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Thread: Sin Cos and Tan

  1. #1 Sin Cos and Tan 
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    Hello all,

    I was wondering if anybody would be able to explain to me, what sin, cos and tan are. Not how they can be used in trigonometry, but why they are what they are. as in, how did they find these numbers in the first place, what is the difference between them, and well, proving the ideas of cos, sin and tan.
    I have been looking online, bu the explanations are confusing. So I have turned to this site, to see if anybody could give me a fairly simple answer.


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  3. #2  
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    Ben, the simplest way I can explain it is in terms of similar triangles. If you have ever taken a class in plane geometry you could do a formal proof, but it is almost obvious. If you have two triangles of different sizes with equal angles, the sides will all be in the same proportion. You can blow the triangle up or shrink it down and the three angles will stay the same. The proportions of each side to the other will also stay the same.

    Trigonometry is just the study of similar triangles where one of the angles is a right angle. This means if you choose a value for one of the other angles, call it Theta, then every right triangle with one of the angles equal to Theta, will look like every other right triangle with angle Theta but just be a different size.

    So if you want to, you can draw a whole bunch of right triangles with different angle Theta and measure the proportion of the three sides – the ratio of the side opposite your angle to the hypotenuse is the sin, the ratio of the side adjacent to the angle to the hypotenuse is the cosine, and the opposite over the adjacent is the tangent. You could record these ratios for each angle Theta in a table of angles, and the ratio would be the same for all sizes of right triangles having the same angle. You would have created a trig table. This is what we actually used to use when I studied trig, before we had calculators.

    Now if you plot your table of angles on a horizontal scale with the sine on the vertical scale, you get a sine wave. The cosine function looks the same but is just shifted 90 degrees out of phase. This is something that turns up very frequently in nature in any kind of vibrating system. The alternating power in your house current is also a sine wave.

    Trig was originally developed by the ancient Greeks, Egyptians and Indians, mainly for use in astronomy. You will use these ratios very frequently if you ever have to do almost any kind of physics or engineering.


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  4. #3  
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    Sin 30 = 0.5, but what is the relation between the angular dimension and 0.5. I know Sin means the ratio between opp. and hyp. But how we convert the figure 30 to 0.5? is there any specific formula for that?
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  5. #4  
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    A value for sine: between -1 to 1, is dependent on an angle from 0 to 360 degrees.

    The relationship is between an angle and the ratio of the length of the long size and the perpendicular, hence: sine'=opposite/hypotenuses. Its as simple as that!
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  6. #5  
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    sak: I imagine you're asking for a "formula" for sine that doesn't require you drawing triangles. There is indeed such a formula... First, let's use radians instead of degrees (to convert degrees to radians, use that each degree is pi/180 radians). Then:

    sin(x) = x - x^3/3! + x^5/5! - x^7/7! +...

    This is the power series expansion of sin. You can't calculate this exactly, but the series converges very quickly and, for positive x, this is an alternating series, so you can get a good approximation by cutting the series off at some point.
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  7. #6  
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    Quote Originally Posted by serpicojr
    sin(x) = x - x^3/3! + x^5/5! - x^7/7! +...
    thanks for the reply, This is what i was looking for.
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  8. #7  
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    Quote Originally Posted by serpicojr
    can i correct the quation like this: sin(x) = x - (x^3)/3! + (x^5)/5! - (x^7)/7! +...
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  9. #8  
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    Yes, that is what i meant.
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  10. #9  
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    Actually there's a fun trick you can do with this. And it is a trick, in the sense that it looks really hard, but is actually pretty easy - just a matter of shuffling definitions around, sort of.

    As serpicojr said, the Taylor series for sin x is

    sin x = x - x<sup>3</sup>/3! + x<sup>5</sup>/5! - x<sup>7</sup>/7! +...

    Now cos x = d/(dx)sin x, so cos x = 1 - 3x<sup>2</sup>/3! + 5x<sup>4</sup>/5! - 7x<sup>6</sup>/7! +... which reduces to

    cos x = 1 - x<sup>2</sup>/2! + x<sup>4</sup>/4! - x<sup>6</sup>/6! +....

