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Sine calcuation

  1. Aug 14, 2010 #1
    hi every one,

    i have one doubt i studied abt trignomentry. there finding the triangle angle or side of the triangle using sine function. if we are taking right angle triangle sine A = opp/hypo, cos A = adj/hypo and tan A=opp/adj. here we are finding angle for A only why we are having three formulas what is the use for that?. and how we are finding sin A. what is the procedure behind that for ex: sin 90 = 1 how this ans come? Pls anybody reply me.
     
  2. jcsd
  3. Aug 14, 2010 #2
    There are more than three formulas. There are six major ones, which are sine, cosine, and tangent, and the reciprocal identities cosecant, secant, and cotangent (respectively). Cosine and sine represent the x and y values (respectively) of a point that lies on a circle whose radius is exactly equal to the hypotenuse line formed by the unique opposite and adjacent lines required to form the trigonometric values. Tangent represents the slope of the hypotenuse. All six functions become very useful when you are studying calculus. For instance, if I were to take the derivative of the tangent function, I would write sec^2(x), which is the same as 1/cos^2(x). Or if I wanted to find the derivative of the sine function, I would write cos(x). Sine and cosine are absolutely essential to any serious, in-depth study of trigonometry.

    In the old days, mathematicians would copy down tables of trig values for as many angles as they could. They did this by using the identities of the trig functions. For instance, if they wanted to find out what sin(pi/4) was, they would construct a triangle whose opposite and adjacent sides were equal, and then divide the opposite by the resultant hypotenuse.

    Nowadays, we have calculators with algorithms for that. Thank goodness! Now we only have to memorize the values for pi/3, pi/4, pi/6, and pi/2.
     
  4. Aug 14, 2010 #3

    CRGreathouse

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    And don't forget the versine, the hacovercosine, the exsecant... thank goodness I'm not an olden-days sailor.
     
  5. Aug 16, 2010 #4
    thnks for reply
     
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