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Trigonometry problem

  1. Jan 22, 2016 #1
    1. The problem statement, all variables and given/known data

    What is the value of ## \frac{1}{cos^2 10^{\circ}}+\frac{1}{cos^2 20^{\circ}}+\frac{1}{cos^2 40^{\circ}}-\frac{1}{cos^2 45^{\circ}} ## ?

    A. 1
    B. 5
    C. 10
    D. 15
    E. 20

    2. Relevant equations

    cos^2 x + sin^2 x = 1
    sin 2x = 2 sin x cos x
    cos 2x = cos^2 x-sin^2 x

    3. The attempt at a solution

    I'm only able to determine cos 45 degrees which is √2 / 2
    I have no idea about cos 10, cos 20, and cos 40 degrees

    Please help me
     
    Last edited: Jan 22, 2016
  2. jcsd
  3. Jan 22, 2016 #2

    Nidum

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    Are you just expected to look up the cos values in the tables or is an analytic solution being asked for ?

    Have you done any basic trigonometry in your classes ?
     
    Last edited: Jan 22, 2016
  4. Jan 22, 2016 #3
    No calculators or tables are allowed in the test.
    I have to do it using trigonometric identities, and formulas
     
  5. Jan 22, 2016 #4

    fresh_42

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    Since it's a multiple choice test you only have to estimate the answer. So how far do you think are cos 40° and cos 45° apart? And where do cos 10° and cos 20° lie? Do you know the graph of cos? Are you sure about the possible answers?
     
  6. Jan 22, 2016 #5

    Samy_A

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    Are you supposed to give the range of that value? As in "between C and D" (just as an example).
    Because none of these five values are anywhere close.

    That works really good to get an estimate of the value.
     
  7. Jan 22, 2016 #6
    Hmm.. you're right.
    It has no right option (in calculator I get 1.86)..
    Sorry for posting this nonsense problem :(
     
  8. Jan 23, 2016 #7

    haruspex

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    Pity it is garbled, there is an interesting puzzle lurking behind it.
    cos(40), cos(80) and -cos(20) are the three roots of 8x3-6x+1. If you change the cos(10) to cos(80) the sum gives 34.
     
  9. Jan 23, 2016 #8

    haruspex

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    Ok, I've found a correction that makes the problem work. Leave the 10 as is but change the 20 and 40 to 50 and 70. That yields one of the given answers.
     
  10. Jan 23, 2016 #9

    SammyS

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    Very good haru !

    Of course, that's the same as ##\displaystyle \ \frac{1}{\cos^2 10^{\circ}}+\frac{1}{\sin^2 20^{\circ}}+\frac{1}{\sin^2 40^{\circ}}-\frac{1}{\cos^2 45^{\circ}} \ ##.
     
  11. Jan 23, 2016 #10

    haruspex

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    How true.
    Terry, as a clue on how to approach such a problem...
    Note that the given angles are awkward in the sense that they are not multiples of 3 degrees. So we need to relate them to angles which are. A very useful formula is ##\cos(3\theta)=4\cos^3(\theta)-3\cos(\theta)##. There's a similar one for sine.
    Since the given angles are multiples of 10 degrees, that gives you cubic equations relating their cosines to well known cosines.
    Using the corrected angles, you discover that all three satisfy the same cubic.
    The next trick is to consider how the sum of the inverse squares of the roots of a cubic relates to the coefficients in the cubic.
     
  12. Jan 24, 2016 #11
    Okay, so now the problem becomes

    ##
    \frac{1}{cos^210^ \circ}+\frac{1}{cos^250^ \circ}+\frac{1}{cos^270^ \circ}-\frac{1}{cos^245^ \circ}
    ##

    I still get stuck..
    How do I use the formula ## cos (3\theta) = 4 cos ^3 (\theta) - 3 cos (\theta) ## ??

    Neither 50 nor 70 are the triple of 10

    Please tell me the steps to solve it, since I don't have any idea at all
     
  13. Jan 24, 2016 #12

    Samy_A

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    cos(3*50°)=?
    cos(3*70°)=?
     
  14. Jan 25, 2016 #13
    cos (3*50) = 4 cos^3(50) - 3 cos(50)
    - √3 / 2 = 4 cos^3(50) - 3 cos(50)

    cos(3*70°)= 4 cos^3(70) - 3 cos(70)
    - √3 / 2 = 4 cos^3(70) - 3 cos(70)

    cos (3*10) = 4 cos^3(10) - 3 cos(10)
    √3 / 2 = 4 cos^3(10) - 3 cos(10)

    Then, I still don't get the idea..
    How to get cos(50), cos(70) and cos(10) ? It's a cubic function..
    And, at first I think cos(50) equals to cos(70) due to the same cos(3x), but they're different..
    Please help
     
  15. Jan 25, 2016 #14

    Samy_A

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    The idea is not to solve the cubic equation, but to use properties of sums/products (and combinations thereof) of its roots.
    I made use of Vieta's formulas, and some more computations. There may well be an easier way that escaped me.
     