    Recall now the Taylor series for e<sup>x</sup> is

    e<sup>x</sup> = ∑ x<sup>n</sup>/n! whereby

    e<sup>iy</sup> = 1 + iy - y<sup>2</sup>2! - i(y<sup>3</sup>/3!) + y<sup>4</sup>/4! + i(x<sup>5</sup>/5!) +.... (since odd powers of i are 1 and -1 alternating, even powers of i are -1 and 1 alternating). Gathering terms, we find

    e<sup>iy</sup> = (1 - y<sup>2</sup>/2! + y<sup>4</sup>/4! -...) + i(y - y<sup>3</sup>/3! + y<sup>5</sup>/5! -...) which from the above is just

    e<sup>iy</sup> = cos y + i sin y. Remembering that e<sup>-x</sup> converges as 1 - x<sup>2</sup>/2! - x<sup>3</sup>/3! +...

    one finds (by taking the appropriate pairs of Taylors) that

    sin y = 1/2(e<sup>iy</sup> - e<sup>-iy</sup>)

    and

    cos y = 1/2(e<sup>iy</sup> + e<sup>-iy</sup>)

    Cool, or what?
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  11. #10  
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    Quote Originally Posted by Guitarist
    (since odd powers of i are 1 and -1 alternating, even powers of i are -1 and 1 alternating).
    Well, well. Nobody spotted this howler?

    Odd powers of i are, of course, -i and i, alternating, and not as I wrote.

    Shame on you all!
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  12. #11  
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    Quote Originally Posted by Guitarist
    Quote Originally Posted by Guitarist
    (since odd powers of i are 1 and -1 alternating, even powers of i are -1 and 1 alternating).
    Well, well. Nobody spotted this howler?

    Odd powers of i are, of course, -i and i, alternating, and not as I wrote.

    Shame on you all!
    I was so busy enjoying your fun trick I entirely failed to spot that howler. Apologies.


    ps. Was that deadpan enough from a mathematical know-nothing? :P

    pps. Perhaps not on this thread (unless it could be seen to be relevant) - but I still have questions about Euler's (Leibniz's?) famous e^πi - 1 = 0 and you seem like the sort of person who might be able to answer them. Yes?
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  13. #12  
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    e^iπ + 1 = 0

    In all honesty, I think Guitarist's proof that e^ix = cos(x) + i sin(x) sums up everything there is to know about the above.
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  14. #13  
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    Quote Originally Posted by serpicojr
    e^iπ + 1 = 0

    In all honesty, I think Guitarist's proof that e^ix = cos(x) + i sin(x) sums up everything there is to know about the above.
    My query concerned the use of radians as measures of angles, and the subsequent necessary (?) connection between pi and e. Or isn't there one?
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  15. #14  
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    Oh, I see. Well, pi radians = 180 degrees. Tell us why! (hint: what is the circumference of a circle?).

    In the "fun trick" just replace y with pi then try to figure out what is cos pi and what is i sin pi.

    It'll fall into your lap! Oh I just noticed - you don't have the Euler (not Leibniz) identity quite right. I suggest you double check, although serpicojr did correct you. Maybe you missed the difference?

    Is there a connection between pi and e? None that I know of, but others may know better.
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  16. #15  
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    I'm trying to think of a good way of describing why π and e must be related, but in the end I can't think of any way to describe is besides deriving the formula e^ix = cos(x) + i sin(x) (from power series as Guitarist did or from differential equations). But both of these methods of deriving the formula rely on the fact that we use radians as our unit of measuring angles--using this unit, we have sin' = cos and cos' = -sin. This relation determines both their power series about 0 and the fact that they satisfy y'' + y = 0 (which are the key steps in the above derivations). And it's no surprise that radians are the correct unit to use the normalize sin and cos so that "they are the derivatives of each other"--radians relate the measure of an angle to a natural notion of length (that of the arc on the unit circle subtended by said angle). If you would like me to go into more depth about any of this, let me know.
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  17. #16  
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    There are some interesting equations over here: http://www.maa.org/editorial/mathgam..._03_15_05.html. They don't actually relate e and pi though, since they're only approximations, but it makes you wonder.
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  18. #17  
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    Quote Originally Posted by serpicojr
    I'm trying to think of a good way of describing why π and e must be related, but in the end I can't think of any way to describe is besides deriving the formula e^ix = cos(x) + i sin(x) (from power series as Guitarist did or from differential equations). But both of these methods of deriving the formula rely on the fact that we use radians as our unit of measuring angles--using this unit, we have sin' = cos and cos' = -sin. This relation determines both their power series about 0 and the fact that they satisfy y'' + y = 0 (which are the key steps in the above derivations). And it's no surprise that radians are the correct unit to use the normalize sin and cos so that "they are the derivatives of each other"--radians relate the measure of an angle to a natural notion of length (that of the arc on the unit circle subtended by said angle). If you would like me to go into more depth about any of this, let me know.
    Serpicojr, thanks for that. This is (I think) the point I was trying to make - that the famous identity may exist only because we measure in radians which itself is necessarily related to pi.

