MHB How do I find cos and sin values without a calculator?

  • Thread starter Thread starter Casio1
  • Start date Start date
Casio1
Messages
86
Reaction score
0
Please refer to the diagram.

I am asked to find cos - 1800 and sin -1800 by plotting a point p on a circle.

OK I don't understand the following before I begin to answer.

How do I know where to start to draw the circles?

I have the x and y axis. I know that anti-clockwise rotation is positive and clockwise is negative.

If I look first at cos - 1800

I know I am going to draw a semi circle in the clockwise direction, which will represent 1800 but show as - 1800 on the diagram.

I am confused how I draw the sin -1800 and am assuming that I draw the semi circle in a anti-clockwise direction labelled also - 1800 to end up with a circle.

I also don't understand from a drawing how I am supposed to know that cos - 1800 = - 1, and sin - 1800 = 0

From first principles and not looking at the calculator, how would I know that these values are right, I could have drawn the semi-circles from x = 2 or 3 etc?

Please advise if you can.

Kind regards

Casio
 

Attachments

  • Sin and cosine definitions.png
    Sin and cosine definitions.png
    2.5 KB · Views: 71
Mathematics news on Phys.org
For simplicity use the unit circle (radius=1) in your diagrams
I'd use parametric equations in this case to split your axes into $x = \cos(t) \text{ and } y = \sin(t)$.

You've correctly put point P on the graph so what is the value of x at -180? What about y? The values will be equal to $\cos(-180)$ and $\sin(-180)$ respectively.
 
SuperSonic4 said:
For simplicity use the unit circle (radius=1) in your diagrams
I'd use parametric equations in this case to split your axes into $x = \cos(t) \text{ and } y = \sin(t)$.

You've correctly put point P on the graph so what is the value of x at -180? What about y? The values will be equal to $\cos(-180)$ and $\sin(-180)$ respectively.

As I am just learning this from first principles my foundation understanding of the subject is a little unstable at the moment, so I would like to keep things as simple as possible if I we can. Parametric equations might be getting a little too involved just at the moment thanks.

So referring to what you advise above and in conjunction with the diagram, point P is positioned at - x, the point would be using your idea of the radius = 1 being cos - 1800 = - 1

The y-axis or y value when x = - 1 must be that y = 0 with reference to the diagram.

So in conclusion;

cos - 1800 = - 1, and sin - 1800 = 0

Thanks

Casio(Smile)
 
Casio said:
As I am just learning this from first principles my foundation understanding of the subject is a little unstable at the moment, so I would like to keep things as simple as possible if I we can. Parametric equations might be getting a little too involved just at the moment thanks.

So referring to what you advise above and in conjunction with the diagram, point P is positioned at - x, the point would be using your idea of the radius = 1 being cos - 1800 = - 1

The y-axis or y value when x = - 1 must be that y = 0 with reference to the diagram.

So in conclusion;

cos - 1800 = - 1, and sin - 1800 = 0

Thanks

Casio(Smile)

Parametric equations is a fancy way of splitting the x and y axes into components. It enables you to say that whatever the x value is then that is equal to the value of $\cos(t)$ (t is just a variable) and the y value is $\sin(t)$

In this case because P is at (-1,0) then $\cos(-180) = x$ and $\sin(-180) = y$

Without knowing what you've studied so far it's hard to suggest a method of working out. For example you (can) use the unit circle to prove periodicity so you can't say that $\sin(-180) = \sin(-180+360)$. Have you covered even and odd functions (see spoilers for what I mean)
  • An even function is where $f(-x) = f(x)$. Examples are $f(x) = x^2$ and, more thematically appropriate $f(x) = \cos(x)$
  • An odd function is where $f(-x) = -f(x)$. Examples include $f(x) = x^3$ and $f(x) = \sin(x)$

You'd use these functions to say that $\cos(-180)= \cos(180)$ and work out the positive angle
 
SuperSonic4 said:
Parametric equations is a fancy way of splitting the x and y axes into components. It enables you to say that whatever the x value is then that is equal to the value of $\cos(t)$ (t is just a variable) and the y value is $\sin(t)$

In this case because P is at (-1,0) then $\cos(-180) = x$ and $\sin(-180) = y$

Without knowing what you've studied so far it's hard to suggest a method of working out. For example you (can) use the unit circle to prove periodicity so you can't say that $\sin(-180) = \sin(-180+360)$. Have you covered even and odd functions (see spoilers for what I mean)
  • An even function is where $f(-x) = f(x)$. Examples are $f(x) = x^2$ and, more thematically appropriate $f(x) = \cos(x)$
  • An odd function is where $f(-x) = -f(x)$. Examples include $f(x) = x^3$ and $f(x) = \sin(x)$

You'd use these functions to say that $\cos(-180)= \cos(180)$ and work out the positive angle

Hi, yes very briefly but a lot of revision to do on it to get a solid foundation.

Kind regards

Casio
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top