- Thread starter
- #1

#### Farmtalk

##### Active member

- Dec 25, 2012

- 42

I've been trying to understand more of the "why" than the "how" of mathematics, and this one is very intriguing to me

- Thread starter Farmtalk
- Start date

- Thread starter
- #1

- Dec 25, 2012

- 42

I've been trying to understand more of the "why" than the "how" of mathematics, and this one is very intriguing to me

- Feb 13, 2012

- 1,704

When Leonhard Euler 'discovered' the relation...

I've been trying to understand more of the "why" than the "how" of mathematics, and this one is very intriguing to me

$$e^{i x} = \cos x + i\ \sin x\ (1)$$

... hi 'banalized' all the previous trigonometric relations. From (1) You derive...

$$\sin x = \frac{e^{i\ x} - e^{- i\ x}}{2\ i}$$

$$\cos x = \frac{e^{i\ x} + e^{- i\ x}}{2}\ (2)$$

Using (2) and a little of algebra You can arrive to the sum and difference identity for sines and cosines...

Kind regards

$\chi$ $\sigma$

- Admin
- #3

To begin, draw a horizontal line segment and label the left end $A$ and the right end $B$ and denote the length of the segment as $c$. Now place a point, labeled $C$ above the horizontal line, which is to the right of $A$ and to the left of $B$. Draw a line segment from $A$ to $C$ and label its length $b$ and draw a line segment from $B$ to $C$ and label its length $a$. We now have an acute triangle. Now, orient a Cartesian coordinate system such that vertex $A$ is at the origin, and segment $\overline{AB}$ lies on the $x$-axis. Next drop a vertical line from vertex $C$ to line segment $\overline{AB}$, dividing $\overline{AB}$ into two parts, the left which we label with the length $u$. The length of the vertical line we label $v$, as in the following sketch:

Thus, the coordinates of the vertices are as follows:

$A=(0,0)$

$B=(c,0)$

$C=(u,v)$

Now, let angle $A$ denote the angle at vertex $A$, and the same with $B$ and $C$.

Thus, we have:

\(\displaystyle \cos(A)=\frac{u}{b}\,\therefore\,u=b\cos(A)\)

\(\displaystyle \sin(A)=\frac{v}{b}\,\therefore\,v=b\sin(A)\)

Thus, the coordinates of vertex $C$ are $(b\cos(A),b\sin(A))$. Now we can use the square of the distance formula to compute the distance from vertex $C$ to vertex $B$ (which is labeled $a$):

\(\displaystyle a^2=(b\cos(A)-c)^2+(b\sin(A)-0)^2\)

\(\displaystyle a^2=b^2\cos^2(A)-2bc\cos(A)+c^2+b^2\sin^2(A)\)

\(\displaystyle a^2=b^2\left(\sin^2(A)+\cos^2(A) \right)-2bc\cos(A)+c^2\)

\(\displaystyle a^2=b^2+c^2-2bc\cos(A)\)

I will leave it to the reader to verify this result when angle $A$ is obtuse. Now we are ready to prove the angle sum/difference identities for sine and cosine.

On the unit circle, place two points on the circle in quadrant 1, label the point closer to the $y$-axis $Q$ and the point closer to the $x$-axis $P$. Now draw three line segments, one from $O$ to $Q$, one from $O$ to $P$ and one from $P$ to $Q$. Label the angle subtended by the $x$-axis and segment $\overline{OP}$ as $t$ and the angle subtended by the $x$-axis and segment $\overline{OQ}$ as $s$. Thus, the angle subtended by segments $\overline{OP}$ and $\overline{OQ}$ is $s-t$. Refer to the following sketch:

Now we will use two methods to compute the square of the distance from point $P$ to point $Q$. Using the law of cosines, we get:

\(\displaystyle \overline{PQ}^2=\overline{OP}^2+\overline{OQ}^2-2\cdot\overline{OP}\cdot\overline{OQ}\cos(s-t)\)

Because we are on the unit circle, $\overline{OP}=\overline{OQ}=1$, so we have:

\(\displaystyle \overline{PQ}^2=1^2+1^2-2\cdot1\cdot1\cos(s-t)\)

\(\displaystyle \overline{PQ}^2=2-2\cos(s-t)\)

