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# Taking Trigonometry Series-ly

Part1:

There are various approaches to this part of the problem, although the answers in radians should look something close to $$ \sin(x)\approx x-\frac{1}{6}x^3\quad\quad \cos(x) = 1-\frac{1}{2}x^2 $$

One approach would be to start using well known properties of trig functions. For example, we know that $a$ must be $0$ and $e$ must be exactly 1 because $\sin(0)=0$ and $\cos(0) = 1$.

We can also use some shortcuts because we know that $\sin(x) = -\sin(-x)$ and $\cos(x)=+\cos(-x)$. So, we can say that

$$

\sin(x)\approx bx +dx^3\quad\quad \cos(x) \approx 1+fx^2

$$

Then look at a very small value of $x$, say $0.0000001$. Unless the other coefficients are very large, we can ignore then. This tells us that, approximately

$$

b\approx \sin(0.0000001) \quad f\approx \sqrt{\frac{\cos(0.0000001)-1}{0.0000001^2}}

$$

We could find a value of $d$ by trying some larger values of $x$.

It should be clear that the numbers in radians are much smaller (and nicer) than in degrees.

Part 2:

$ \frac{d^2 f}{dx^2}+x =0 $

Substitute $ f(x) = sinx, f ' (x) = cosx , f ''(x) = -sinx $

$ \frac{d^2 f}{dx^2} =-x $

$\frac{df}{dx}= cosx =\int -x = -\frac{x^2}{\frac{x^5}{5!}2} + c$ (sub cos0 = 1 to find c=1)

$f(x)=sinx = \int cosx = \int -\frac{x^2}{2} + 1 = -\frac{x^3}{3!} + x + c $ (sub sin(0) = 0 to find c=0)

$\int sinx = -cosx = \int-\frac{x^3}{6!} + x = -\frac{x^4}{4!} + \frac{x^2}{2!} + c$ (-cos(0) = c =-1)

$\int -cos = -sinx = \int -\frac{x^4}{4!} + \frac{x^2}{2!} -1 = -\frac{x^5}{5!} + \frac{x^3}{3!} -x$

$\int -sinx = cosx = \int -\frac{x^5}{5!} + \frac{x^3}{3!} -x = -\frac{x^6}{6!} + \frac{x^4}{4!}-\frac{x^2}{2!} + 1 $

Hence cosx = $1 -\frac{x^2}{2!}+ \frac{x^4}{4!} -\frac{x^6}{6!}$

And Sinx = $x - \frac{x^3}{3!} +\frac{x^5}{5!}$

Extension:

Infinite power series of sinx =$x - \frac{x^3}{3!} +\frac{x^5}{5!} + ... + \frac{x^n}{n!}$ for odd n

Infinite series of cosx =$1 -\frac{x^2}{2!}+ \frac{x^4}{4!} -\frac{x^6}{6!} +... \frac{x^n}{n!} $for even n

Here we make a comparison of the evaluation of \frac{\pi}{2} by the series approxomations and the actuall functions.

$sin(\frac{\pi}{2}) = 1$

approximation :$sin(\frac{\pi}{2}) = 1.004524856$

$cos (\frac{\pi}{2}) =0 $

appromimation: $cos (\frac{\pi}{2}) =-8.945230096 x 10^{-4}$

Calculators and computers do not store a list of results for each and every possible function. A much more efficient method is to store a polynomial expression for each funtion using the taylor/maclaurin series.

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Age 16 to 18

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Part1:

There are various approaches to this part of the problem, although the answers in radians should look something close to $$ \sin(x)\approx x-\frac{1}{6}x^3\quad\quad \cos(x) = 1-\frac{1}{2}x^2 $$

One approach would be to start using well known properties of trig functions. For example, we know that $a$ must be $0$ and $e$ must be exactly 1 because $\sin(0)=0$ and $\cos(0) = 1$.

We can also use some shortcuts because we know that $\sin(x) = -\sin(-x)$ and $\cos(x)=+\cos(-x)$. So, we can say that

$$

\sin(x)\approx bx +dx^3\quad\quad \cos(x) \approx 1+fx^2

$$

Then look at a very small value of $x$, say $0.0000001$. Unless the other coefficients are very large, we can ignore then. This tells us that, approximately

$$

b\approx \sin(0.0000001) \quad f\approx \sqrt{\frac{\cos(0.0000001)-1}{0.0000001^2}}

$$

We could find a value of $d$ by trying some larger values of $x$.

It should be clear that the numbers in radians are much smaller (and nicer) than in degrees.

Part 2:

$ \frac{d^2 f}{dx^2}+x =0 $

Substitute $ f(x) = sinx, f ' (x) = cosx , f ''(x) = -sinx $

$ \frac{d^2 f}{dx^2} =-x $

$\frac{df}{dx}= cosx =\int -x = -\frac{x^2}{\frac{x^5}{5!}2} + c$ (sub cos0 = 1 to find c=1)

$f(x)=sinx = \int cosx = \int -\frac{x^2}{2} + 1 = -\frac{x^3}{3!} + x + c $ (sub sin(0) = 0 to find c=0)

$\int sinx = -cosx = \int-\frac{x^3}{6!} + x = -\frac{x^4}{4!} + \frac{x^2}{2!} + c$ (-cos(0) = c =-1)

$\int -cos = -sinx = \int -\frac{x^4}{4!} + \frac{x^2}{2!} -1 = -\frac{x^5}{5!} + \frac{x^3}{3!} -x$

$\int -sinx = cosx = \int -\frac{x^5}{5!} + \frac{x^3}{3!} -x = -\frac{x^6}{6!} + \frac{x^4}{4!}-\frac{x^2}{2!} + 1 $

Hence cosx = $1 -\frac{x^2}{2!}+ \frac{x^4}{4!} -\frac{x^6}{6!}$

And Sinx = $x - \frac{x^3}{3!} +\frac{x^5}{5!}$

Extension:

Infinite power series of sinx =$x - \frac{x^3}{3!} +\frac{x^5}{5!} + ... + \frac{x^n}{n!}$ for odd n

Infinite series of cosx =$1 -\frac{x^2}{2!}+ \frac{x^4}{4!} -\frac{x^6}{6!} +... \frac{x^n}{n!} $for even n

Here we make a comparison of the evaluation of \frac{\pi}{2} by the series approxomations and the actuall functions.

$sin(\frac{\pi}{2}) = 1$

approximation :$sin(\frac{\pi}{2}) = 1.004524856$

$cos (\frac{\pi}{2}) =0 $

appromimation: $cos (\frac{\pi}{2}) =-8.945230096 x 10^{-4}$

Calculators and computers do not store a list of results for each and every possible function. A much more efficient method is to store a polynomial expression for each funtion using the taylor/maclaurin series.