Welcome to our community

Be a part of something great, join today!

Trigonometry Prove Sum and Difference Sine?

Farmtalk

Active member
Dec 25, 2012
42
I was wondering if someone could take the time to show a proof for the sum and difference identity for sine. I've seen and learned to understand some other identities, but never this one.

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

chisigma

Well-known member
Feb 13, 2012
1,704
I was wondering if someone could take the time to show a proof for the sum and difference identity for sine. I've seen and learned to understand some other identities, but never this one.

I've been trying to understand more of the "why" than the "how" of mathematics, and this one is very intriguing to me :cool:
When Leonhard Euler 'discovered' the relation...

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

... 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

$\chi$ $\sigma$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Let's now prove the angle sum/difference identities for sine, cosine and tangent. First, we need to prove the law of cosines. I trust the reader will have available pencil and paper to make the drawings I describe. (Whew)

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:

anglesd01.png

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:

anglesd02.png

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:

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

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

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

Farmtalk

Active member
Dec 25, 2012
42
I wish I could give you 5 thanks a piece! I know it probably took some time for each of you to post! I just wanted to say I'm real grateful that you both posted! Thanks a ton!!!!!:D
 

HallsofIvy

Well-known member
MHB Math Helper
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).
 
Last edited:

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Yet, another way! We can define sine and cosine as power series.

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 uniformly on all closed an bounded intervals and, particularly, that they can be differentiated "term by term" at any x.

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.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,708
Here is yet another method, which relies on knowing a bit of linear algebra. The matrix $\begin{bmatrix}\cos a & -\sin a \\ \sin a & \cos a\end{bmatrix}$ represents the operation of rotation through an angle $a$. In other words, if you rotate the point $\begin{bmatrix}x \\ y\end{bmatrix}$ through an angle $a$ around the origin, it gets moved to the point $\begin{bmatrix}\cos a & -\sin a \\ \sin a & \cos a\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}$.

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.
 

soroban

Well-known member
Feb 2, 2012
409
Hello, Farmtalk!

I assume you are seeking a basic proof.


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]a = \angle QOR.\;QR \perp OR[/tex]

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]