42. freckles
    • one year ago
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    I honestly haven't given it a try

  43. anonymous
    • one year ago
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    An aero plane flying at 1 km horizontally above the ground make an angle of elevation 60 after 10 seconds its 30 find the speed of the aerplane in km/hr

  44. freckles
    • one year ago
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    wait what does that mean what is the angle of elevation 60 or 30?

  45. freckles
    • one year ago
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    and what is 30?

  46. anonymous
    • one year ago
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    from 60 to 30 in secs from the smae point

  47. freckles
    • one year ago
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    oh the angle of elevation changed from 60 to 30 in 10 sec gotcha

  48. freckles
    • one year ago
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    |dw:1436680212196:dw|

  49. freckles
    • one year ago
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    so to find the speed of the airplane over that 10 sec we will need to find the distance it traveled for both angle of elevations...

  50. freckles
    • one year ago
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    |dw:1436680529581:dw|

  51. freckles
    • one year ago
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    wait my drawing is off

  52. freckles
    • one year ago
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    I will just make two separate drawings |dw:1436680577539:dw| |dw:1436680738212:dw|

  53. freckles
    • one year ago
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    We need to find the distance the plane has flown for both pictures

  54. freckles
    • one year ago
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    notice that is x for the first drawing and B for the second drawing

  55. freckles
    • one year ago
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    then once you find the distances for both pictures you will find the difference of those distances and that resulting distance/10 secs will be the speed of the airplane in that 10 secs you have that change

  56. anonymous
    • one year ago
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    How to find the dist?

  57. freckles
    • one year ago
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    you can use a trig ratio

  58. freckles
    • one year ago
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    for example you are given the height which is 1km (the opposite side of the angle of elevation)

  59. freckles
    • one year ago
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    and you are wanting to find the length that airplane has flown for both (which is the adjacent side )

  60. freckles
    • one year ago
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    opp/adj is the tangent ratio

  61. freckles
    • one year ago
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    once you computer x and B (don't ask why I called them that; I wasn't thinking too much about the letters to use) \[\text{ the speed over that 10 sec change }=\frac{\text{ distance traveled in that 10 sec }}{ 10 \sec} \\ \frac{|x-B|}{10 \sec}\]

  62. anonymous
    • one year ago
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    Thanks. I will manage the rest...

  63. freckles
    • one year ago
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    when you are done with that computation your units should by km/sec but if you don't like how small the number is next to those units you can always covert the sec part to either minutes or hours

  64. freckles
    • one year ago
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    oh your question says to put it an hours

  65. freckles
    • one year ago
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    lol

  66. freckles
    • one year ago
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    just let me know if you need any further help with this one

  67. freckles
    • one year ago
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    or if you want me to check your answer

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