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sin60° as a fraction

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    Can someone please help me on this question. I need to write sin60° as a fraction, but can't work out how. I know that it involves knowing that sin30° is 1/2, but I don't know where to go from there - please help!
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    sin60° = √3/2
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    use this diagram to help you:


    bearing in mind that this is an equilateral triangle.
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    I' really sorry for being so dumb (maybe its tiredness) but I still don't get it. Could you maybe explain it in words (I don't learn well from diagrams)
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    Yeah it's just one of those things you need to know, apparently.

    I have no idea how I'm going to remember the different values for sin cos and tan of 30, 45 and 60, I'm doomed I tell ya.

    [EDIT] In visesh's diagram remember sohcahtoa. Sin = Opp / Hyp. As for where the diagram itself comes from, it's just one of two triangles you need to have burned into the back of you brain. I'm sure if visesh is kind enough, he'll post the other one for us.
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    sin (A+B)=sin A cos B + cos A sin B

    sin 60°= sin (30°+30°)
    = sin 30° cos 30° + cos 30° sin 30°
    = 2 sin 30° cos 30°
    = 2 (1/2) ((√3)/2)
    = (√3)/2
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    (Original post by wacabac)
    sin (A+B)=sin A cos B + cos A sin B

    sin 60°= sin (30°+30°)
    = sin 30° cos 30° + cos 30° sin 30°
    = 2 sin 30° cos 30°
    = 2 (1/2) ((√3)/2)
    = (√3)/2
    Well if you know that cos30 = root(3)/2 then you can just use:

    sin(90 - x) = cosx

    which incidently is proved by your identity.
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    OK then, so what about cos45°?
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    Don't try understanding it! Just learn them...

    Look:

    sin30 = 1/2
    sin60 = √3/2
    cos60 = 1/2
    cos30 = √3/2
    tan30 = 1/√3
    tan60 = √3

    You want more? ...
    --------------
    (Original post by ~Bex~)
    OK then, so what about cos45°?
    cos45° = 1/√2
    sin45° = 1/√2
    tan45° = 1
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    AC = √ 3 because it is a non-hypotenuse side of a triangle with sides 1 and hypotenuse 2, and √ (2²-1²) = √3 by pythagoras.

    BC = 1 because AB = 2 and the white + red triangles together make an equilateral triangle.

    angle BAC = 30 degrees for the same reason.


    In order to work out sin 60, use AC as the hypotenuse and AB as the opposite side.
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    I think that you might want to upgrade your calculator. Casio now offer "Natural Display" calculators (allowed in exams) that will show root(3)/2 (but it shows it how you would write it) when you key in sin60, and all of the other ones listed above.

    See the product page on the casio website.

    http://www.casio.co.uk/prod/product.asp?ID=1749

    I bought mine for about £5 at WHSmith.
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    (Original post by Joe_87)
    I have no idea how I'm going to remember the different values for sin cos and tan of 30, 45 and 60, I'm doomed I tell ya.
    You could get away with just learning sin, the cos values are just sin values backwards, and tan is just sin/cos, or just remember:

    Angle------0°----------30°-------45°----------60°--------90°
    Sin--------0-----------1/2------1/(√2)-------(√3)/2--------1
    Cos--------1---------(√3)/2-----1/(√2)--------1/2---------0
    Tan--------0---------1/(√3)-------1-----------√3----------∞
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    (Original post by samd)
    I think that you might want to upgrade your calculator. Casio now offer "Natural Display" calculators (allowed in exams) that will show root(3)/2 (but it shows it how you would write it) when you key in sin60, and all of the other ones listed above.

    See the product page on the casio website.

    http://www.casio.co.uk/prod/product.asp?ID=1749

    I bought mine for about £5 at WHSmith.

    This is for C1 (non calc)
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    (Original post by wacabac)
    You could get away with just learning sin, the cos values are just sin values backwards, and tan is just sin/cos, or just remember:

    0° 30° 45° 60° 90°
    Sin 0 1/2 1/(√2) (√3)/2 1
    Cos 1 (√3)/2 1/(√2) 1/2 0
    Tan 0 1/(√3) 1 √3 ∞
    I just used the triangle...i have a better "visual" memory than a memory for letters and numbers
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    (Original post by ~Bex~)
    This is for C1 (non calc)
    Ruined.

    I'd still recommend it for the later modules.
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    What??? There isn't any trig in C1 is there? Not for Edexcel anyway.
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    sin75° = (√3 + 1)/2√2
    cos75° = (√3 - 1)/2√2
    tan75° = (√3 + 1)/(√3 - 1)

    sin15° = (√3 - 1)/2√2
    cos15° = (√3 + 1)/2√2
    tan15° = (√3 - 1)/(√3 + 1)

    Anymore? ...
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    (Original post by Joe_87)
    What??? There isn't any trig in C1 is there? Not for Edexcel anyway.
    Ruined.
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    OK, now can people maybe help me with another question on the same topic?

    Having found that tanB=1.732, I need to show that:
    1+tan²B=sec²B
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    simple.

    sin²x+cos²x=1

    divide through by cos²x
 
 
 
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