    I am approaching this from a dilletante/philosophy point of view because I had a long debate with a (now alas dead) friend who claimed that pi popped up in all sorts of relations, many/most of which are not connected to its function as circumference/diameter. He cited this one and my point was that it seems obvious why pi is in this one - pi radians is, as you say, a straight line (180 degrees) - and therefore in an equation such as this - using i, could easily be seen to be converting +1 into -1 (whichever way the signs go in the equation - and yes, I did get it wrong).

    Again, if I've read too much into it, any maths brains' input gratefully received.

    cheer

    the sunshine warrior
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  19. #18  
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    The fact that e^iπ = -1 is completely independent of the fact that we like to use radians. In fact, it is relations like these which indicate that radians are the natural unit of measuring angles. And this is a really special thing! How many measured quantities have mathematically natural units?

    π appears in many other places, many of which you wouldn't expect--I'd love to discuss further, but I've got to get to work, so I'll fill you in later.
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  20. #19  
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    Quote Originally Posted by serpicojr
    The fact that e^iπ = -1 is completely independent of the fact that we like to use radians. In fact, it is relations like these which indicate that radians are the natural unit of measuring angles. And this is a really special thing! How many measured quantities have mathematically natural units?

    π appears in many other places, many of which you wouldn't expect--I'd love to discuss further, but I've got to get to work, so I'll fill you in later.
    Ta

    Look forward to that.
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  21. #20  
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    Hello people,

    Well, since the making of this post, I have recently purchased a book on the world of trigonometry. From arcsin to cotangents, I plan to read it all.

    Never the less, I have hit a slight rut.

    Page 29, where it says

    sin -(angle) = - sin (angle)
    cos -(angle) = cos (angle)
    tan -(angle) = - tan (angle)

    I have difficulty understanding things unless I understand their roots. With this question, I don't see why some questions which have a negative angle, end up negative, whereas some end up positive.

    Bar from tan, as if tan is sin/cos, then I can see that if sin ends up negative and cos positive, then it is just -/+, which is minus.
    But I don't understand the differences of the sin and cos.
    Or am I missing something obvious? Any help if greatly appreciated.
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  22. #21  
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    Well, what is the sine or cosine of a negative angle? What does a negative angle even mean? To make sense of this, it is first useful to view sine and cosine in terms of the unit circle.

    The unit circle is the circle of radius 1 centered at the origin in the place. For those who prefer algebra, it is the locus of points satisfying the equation x^2 + y^2 = 1. Now let t be an angle between 0 and π/2 (we're using radians--if you still prefer angles, make the switch now), and draw a line segment from the origin to the unit circle in the first quadrant so that the segment and the x-axis meet at an angle of t. This segment has length 1 (why?). Dropping a vertical segment from the intersection of the segment and the circle to the x-axis, and the making a horizontal line from this point on the x-axis back to the origin, we have described a right triangle. The hypotenuse is the segment from the origin to the circle and hence has length 1. Now sin(t) is the ratio between the side opposite the angle at the origin (the vertical side) and the hypotenuse. The vertical side has length equal to the y-coordinate of the point on the unit circle, and so letting this quantity be y, we have sin(t) = y/1 = y. cos(t) is the ratio between the side adjacent the angle at the origin (the horizontal side) and the hypotenuse. The horizontal side has length equal to the x-coordinate of the endpoint of the vertical line on the x-axis, and this is the same as the x-coordinate of the point on the unit circle (since they lie on the same vertical line). Calling this quantty x, we have cos(t) = x/1 = x. Now forgetting about the triangle, we see that we could have defined sine and cosine as follows: draw a line segment in the first quadrant from the origin to the unit circle making an angle of t with the x-axis; then the point of intersection of the segment and the unit circle is (cos(t), sin(t)).

    Now it is easy to extend the definition of sine and cosine to angles greater than π/2. But what about negative angles? Well, we get positive angles by rotating a line segment about the origin in the counterclockwise direction, so we should get negative angles by rotating a line segment about the origin in the clockwise direction. So it makes sense that, for t > 0, the angle -t < 0 is obtained by forming the angle t as the angle between a line segment and the x-axis and then flipping said segment about the x-axis. Flipping about the x-axis preserves the x-coordinate and negates the y-coordinate. And thus...

    (cos(-t), sin(-t)) = (cos(t), -sin(t))
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  23. #22  
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    ten_ben, if you read and understood serpicojr's answer there is no reason for you to read further. But maybe you need a dumbed down version at this point in your studies, just to understand the sign conventions.