Now, using the distance formula, we find that:

\(\displaystyle \overline{PQ}^2=(\cos(s)-\cos(t))^2+(\sin(s)-\sin(t))^2\)

\(\displaystyle \overline{PQ}^2=\cos^2(s)-2\cos(s)\cos(t)+\cos^2(t)+\sin^2(s)-2\sin(s)\sin(t)+\sin^2(t)\)

\(\displaystyle \overline{PQ}^2=\left(\cos^2(s)+\sin^2(s) \right)+\left(\cos^2(t)+\sin^2(t) \right)-2\cos(s)\cos(t)-2\sin(s)\sin(t)\)

\(\displaystyle \overline{PQ}^2=2-2\cos(s)\cos(t)-2\sin(s)\sin(t)\)

Now, equating the two expressions for $\overline{PQ}^2$, we obtain:

\(\displaystyle 2-2\cos(s-t)=2-2\cos(s)\cos(t)-2\sin(s)\sin(t)\)

Subtract through by 2:

\(\displaystyle -2\cos(s-t)=-2\cos(s)\cos(t)-2\sin(s)\sin(t)\)

Divide through by -2:

\(\displaystyle \cos(s-t)=\cos(s)\cos(t)+\sin(s)\sin(t)\)

This is the angle-difference identity for cosine.

Now, we may write the left side as:

\(\displaystyle \cos(s+(-t))=\cos(s)\cos(t)+\sin(s)\sin(t)\)

Using the identities $\cos(-x)=\cos(x)$ and $\sin(-x)=-\sin(x)$ we have:

\(\displaystyle \cos(s+(-t))=\cos(s)\cos(-t)-\sin(s)\sin(-t)\)

Replacing $-t$ with $t$, we get:

\(\displaystyle \cos(s+t)=\cos(s)\cos()-\sin(s)\sin(t)\)

This is the angle-sum identity for cosine.

Now, for sine we can use the co-function identity $\sin(x)=\cos\left(\dfrac{\pi}{2}-x \right)$ which of course may be derived from what we have already found:

\(\displaystyle \sin(s+t)=\cos\left(\frac{\pi}{2}-(s+t) \right)=\cos\left(\left(\frac{\pi}{2}-s \right)-t \right)\)

Using the angle difference for cosine which we've already derived, we get:

\(\displaystyle \sin(s+t)=\cos\left(\frac{\pi}{2}-s \right)\cos(t)+\sin\left(\frac{\pi}{2}-s \right)\sin(t)\)

Using the co-function identities, this gives us:

\(\displaystyle \sin(s+t)=\sin(s)\cos(t)+\cos(s)\sin(t)\)

This is the angle-sum identity for sine.

Now, if we write the left side as:

\(\displaystyle \sin(s+(-t))=\sin(s)\cos(-t)+\cos(s)\sin(-t)\)

Applying the negative angle identities, we have:

\(\displaystyle \sin(s-t)=\sin(s)\cos(t)-\cos(s)\sin(t)\)

This is the angle-difference identity for sine.

Now, for the tangent function we can use $\tan(x)\equiv\dfrac{\sin(x)}{\cos(x)}$:

\(\displaystyle \tan(s+t)=\frac{\sin(s+t)}{\cos(s+t)}=\frac{\sin(s)\cos(t)+\cos(s)\sin(t)}{\cos(s)\cos(t)-\sin(s)\sin(t)}\)

Divide each term in the numerator and denominator by $\cos(s)\cos(t)$:

\(\displaystyle \tan(s+t)=\frac{\tan(s)+\tan(t)}{1-\tan(s)\tan(t)}\)

This is the angle-sum identity for tangent. For the angle-difference identity, we may write:

\(\displaystyle \tan(s-t)=\frac{\sin(s-t)}{\cos(s-t)}=\frac{\sin(s)\cos(t)-\cos(s)\sin(t)}{\cos(s)\cos(t)+\sin(s)\sin(t)}\)

Divide each term in the numerator and denominator by $\cos(s)\cos(t)$:

\(\displaystyle \tan(s-t)=\frac{\tan(s)-\tan(t)}{1+\tan(s)\tan(t)}\)

Thus, we have proven the following identities:

