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Exact value of tan10?

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    How do you go about finding the exact value of tan10?

    I know the 2 triangles for values such as sin45, sin 30, sin60 and the ones for cosine, but how do you find exact values for tan?

    Thanks!
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    Chances are it's irrational, so the exact value is tan10, or whatever letter you assign for tan10.

    You should also stress that you're working in degrees too.
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    (Original post by Manitude)
    Chances are it's irrational, so the exact value is tan10, or whatever letter you assign for tan10.

    You should also stress that you're working in degrees too.
    Ah ok.

    How would I go about solving this then:

    10. A pilot plans to fly an aeroplane from his base to another airport 300 km due north. His aeroplane can travel at a speed of 170 km/h relative to the air.

    (b) After flying for one hour he discovers that he is only 153 km from his starting point and has travelled along a line bearing 010◦. As- suming the wind has been steady throughout the journey calculate its speed. [5]

    any ideas?
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    (Original post by Benniboi1)
    Ah ok.

    How would I go about solving this then:

    10. A pilot plans to fly an aeroplane from his base to another airport 300 km due north. His aeroplane can travel at a speed of 170 km/h relative to the air.

    (b) After flying for one hour he discovers that he is only 153 km from his starting point and has travelled along a line bearing 010◦. As- suming the wind has been steady throughout the journey calculate its speed. [5]

    any ideas?
    Draw a diagram! Remember that the 170 km/h is relative to the air, which could be on a funny angle compared to his flight direction.

    Also you have to remember the difference between a scalar and a vector. speed is a scalar - it only has a value but no direction. The 153km has value and direction (bearing 010).
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    (Original post by Benniboi1)
    How do you go about finding the exact value of tan10?

    I know the 2 triangles for values such as sin45, sin 30, sin60 and the ones for cosine, but how do you find exact values for tan?

    Thanks!
    Based on the info given I will assume you picked 10 as an example.
    You can, quite easily find tan15 or tan75 for example using an additional formulae or half angle identities.
    I kinda get the impression though that is not what you are asking.
    Do you know what the definition of tan and how you can use the info you already know to find the same values for tan as you have with sin and cos?
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    (Original post by Manitude)
    Draw a diagram! Remember that the 170 km/h is relative to the air, which could be on a funny angle compared to his flight direction.

    Also you have to remember the difference between a scalar and a vector. speed is a scalar - it only has a value but no direction. The 153km has value and direction (bearing 010).
    Surely then the wind could be a range of different directions with different speeds that would knock him off course by this much?

    by the question saying find the speed and not velocity, how can I work out the direction and magnitude of the wind if I do not know one of the 2?
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    (Original post by MissBrown)
    Based on the info given I will assume you picked 10 as an example.
    You can, quite easily find tan15 or tan75 for example using an additional formulae or half angle identities.
    I kinda get the impression though that is not what you are asking.
    Do you know what the definition of tan and how you can use the info you already know to find the same values for tan as you have with sin and cos?
    I don't think I know how to work out tan15 and 75, I can do 30, 45 and 60.

    It wasn't what I was asking but I want to know now anyway! :rolleyes:

    so yeah I can find 30, 45 and 60 for all 3 but I cant find any other angles for any of Sin, Cos or Tan, any links to a website that explains it would be helpful!
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    (Original post by Benniboi1)
    I don't think I know how to work out tan15 and 75, I can do 30, 45 and 60.

    It wasn't what I was asking but I want to know now anyway! :rolleyes:

    so yeah I can find 30, 45 and 60 for all 3 but I cant find any other angles for any of Sin, Cos or Tan, any links to a website that explains it would be helpful!
    tan(15) = tan(45-30)
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    (Original post by Benniboi1)
    Ah ok.

    How would I go about solving this then:

    10. A pilot plans to fly an aeroplane from his base to another airport 300 km due north. His aeroplane can travel at a speed of 170 km/h relative to the air.

