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How is sine of 90 degree possible

  1. Apr 27, 2012 #1
    I don't know how can a triangle be formed with two angles each 90 degree. If it is not possible, then how do we calculate sin 90 degree = 1 ?
     
  2. jcsd
  3. Apr 27, 2012 #2
    There's a more general definition of sin and cosine.

    Consider the unit circle, and imagine a ray emanating from the origin making an angle of [itex]\theta[/itex] with the positive x-axis. You can see right away that this is perfectly well defined for all angles.

    Now the ray intersects the unit circle at a point (x,y). We call the x-coordinate [itex]cos \ \theta[/itex], and we call the y-coordinate [itex]sin \ \theta[/itex].

    You can (and should) convince yourself that this definition corresponds to the usual triangle-based definitions from trig; and that sin is now perfectly well defined for 90 degrees.

    One nice feature of this way of defining the trig functions is that you don't have to memorize anything about quadrants. You just read off the x-y coordinates for any angle, regardless of quadrant.
     
  4. Apr 27, 2012 #3
    300px-Unit_circle_angles_color.svg.png
    My favorite presentation of the above facts.
     
  5. Apr 29, 2012 #4
    But, this circle shows sin and cos only. What about tan, cot, sec, cosec ?
     
  6. Apr 29, 2012 #5

    Office_Shredder

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    Those can all be defined in terms of sin and cos, for example tan(x)=sin(x)/cos(x)
     
  7. Apr 29, 2012 #6

    arildno

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    You can readily extend that diagram to find length segments that equal those other trig values.
    For example, draw a vertical line tangent to the circle at the intersection point with the horizontal axis.
    Now, if you are to find tan(a), where "a" is some angle made to the horizontal axis and the hypotenuse, just extend that hypotenuse line until it intersects with the vertical line just drawn.
    The length of the vertical line segment as defined between A) the tangent point with the circle and B) the intersection point with the hypotenuse line, equals tan(a).*
    (The length of the hypotenuse line segment in this construction is sec(a), by the way)

    This is actually WHY the trig function "tangent" is called "tangent" in the first place. :smile:
    (In order to find the co-tangent, you proceed similarly by drawing a HORIZONTAL line that is tangent to the circle at the y-axis.)

    * Try to figure out WHY this must be so, using your ideas about how the tangent to an angle is defined for a right-angled triangle!
     
    Last edited: Apr 29, 2012
  8. Apr 29, 2012 #7
    OK, trigonometry is clear. Please tell why is it so ?

    n(n-1)(n-2) ..... (n-r+1) = n!/(n-r)!

    Thanks in advance....
     
  9. Apr 29, 2012 #8

    Mentallic

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    [tex]\frac{n!}{(n-r)!}=\frac{n(n-1)(n-2)...(n-r+1)(n-r)(n-r-1)...3.2.1}{(n-r)(n-r-1)...3.2.1}[/tex]

    Is it now clear why that equality holds?
     
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