    Picture a set of x-y axes and pick a point where x and y are both positive. If you draw a line from the origin of the graph to that point, your line forms an angle to the x axis. It forms an angle with the y axis as well, but by convention we measure angles counterclockwise from the x axis. Then the sine is y divided by the hypotenuse and is positive. The cosine is x divided by the hypotenuse and is positive.

    For negative angles less than 90 degrees (clockwise), y is negative. This makes the sine and tangent negative, but x is still positive, which makes the cosine positive.
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  24. #23  
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    Thank you both greatly. Now I have the explanation, it seems fairly obvious! Thank you both anyway, I appreciate it.
    Now, I apologise, but I have some more questions (My book has come onto a page of formulae, but I find it hard to accept these things unless I know why, and some things take explaining!).
    Ok, my next question, why is...

    sin 2(angle) = 2 sin(angle) cos(angle)

    cos 2(angle) = 1-(2 sin^2 (angle)) or (2 cos ^2 angle) - 1
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  25. #24  
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    Unfortunately, the easiest way I can think of to prove this off the top of my head is to use linear algebra. You wouldn't happen to know anything about matrix multiplication, would you?
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  26. #25  
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    Just researched matrix multiplications online, and alas, I unfortionately do not know anything about them. Sorry, is there no other way to explain it, or should I try to get my head around matrixes?
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  27. #26  
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    There's gotta be a geometric way of proving these. But the algebraic way of thinking about these really requires matrices. I suppose I could try to explain it without appealing to matrices explicitly.

    So let's rotate the plane about the origin by the angle t. When we do this, we send the point (1,0) to the point (cos t, sin t) and the point (0,1) to the point (-sin t, cos t): draw yourself a picture to convince yourself of this. Now if we applied this rotation twice, we'd just be rotating by 2t, and the effect would be sending (1,0) to (cos 2t, sin 2t) and (0,1) to (-sin 2t, cos 2t).

    Now here's the key part... rotations about the origin are linear. This means that if the point (x,y) gets mapped to (z,w), then the point (ax,ay) gets mapped to (az,aw) for any real number a, and if the point (x',y') gets mapped to the point (z',w'), then (x+x',y+y') gets mapped to (z+z',w+w').

    Now, as stated above, (1,0) gets mapped to (cos t, sin t) under one application of the rotation (and (0,1) gets mapped to (-sin t, cos t)). Then using properties of linearity, we see:

    -(cos t, 0) gets mapped to (cos t cos t, cos t sin t)
    -(0, sin t) gets mapped to (-sin t sin t, cos t sin t)
    -(cos t, sin t) gets mapped to (cos t cos t - sin t sin t, 2 cos t sin t)

    So, under two applications of the rotation, (1,0) gets mapped to ((cos t)^2 - (sin t)^2, 2 cos t sin t). We already said it got mapped to (cos 2t, sin 2t), so we've deduced the relations:

    -cos 2t = (cos t)^2 - (sin t)^2
    -sin 2t = 2 cos t sin t

    Now using the fact that (cos t)^2 + (sin t)^2 = 1 (this is just the Pythagorean theorem!), we can change the equality for cos 2t into the equivalent equalities you listed.
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  28. #27  
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    The geometry case is just simple triangle chasing - draw two right angle triangles on top of each other and off you go. If you want i can do the sketch and post it later.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    Hey
    I wouldnt mind getting that picture drawn "river_rat". Apologies to "serpicojr", but I'm havign a tad of difficulty getting my head around what you have said. so that diagram would be very helpful.
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  30. #29  
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    Just did this quickly in wingeom :



    You need to use the fact that AC = 1 and then derive the trig ratios i gave you on the pic. Then work out what sin(alpha + beta) is and then derive the formula for sin(2 theta) and cos(2 theta) which is what you wanted.

    If you need some help with the steps just shout.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  31. #30  
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    A radian is the ratio of an arc's length to it's distance from the point about which the arc is rotated. This allows you derive the circumference of a circle or the length of an arc, a portion of the circlle's outer edge, based on the radian measurement. i.e. 2pi * r is the circumference of a circle because you are multiplying te radius r by a full 2pi rotation about the circle. In complex analysis, e is used as a base to describe the direction of a point in the complex plane. Because e^iy always has a magnitude of one, it can be used as a unit direction vector if you'd like to think of it that way. ie, in complex z = |z| * e ^ ( i * theta ) = sqrt( x^2 + y^2 ) * [ cos( Atan( y / x) ) + i sin( Atan( y / x ) ] provided x != 0.
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