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Question about sine values for angles 0-180

  1. Apr 22, 2015 #1
    So my math teacher told me that as long as an angle is less than 90 degrees. The sine value of that angle will also be equal to 180 subtracted by that angle. Why is it this way? I just don't understand why it is true spesifically for sine values. Is there are a way to compare this to triangles in general? Would be really grateful for any answers, im struggling to understand this concept.
  2. jcsd
  3. Apr 22, 2015 #2
    Easiest way to understand it is to look at a unit circle, where the sine is the y-coordinate. From there it should be easy to see that [itex]\sin(x) = \sin(\pi - x)[/itex] not just for angles less than 90 degrees, but for all angles. Even better when you note that [itex]\sin(\pi - x)[/itex] is just a reflection across the y-axis.
  4. Apr 22, 2015 #3
    it is an identity, you can prove it by making one side of the equation look like the opposite side of the equation.
  5. Apr 22, 2015 #4
    If you are looking for a mathematical proof, a strict, rigorous proof, I'm afraid there isn't one, because it is just so by definition, I believe. If you ask why do we define it in such a way and not in any other ways, eg., in the same way for all sine, cosine and tangent (ie., ##\cos(\pi-x)=\cos x## and ##\tan(\pi-x)=\tan x##). The shortest and most direct answer is that this definition is one that is useful in a wide range of applications, one that appears over and over again, so often that we give it the name sine function. In a sense, the question is similar to the question: "Why do we name an apple 'apple'?" Well, you can name it by another word if you want. The French names it "pomme", for example. Why can't ##\sin(\pi-x)## be ##2\sin(x)## or ##-\sin(x)##, you may ask. In fact, it can. It's just that that sine function wouldn't be the sine function the rest of the world is talking about. It will just be your own little sine function, and known only by you, if you like it to be so.
  6. Apr 22, 2015 #5


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    [itex]sin(\pi - x)=sin(\pi )cos(x)-cos(\pi )sin(x),\ sin(\pi )=0,\ cos(\pi)=-1,\ therefore\ sin(\pi -x)=sin(x)[/itex]
  7. Apr 22, 2015 #6


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    No, it's that way because the math works out that way, not by definition, and there IS a proof as mathman has demonstrated.
  8. Apr 22, 2015 #7


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    The only definitions here are of the trig functions themselves, usually defined in terms of lengths of sides of reference triangles in the unit circle (sine and cosine) or ratios of lengths of sides (tangent and the rest). All of the other trig identities can be proven, including those you listed, so are not merely definitions.
    As someone else in this thread already said, ##\sin(\pi - x)## can be shown by the use of the addition formula as being identically equal to ##\sin(x)##. This is not a definition.
  9. Apr 22, 2015 #8


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    To illustrate, this diagram should help:

    The trigonometric functions, sine, cosine, tangent, etc., are also known as circular functions, since their values can be calculated by constructing a circle with unit radius and applying the Pythagorean Theorem.

    If you plot the values of sine θ versus θ, you should get the following curve:

    As you can see, the values of sine θ repeat after the value of θ increments by 2π. Because the values of sine (and the other trig functions) repeat every so often, these functions are also called periodic. For the sine, sine (θ + 2π) = sine (θ) and sine (θ + π) = -sine (θ), as shown on the diagram above.
  10. Apr 22, 2015 #9
    I agree with you, there is clearly a proof that proves this identity for all angles.
  11. Apr 22, 2015 #10


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    Yeah, Walter Cronkite ("And that's the way it is ...") was no mathematician. :wink:

  12. Apr 22, 2015 #11
    Here is a graph of y=sin(x) that may help you to understand more about relation sin(x)=sin(pi-x), 0 =< x <= pi/2.

    It shows some chosen values of x which are pi/6 and pi/4.
    Sin (5pi/6) = sin (pi - (pi/6)) = sin (pi/6) = 1/2

    Attached Files:

  13. Apr 22, 2015 #12
    If you say that ##\sin(\pi-x)=\sin(x)## is not by definition, but it's proven by the addition identity, that means that the left-hand side (LHS) of the addition identity ( ##\sin(\pi-x)=\sin(\pi)\cos(x)-\cos(\pi)\sin(x)## ) has no meaning by itself but gets its meaning from the right-hand side (RHS) of the addition identity.

    Then, shouldn't we call the addition identity "the addition definition" or "the defining statement for the addition of angles in trigonometrical functions", instead of calling it an identity?

    Just like the Pythagoras's theorem, both sides of an trigonometrical identity should already be defined. Pythagoras's theorem: ##c^2=a^2+b^2##, where ##c## is the longest length of a right-angle triangle while ##a## and ##b## are its other two lengths. So the LHS is already defined; it already has a meaning and does not get its meaning from the RHS.
    Last edited: Apr 23, 2015
  14. Apr 23, 2015 #13


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    I think you might be mistaking an identity for a definition. They are not the same thing. In the first identity above, we are not defining sin(x): it is already defined in terms of a coordinate of a reference point on a unit circle. By means of geometry or the use of the addition formula, one can prove that sin(π - x) and sin(x) represent the same number, independent of x. Definitions are accepted (or not), but they aren't proved.

    In contrast, an identity is an equation that is true for all values of the variables that are involved. As an example, the equation ##x^2 + 2x + 1 = (x + 1)^2## does not define either ##x^2 + 2x + 1## or ##(x + 1)^2##. It allows you to replace either expression with the other.
    Although the addition formulas give definitions for sine (etc.) of a sum of angles, I've never read of anyone calling this a definition. These formulas are invariably proved. Can you cite any references where they are called definitions?
    Well, of course. An identity is meaningless if we have no idea what either side means. I don't get your point.
  15. Apr 23, 2015 #14
    Im having trouble seeing the practical application of sin(pi-x)=sinx. I have just started with trigonometry this year and i only know the very basics. How would it be if you used a not right triangle with an angle bigger than 90 degrees as an example. How could you prove that sine also works for angles bigger than 90 degrees?
  16. Apr 23, 2015 #15


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    See Post #8 above.

    As you move thru trigonometry, you'll also learn the law of sines and the law of cosines, which can be used to solve many different types of triangle problems.

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