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    Hi. I have a task: Using exact ratios, show that
    sin^2 60*sec60* - cosec60*= (9-4√3)/6

    Could please explain what "Using exact ratios" means?

    Cheers
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    (Original post by Brutal)
    Hi. I have a task: Using exact ratios, show that
    sin^2 60*sec60* - cosec60*= (9-4√3)/6

    Could please explain what "Using exact ratios" means?

    Cheers
    Trigonometric values for angles 30, 45, 60, 90... Can be found exactly by a drawing of an equilateral or right isosceles triangle and using Pythagoras.
    Clue: Find the vertical height of an equilateral triangle of side unit length.
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    (Original post by Brutal)
    Hi. I have a task: Using exact ratios, show that
    sin^2 60*sec60* - cosec60*= (9-4√3)/6

    Could please explain what "Using exact ratios" means?

    Cheers
    sin 60 = sqrt(3)/2, cos 60 = 1/2. Thus the expression becomes:
    3/4 * 2 - 2/sqrt(3)
    = 3/2 - 2sqrt(3)/3 by rationalising the denominator
    = (9-4sqrt(3))/6 since 3/2 = 9/6 and 2sqrt(3)/3 = 4sqrt(3)/6.
    QED
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    Don't even get me started on exact trig values. They really interest me for some reason. You can find all sine values for integer angles (degrees). Even  \sin 1^{\circ} can be found exactly and from there you can find them all.
    But yeah for these questions just use the exact value of sin(60) and cos(60) and you'll have to rationalise the denominator along the way until you get it into the very attractive form on the right hand side as you posted.
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    (Original post by Ano123)
    Don't even get me started on exact trig values. They really interest me for some reason. You can find all sine values for integer angles (degrees). Even  \sin 1^{\circ} can be found exactly and from there you can find them all.
    But yeah for these questions just use the exact value of sin(60) and cos(60) and you'll have to rationalise the denominator along the way until you get it into the very attractive form on the right hand side as you posted.
    Have you read through ptolemy's construction of trigonometric tables. I believe that's where we get Ptolemy's Theorem from.
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    (Original post by EricPiphany)
    Have you read through ptolemy's construction of trigonometric tables. I believe that's where we get Ptolemy's Theorem from.
    I have briefly.
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    Thanks to all. This was very helpful
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    Brutal EricPiphany



    The Right angled triangle on the left is one half of an equilateral triangle.
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    (Original post by Ano123)
    Even  \sin 1^{\circ} can be found exactly.
    Here it is.



    Unfortunately, the answer involves the cube roots of complex numbers. If you are looking for 'nice' algebraic forms for trigonometric functions in degrees you need to stick to the multiples of 3.
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    (Original post by Mr M)
    Here it is.



    Unfortunately, the answer involves the cube roots of complex numbers. If you are looking for 'nice' algebraic forms for trigonometric functions in degrees you need to stick to the multiples of 3.
    It's beautiful.
 
 
 
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