  16. Jan 25, 2016 #15

    haruspex

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    Yes, in my post #30 I missed a step out.
    You need the cosines to be the three roots of the same cubic, but the three cubics you found are not quite the same.
    Since the cosines are all squared in the target formula, cos(50) will be equivalent to -cos(50)=cos(130) etc. That works, because 3*130=390=360+30, giving the same cubic as for cos(10). Do the same to find a substitute for 70.
    As you say, the cosines of 10, 130 and the third angle are all different, but they satisfy the same cubic, so they must be the three roots of that cubic.
    Having done that, our target expression, the sum of the inverses of the squares of those cosines, is a-2+b-2+c-2 where a, b and c are all roots of the cubic 4x3-3x-cos(30)=0.
    In general, if a cubic is px3+qx2+rx+s=0, there are several simple formulae relating the roots to the coefficients. s/p=abc, for example. Using these, can you find a way to express a-2+b-2+c-2 in terms of p, q, r and s?
     
  17. Jan 26, 2016 #16

    cos (3*50) = 4 cos^3(50) - 3 cos(50)
    - √3 / 2 = 4 cos^3(50) - 3 cos(50)

    cos(3*70)= 4 cos^3(70) - 3 cos(70)
    - √3 / 2 = 4 cos^3(70) - 3 cos(70)

    cos(3*170)= 4 cos^3(170) - 3 cos(170)
    - √3 / 2 = 4 cos^3(170) - 3 cos(170)

    4x^3 - 3x + √3 / 2 = 0
    Which means that I can apply Vieta's Formula to get all the sums or products of cos(50), cos(70), or cos(170)

    cos (3*10) = 4 cos^3(10) - 3 cos(10)
    √3 / 2 = 4 cos^3(10) - 3 cos(10)

    cos (3 *110) = 4 cos^3(110) - 3 cos(110)
    √3 / 2 = 4 cos^3(110) - 3 cos(110)

    cos (3*130) = 4 cos^3(130) - 3 cos(130)
    √3 / 2 = 4 cos^3(130) - 3 cos(130)

    4x^3 - 3x - √3 / 2 = 0
    Which means that I can apply Vieta's Formula to get all the sums or products of cos(10), cos(110), or cos(130)

    But, cos 10, cos 50, and cos 70 is not roots of the same equation.. I can't get the products nor the sums using Vieta..
    Please help.. I still don't get it
     
  18. Jan 26, 2016 #17

    Samy_A

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    I'm confused too.

    You had (I clipped part of the post):
    Look at the cubic equation 4x³-3x+√3 / 2=0. What you have are the three roots of that equation: cos(50), cos(70) and -cos(10).
    Only squares appear in the expression to solve, so the minus in -cos(10) is not a problem.
     
  19. Jan 26, 2016 #18

    haruspex

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    As I posted
    So you can change the cos(50) and cos(70) in the target expression to -cos(50)=cos(130) and -cos(70)=cos(110).
     
  20. Jan 27, 2016 #19
    cos (3*10) = 4 cos^3(10) - 3 cos(10)
    √3 / 2 = 4 cos^3(10) - 3 cos(10)

    cos (3 *110) = 4 cos^3(110) - 3 cos(110) ....... (which is the same as -cos(70))
    √3 / 2 = 4 cos^3(110) - 3 cos(110)

    cos (3*130) = 4 cos^3(130) - 3 cos(130) ........ (which is the same as - cos(50))
    √3 / 2 = 4 cos^3(130) - 3 cos(130)

    4x^3 - 3x - √3 / 2 = 0

    cos 10 (- cos 50 )(- cos 70) = √3 / 8
    (cos 10 (- cos 50 )(- cos 70) )^2 = 3/64

    Since (cos 50)^2 = (-cos 50)^2 and (cos 70)^2 = (-cos 70)^2, then

    cos^2 10 cos^2 50 + cos^2 50 cos^2 70 + cos^2 70 cos^2 10
    = cos^2 10 (- cos 50)^2 + (-cos 50)^2 (-cos 70)^2 + (-cos 70)^2 cos^2 10
    = (- cos 10 cos 50 + cos 50 cos 70 - cos 70 cos 10)^2 - 2 (- cos 10 cos 50 + cos 50 cos 70 - cos 70 cos 10)
    = (-3/4)^2 - 2(-3/4)
    = (9/16) + (3/2)
    = (33/16)

    So the answer I get is

    ## \frac{\frac{33}{16}}{\frac{3}{64}} - \frac{1}{cos^2(45^\circ))} = 44 - 2 = 42 ##

    But, it doesn't yield any of the options :(
    Please tell me where I got wrong..
     
    Last edited: Jan 27, 2016
  21. Jan 27, 2016 #20

    Samy_A

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    Verify this formula.

    It may be easier just to work with symbols: compute (xy+yz+zx)² and see what it gives.

    And don't worry about the signs. In the end you will use Vieta's formulas, not the values of the roots.
     
    Last edited: Jan 27, 2016
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