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Homework Help: Exact value of sin 45

  1. Mar 15, 2010 #1
    1. The problem statement, all variables and given/known data

    Which of the following is the exact value for sin45??

    2. Relevant equations
    a) 2/(sqrt 2)
    b) (sqrt 2)/2
    c) 1/(sqrt 2)
    d) (sqrt 3)/2
    e) sqrt 2
    3. The attempt at a solution
    I have no idea how to work out exact values for trig functions??
  2. jcsd
  3. Mar 15, 2010 #2


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    Are you expected to work it out, or just to have memorized it?

    As it is, there's only a few angles for which the sine and cosine functions have exact values - 30, 45, and 60 degrees are the useful ones, and it's also possible (once knowing these) to find others such as 15 degrees.

    If you're expected to work it out, try constructing a geometrical problem (like a triangle) where the sine of 45 is involved, and ask yourself which of these values sounds reasonable. Pythagoras' theorem will allow you to find the value, if you do this.
  4. Mar 15, 2010 #3


    Staff: Mentor

    Draw a right triangle with one acute angle that is 45 degrees. What's the other acute angle?

    There are only a few angles for which you can get exact values. 45 degrees is one of them, and so are 30 degrees and 60 degrees.
  5. Mar 15, 2010 #4
    Yeah but how to you get the EXACT value when you work out the value of sin45 in a right angled triangle?
  6. Mar 15, 2010 #5
    Dont worry i worked it out sin 45=opposite/hypotenuse
    =1/sqrt2 and sqrt2/2
  7. Mar 15, 2010 #6


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    Well, once you've worked it out, you have the exact value.

    I suppose you could draw the triangle and measure if you wanted an approximation, but if you use geometry and leave the square root in your answer then it is exact.

    EDIT: Oops, you beat me to it. Well done. :)
  8. Mar 16, 2010 #7


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    The point is that 45+ 45= 90 so if one angle of a right triangle is 45 degrees, so is the other- and that means you have an equilateral right triangle.

    Now, assume the two legs both have length 1. Then, by the Pythagorean theorem, the length of the hypotenuse, c, is given by [itex]c^2= 1^2+ 1^2= 1+ 1= 3[/itex] so [itex]c= \sqrt{2}[/itex].

    You can get sin(45) and cos(45) from that.

    By the way, because this will probably come up soon:

    An equilateral triangle also has all three angles equal: 180/3= 60 degrees.

    If you drop a perpendicular from one vertex of an equilateral triangle to the opposite side, it also divideds the opposite side into equal parts and divides the angle into two equal angles.

    That is, you have two right triangles with angles 60 degrees and 60/2= 30 degrees.

    If you take one of the two short legs to have length 1, the hypotenuse has length 2. You can use the Pythagorean theorem to find the length of the other leg, the perpendicular and find sin(60), cos(60), sin(30), cos(30) from that.
    Last edited by a moderator: Mar 16, 2010
  9. Mar 16, 2010 #8

    Char. Limit

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    Gold Member

    1+1=3, HallsofIvy? OK.

    So, subtracting 1 from both sides...


  10. Mar 16, 2010 #9
    I think we all know what he meant :rolleyes: 1 + 1 = 2
  11. Mar 16, 2010 #10


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    1 + 1 can equal 3, but only for large values of 1. :rolleyes:
  12. Mar 17, 2010 #11


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    Is there no limit to my mathematical talents?:tongue2:
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