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First I had the simple notion that the trignometric ratios are ratios of sides of a triangle.

But when $\sin(0) = 0 , \cos(0) = 1 $ what do they mean?

I had read about the unit circle definition and I thought that the unit circle definition was just a nice way to represent the triangle inside a circle and in the process simplify the calculation of ratios at various angles.

Am I wrong about the unit circle definition?

Edit 1: After reading the first answer,the next question that came to my mind was, why was the movement of triangles defined that way luckily this was asked before here.

How does the unit circle work for trigonometric ratios of non-acute angles?

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    $\begingroup$ You can think of $\sin(\theta)$ as the 'y value' and $\cos(\theta)$ as the 'x value' of the point on the unit circle at angle $\theta$. So the y value at angle 0 (deg or rads) - aka sin - is zero and the x value - aka cos - at angle 0 is 1. $\endgroup$ – tangrs May 30 '14 at 13:26
  • $\begingroup$ You are essentially correct. $\endgroup$ – Dan Christensen May 30 '14 at 13:26
  • $\begingroup$ wow !! that was quick. ok so when should I be considering $ \sin $ and $\cos $ to be sides of a triangle ? or is the unit circle definition the only correct way ? $\endgroup$ – Ramkumar May 30 '14 at 13:33
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You can use either method to analyse trigonometric functions (with unit circles, or with triangles), but just keep in mind that (really) trigonometric functions have nothing to do with either. trigonometric functions are not functions of triangles, nor are they functions of circles. Trigonometric functions are functions of angles.

I guess my point is that you can use any geometric definition that is convenient, but in the end, we are left with a function and an argument.

I could just as easily have defined $\sin x$ as $$\sin:R\rightarrow R\:|\sin x = \sum_{k=0}^{\infty} \frac{(-1)^kx^{1+2k}}{(1+2k)!}$$ and i wouldn't be wrong.

I could also define $\sin x$ as $Im(e^{ix})$

None of these definitions have anything to do with geometry; they're numeric definitions. But practically, $\sin$ has a geometric interpretation. An angle is the spread between 2 rays, and may have a one to one correspondence with arc length on a unit circle, but still is not the arc length of a unit circle. it's an angle.

Unfortunately, my answer has become way too abstract. What I initially intended to say is that $\sin$ is a function that can be analysed via either triangles or unit circles, or any other way that the student sees fit.

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  • $\begingroup$ can you please elaborate on the idea ? how are they only functions of angles ? what do they actually calculate with the angles then ? $\endgroup$ – Ramkumar May 31 '14 at 2:45
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They still make sense for $\theta = 0$.

A right-angled "triangle" with an angle of $\theta = 0$ will have the property that the adjacent and hypotenuse are the same length, and the opposite side has zero length.

In that case, the ratio of the opposite side by the hypotenuse would be $0/h = 0$. Moreover, the ratio of the opposite side by the adjacent side (which have the same length) would be $h/h=1$.

The circle idea that you have in-mind is fine. Starting at the origin, some triangles are in the first quadrant and are made by going "right and then up", in the second quadrant by going "left and then up", in the third quadrant by going "left and then down" and in the fourth quadrant by going "right and then down". In between these, you have the odd triangles where two sides coincide and the third has zero length.

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  • $\begingroup$ I am a little confused by the directions given in your answer .in the first quadrant isn't it left and up ? ir cos is 1 and the triangle moves up and in a anti clockwise(left) direction ? $\endgroup$ – Ramkumar May 30 '14 at 13:51
  • $\begingroup$ math.com/school/subject2/images/S2U4L1DP.gif $\endgroup$ – Fly by Night May 30 '14 at 15:54
  • $\begingroup$ i saw the picture with the quadrants ,but still in the first quadrant the hypotenuse is moving left and up or if it starts at 90 degrees right and down i inferred it from the interactive example here d2gbom735ivs5c.cloudfront.net/m/algebra/images/circle-trig.swf .I may be confused a little can you please explain it a bit more ? $\endgroup$ – Ramkumar May 31 '14 at 3:17
  • $\begingroup$ @Ramkumar Start at $O(0,0)$, move right to the point $A(3,0)$. Move up (from there) to the point $B(3,4)$. The hypotenuse $OB$ has vector $3{\bf i}+4{\bf j}$, i.e. $3$ right and then $4$ up. $\endgroup$ – Fly by Night Jun 1 '14 at 12:09

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