\(\displaystyle \sin(s\pm t)=\sin(s)\cos(t)\pm\cos(s)\sin(t)\)

\(\displaystyle \cos(s\pm t)=\cos(s)\cos(t)\mp\sin(s)\sin(t)\)

\(\displaystyle \tan(s\pm t)=\frac{\tan(s)\pm\tan(t)}{1\mp\tan(s)\tan(t)}\)

- Thread starter
- #4

- Dec 25, 2012

- 42

- Jan 29, 2012

- 1,151

One of the difficulties with something like this is that it depends upon exactly how you have **defined** the functions. Chisigma showed how using the "[tex]sin(x)= \frac{e^{ix}- e^{-ix}}{2i}[/tex] definition and MarkFL showed how using the definition of sine and cosine in terms of coordinates of a point on the unit circle.

Another definition of sine and cosine that I like is:

(Admittedly, this is no longer "Pre-Calculus".)

"y= cos(x) is the unique function satisfying the initial value problem y''= -y with y(0)= 1, y'(0)= 0."

and

"y= sin(x) is the unique function satisfying the initial value problem y''= -y with y(0)= 0, y'(0)= 1."

It is easy to show that sine and cosine, as defined above, are**independent** functions and so any solutions to the differential equation y''= -y can be written as a linear combination of those two functions. In fact it is easy to see that the solution to the initial value problem y''= -y, y(0)= A, y'(0)= B is exacdtly

y(x)= A cos(x)+ B sin(x). That is, the coefficients of cosine and sine**are** the initial values of y and y'.

It is then easy to see that if y(x)= (sin(x))', the derivative of sin(x) then y'(x)= (sin(x))''= - sin(x) and, differentiating again, that y''(x)= - (sin(x))'= -y. That is, this new function satisfies the same differential equation while satisfying y(0)= (sin(x))' at 0 which, by definition, is 1 and y'(0)= - sin(0)= 0. That is, the derivative of sin(x) satisfies exactly the same initial value problem as cos(x) and so, by uniqueness, (sin(x))'= cos(x).

Similarly, it is easy to see that if y(x)= (cos(x))', y satisfies the same differential equation except that now we have y'(0)= - cos(0)= -1 so that (cos(x))'= -sin(x).

Now, let y(x)= sin(x+ a) for some number a. Using the chain rule, since the derivative of x+ a is 1, it is easy to see that y again satisfies y''= -y but now y(0)= sin(a) and y'(0)= cos(a). That tells us that

sin(x+ a)= sin(a)cos(x)+ cos(a)sin(x).

Taking x= b, we have

sin(a+ b)= sin(b+ a)= sin(a)cos(b)+ cos(a)sin(b).

Similarly, if we let y= cos(x+ a), we have y''= -y but now y(0)= cos(a), y'(0)= -sin(a) so that

cos(x+ a)= cos(a)cos(x)- sin(a)sin(x) and so

cos(a+ b)= cos(a)cos(b)- sin(a)sin(b)

Of course, replacing "b" by "-b" and using the fact that sine is an odd function, cosine is an even function

sin(a- b)= sin(a)cos(-b)+ cos(a)sin(-b)= sin(a)cos(b)- cos(a)sin(b).

Another definition of sine and cosine that I like is:

(Admittedly, this is no longer "Pre-Calculus".)

"y= cos(x) is the unique function satisfying the initial value problem y''= -y with y(0)= 1, y'(0)= 0."

and

"y= sin(x) is the unique function satisfying the initial value problem y''= -y with y(0)= 0, y'(0)= 1."

It is easy to show that sine and cosine, as defined above, are

y(x)= A cos(x)+ B sin(x). That is, the coefficients of cosine and sine

It is then easy to see that if y(x)= (sin(x))', the derivative of sin(x) then y'(x)= (sin(x))''= - sin(x) and, differentiating again, that y''(x)= - (sin(x))'= -y. That is, this new function satisfies the same differential equation while satisfying y(0)= (sin(x))' at 0 which, by definition, is 1 and y'(0)= - sin(0)= 0. That is, the derivative of sin(x) satisfies exactly the same initial value problem as cos(x) and so, by uniqueness, (sin(x))'= cos(x).