    (b) After flying for one hour he discovers that he is only 153 km from his starting point and has travelled along a line bearing 010◦. As- suming the wind has been steady throughout the journey calculate its speed. [5]

    any ideas?
    Your question does not state that you cannot use a calculator nor does it state that you have to use an exact value for tan(10). Tan(10degrees)= 0.176326.......
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    You can't find an exact value for tan(10) in terms of surds as the octadecagon is non-constructible.
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    (Original post by gdunne42)
    Your question does not state that you cannot use a calculator nor does it state that you have to use an exact value for tan(10). Tan(10degrees)= 0.176326.......
    It's a non-calculator paper, the question just made me wonder whether it was possible to calculate other values of sin, cos and tan
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    Well I guess there is one way you could go about it if you really wanted to find an exact value:

    (cos 10+isin10)^3=cos30+isin30

    (cos10)^3+3i(cos10)^2sin10-3cos10(sin10)^2-i(sin10)^3=0.5sqrt3+0.5i

    thus (cos 10)^3-3(cos10)(sin10)^2=0.5sqrt3 and 3(cos10)^2sin10-(sin10)^3=0.5

    For the first equation if we simplify and let u=cos 10 we get 4u^3-3u=0.5sqrt3

    and for the second if you let u=sin10 then 3u-4u^3=0.5

    Not too familiar with how you'd get an exact value from those equations as I don't know the cubic equation(s) too well :/

    Either way its too hairy to be of use to you :tongue:
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    (Original post by Benniboi1)
    It's a non-calculator paper, the question just made me wonder whether it was possible to calculate other values of sin, cos and tan
    Multiples of 3 degrees can be expressed in terms of surds (plus heaps of non-integer angles).
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    (Original post by Jtking3000)
    Well I guess there is one way you could go about it if you really wanted to find an exact value:

    (cos 10+isin10)^3=cos30+isin30

    (cos10)^3+3i(cos10)^2sin10-3cos10(sin10)^2-i(sin10)^3=0.5sqrt3+0.5i

    thus (cos 10)^3-3(cos10)(sin10)^2=0.5sqrt3 and 3(cos10)^2sin10-(sin10)^3=0.5

    For the first equation if we simplify and let u=cos 10 we get 4u^3-3u=0.5sqrt3

    and for the second if you let u=sin10 then 3u-4u^3=0.5

    Not too familiar with how you'd get an exact value from those equations as I don't know the cubic equation(s) too well :/

    Either way its too hairy to be of use to you :tongue:
    wow! i'd love to try and calculate that in the middle of a 2 hour exam! :eek:
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    (Original post by Mr M)
    Multiples of 3 degrees can be expressed in terms of surds (plus heaps of non-integer angles).
    ohh that helps

    any ideas how you would do the question I posted above without a calculator?
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    (Original post by Jtking3000)
    Well I guess there is one way you could go about it if you really wanted to find an exact value:

    (cos 10+isin10)^3=cos30+isin30

    (cos10)^3+3i(cos10)^2sin10-3cos10(sin10)^2-i(sin10)^3=0.5sqrt3+0.5i

    thus (cos 10)^3-3(cos10)(sin10)^2=0.5sqrt3 and 3(cos10)^2sin10-(sin10)^3=0.5

    For the first equation if we simplify and let u=cos 10 we get 4u^3-3u=0.5sqrt3

    and for the second if you let u=sin10 then 3u-4u^3=0.5

    Not too familiar with how you'd get an exact value from those equations as I don't know the cubic equation(s) too well :/

    Either way its too hairy to be of use to you :tongue:
    More directly tan(10) is one of the solutions of 3x^6-27x^4+33x^2-1=0 which is a disguised cubic.
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    (Original post by Mr M)
    More directly tan(10) is one of the solutions of 3x^6-27x^4+33x^2-1=0 which is a disguised cubic.
    Thats actually pretty cool :O where does that equation come from?
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    (Original post by Jtking3000)
    Thats actually pretty cool :O where does that equation come from?
    Do you know about the roots of cubic equations?
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    Derive the triple-angle formulas for sin(\theta) and cos(\theta) and use them twice. Which is exactly what mr.M has done I think to get that equation (or something to that effect anyway)
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    (Original post by Tobedotty)
    Which is exactly what mr.M has done I think to get that equation (or something to that effect anyway)
    Nope. I just used my knowledge of obscure trig identities (I like regular polygons).

    \alpha=\tan^2 10

    \beta=\tan^2 50

    \gamma=\tan^2 70

    Play around with \sum \alpha, \sum \alpha \beta and \alpha \beta \gamma

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