Similarly, it is easy to see that if y(x)= (cos(x))', y satisfies the same differential equation except that now we have y'(0)= - cos(0)= -1 so that (cos(x))'= -sin(x).

Now, let y(x)= sin(x+ a) for some number a. Using the chain rule, since the derivative of x+ a is 1, it is easy to see that y again satisfies y''= -y but now y(0)= sin(a) and y'(0)= cos(a). That tells us that

sin(x+ a)= sin(a)cos(x)+ cos(a)sin(x).

Taking x= b, we have

sin(a+ b)= sin(b+ a)= sin(a)cos(b)+ cos(a)sin(b).

Similarly, if we let y= cos(x+ a), we have y''= -y but now y(0)= cos(a), y'(0)= -sin(a) so that

cos(x+ a)= cos(a)cos(x)- sin(a)sin(x) and so

cos(a+ b)= cos(a)cos(b)- sin(a)sin(b)

Of course, replacing "b" by "-b" and using the fact that sine is an odd function, cosine is an even function

sin(a- b)= sin(a)cos(-b)+ cos(a)sin(-b)= sin(a)cos(b)- cos(a)sin(b).

Last edited:

- Jan 29, 2012

- 1,151

That is, we define [tex]cos(x)= \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} x^{2n}[/tex] and define [tex]sin(x)= \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n}[/tex].

We can show, perhaps by using the ratio test, that those sums converge for all x. It follows from that that they converge

From that we find that (sin(x))'= cos(x) and (cos(x))'= - sin(x). Differentiating again, (sin(x))''= -sin(x) and (cos(x))''= - cos(x). That is, sin(x) and cos(x) both satisfy y''= -y and it is easy, by evaluating those sums at x= 0, to show that they also satisfy the initial conditions we need to use the previous proof.

- Moderator
- #7

- Feb 7, 2012

- 2,806

Now if you rotate through an angle $a$ and then through an angle $b$, those combined operations are equal to a rotation through an angle $a+b$. Therefore $$\begin{bmatrix}\cos (a+b) & -\sin (a+b) \\ \sin (a+b) & \cos (a+b)\end{bmatrix} = \begin{bmatrix}\cos b & -\sin b \\ \sin b & \cos b\end{bmatrix}\begin{bmatrix}\cos a & -\sin a \\ \sin a & \cos a\end{bmatrix}.$$ Multiply the matrices on the right side of that equation, using the standard procedure for matrix multiplication, and you find that $$\begin{bmatrix}\cos (a+b) & -\sin (a+b) \\ \sin (a+b) & \cos (a+b)\end{bmatrix} = \begin{bmatrix}\cos a \cos b - \sin a\sin b & -\sin a \cos b - \cos a\sin b \\ \sin a \cos b + \cos a\sin b & \cos a \cos b - \sin a\sin b\end{bmatrix}.$$ Compare corresponding entries in those two matrices and you get the addition formulae for cos and sin.

I assume you are seeking a

I was wondering if someone could take the time to show

a proof for the sum and difference identity for sine.

Code:

```
P
. . . *
. . *|*
. * |a*
* | *
* | *
. . . 1 * | *
. . * S*-----* Q
. * | * |
* * |
* * | |
* b * | |
* * a | |
O * * * * * * *
T R
```

Let [tex]b = \angle POQ.\;PQ \perp OQ.\;OP = 1.[/tex]

Draw [tex]PT \perp OR.[/tex]

Draw [tex]QS \parallel OR.[/tex]

Note that: [tex]\angle QPS = a.[/tex]

We find that:.[tex]PQ = \sin b[/tex]

Hence:.[tex]PS \:=\: PQ\cdot\cos a \:=\:\sin b\cos a[/tex]

We find that:.[tex]OQ = \cos b[/tex]

Hence: .[tex]ST \,=\,QR \,=\,OQ\cdot\sin a \,=\,\cos b\sin a[/tex]

Then:.[tex]\sin(a+b) \:=\:\frac{PT}{OT} \:=\:\frac{PS+ST}{OP}[/tex]

. . . . . . . . . . . . [tex]=\:\frac{\sin b\cos a + \cos b\sin a}{1} [/tex]

Therefore: .[tex]\sin(a+b) \:=\:\sin a\cos b + \sin b\cos a[/tex]