# Derivation of Sum and Difference Angles

• Mar 10th 2006, 07:36 PM
c_323_h
Derivation of Sum and Difference Angles
$\displaystyle \sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$
$\displaystyle \sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta$
$\displaystyle \cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$
$\displaystyle \cos(\alpha - \beta) = \cos\alpha\cos\beta + sin\alpha\sin\beta$
and then there is $\displaystyle \tan$.
could someone tell me how these sum and difference angle formulas are derived?
i've tried working it out but no luck. you don't have to do all of them, just one will suffice.
thanks
• Mar 11th 2006, 08:34 AM
topsquark
Quote:

Originally Posted by c_323_h
$\displaystyle \sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$
$\displaystyle \sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta$
$\displaystyle \cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$
$\displaystyle \cos(\alpha - \beta) = \cos\alpha\cos\beta + sin\alpha\sin\beta$
and then there is $\displaystyle \tan$.
could someone tell me how these sum and difference angle formulas are derived?
i've tried working it out but no luck. you don't have to do all of them, just one will suffice.
thanks

I'm going to try this...It would go smoother if I could post a diagram, but my scanner is still acting like it's got a hangover.

The picture:
On a typical xy coordinate system, sketch a line segment of unit length starting at the origin and at an angle $\displaystyle \beta$. For convenience of sketching, place this segment in the first quadrant. Label the endpoint of this segment to be the point Q. $\displaystyle Q=(cos\beta, sin\beta)$. (Note: I don't know why this isn't LaTeX!)

Sketch another line segment of unit length in the second quandrant (again for convenience) starting from the origin and label its endpoint P. This will be at an angle $\displaystyle \alpha$ measured from the +x axis. $\displaystyle P=(cos\alpha,sin\alpha)$. (Note that $\displaystyle \pi /2 < \alpha < \pi$ so I don't need extra negative signs on the components.)

Finally, connect PQ with a line segment. This gives a triangle POQ (O is the origin). The angle POQ is $\displaystyle \alpha-\beta$.

The calculations:
We are going to get the length of PQ in two different ways. The first uses the above diagram in the given coordinate system. Label the coordinates of point P as $\displaystyle (x_1,y_1)$ and Q as $\displaystyle (x_2,y_2)$ and simply use the distance formula.
$\displaystyle PQ^2=(x_1-x_2)^2+(y_1-y_2)^2$
$\displaystyle PQ^2=(1*cos\alpha-1*cos\beta)^2+(1*sin\alpha-1*sin\beta)^2$
$\displaystyle PQ^2=cos^2\alpha-2*cos\alpha \, cos\beta + cos^2\beta$
$\displaystyle +sin^2\alpha-2*sin\alpha \, sin\beta + sin^2\beta$

Finally we get:
$\displaystyle PQ^2=2-2(cos\alpha \, cos\beta + sin\alpha \, sin\beta)$.

New diagram:
More precisely, new coordinate system. Make a new set of x'y' axes, with the x' axis along the line segment OQ. (This merely has the effect of rotating the previous diagram so that OQ now lies along the x axis.) The coordinates of Q are now (1,0) and P are now $\displaystyle (cos(\alpha-\beta), sin(\alpha-\beta))$. Again we find the length of PQ. (The line segments OQ, OP, and QP all have the same lengths as they did in the original coordinate system, of course.)
$\displaystyle PQ^2=(x'_1-x'_2)^2+(y'_1-y'_2)^2$
$\displaystyle PQ^2=(cos(\alpha-\beta)-1)^2+(sin(\alpha-\beta)-0)^2$

Which eventually comes to:
$\displaystyle PQ^2=2-2cos(\alpha-\beta)$

Equating our two values for PQ gives us:
** $\displaystyle cos(\alpha-\beta)=cos\alpha \, cos\beta + sin\alpha \, sin\beta$.

You can use this formula to be clever and find a number of relationships.
1. By setting $\displaystyle \alpha = \pi /2$ in ** we get $\displaystyle cos(\pi /2 -\beta)=sin\beta$.

2. Letting $\displaystyle \beta \rightarrow \pi /2 - \beta$ in the expression in 1 gives $\displaystyle cos\beta = sin(\pi /2 - \beta)$.

3. Putting $\displaystyle \alpha = 0$ into ** we get $\displaystyle cos(-\beta)=cos\beta$.

4. Expanding the cosine term in 1 using ** and putting $\displaystyle \beta \rightarrow -\beta$ we get $\displaystyle sin(-\beta)=-sin\beta$.

We can now use these to get the other sine and cosine relations:
$\displaystyle cos(\alpha+\beta)=cos(\alpha-(-\beta))$
Using ** to expand the cosine on the RHS and relations 3 and 4 give:
$\displaystyle cos(\alpha+\beta)=cos\alpha \, cos\beta - sin\alpha \, sin\beta$.

Starting with 1: $\displaystyle sin\beta=cos(\pi /2 -\beta)$ and putting $\displaystyle \beta \rightarrow \alpha+\beta$ gives:
$\displaystyle sin(\alpha+\beta)=cos(\pi /2 - \alpha - \beta)=cos([\pi /2 - \alpha] - \beta)$.
Now we use ** to expand the RHS:
$\displaystyle sin(\alpha+\beta)=cos[\pi /2 - \alpha] \, cos\beta + sin[\pi /2 -\alpha] \, sin\beta$.
And finally, we use 1 and 2 to get:
$\displaystyle sin(\alpha+\beta)=sin\alpha \, cos\beta + cos\alpha \, sin\beta$.

For the last formula:
$\displaystyle sin(\alpha-\beta)=sin(\alpha+(-\beta))$
We use the above expression for sine to expand this and 3 and 4 to get:
$\displaystyle sin(\alpha-\beta)=sin\alpha \, cos\beta - cos\alpha \, sin\beta$.

The tangent formulas come from using $\displaystyle tan(\alpha \pm \beta)=\frac{sin(\alpha \pm \beta)}{cos(\alpha \pm \beta)}$ and dividing both the numerator and denominator by $\displaystyle cos\alpha \, cos\beta$.

-Dan
• Mar 11th 2006, 01:58 PM
TD!
If you know how the trigonometric functions are defined by using complex exponentials, showing these formulas becomes easy.

We have $\displaystyle e^{i\left( {\alpha + \beta } \right)} = \cos \left( {\alpha + \beta } \right) + i\sin \left( {\alpha + \beta } \right)$

But because of the exponent laws and properties, we can rewrite

$\displaystyle e^{i\left( {\alpha + \beta } \right)} = e^{i\alpha } e^{i\beta } = \left( {\cos \alpha + i\sin \alpha } \right)\left( {\cos \beta + i\sin \beta } \right)$

Working out and using iČ = -1 gives

$\displaystyle \cos \alpha \cos \beta + i\cos \alpha \sin \beta + i\sin \alpha \cos \beta - \sin \alpha \sin \beta$

Grouping real and imaginary parts

$\displaystyle \left( {\cos \alpha \cos \beta - \sin \alpha \sin \beta } \right) + i\left( {\cos \alpha \sin \beta + \sin \alpha \cos \beta } \right)$

Then compare with what we started with

$\displaystyle \cos \left( {\alpha + \beta } \right) + i\sin \left( {\alpha + \beta } \right)$

This yields

$\displaystyle \begin{array}{l} \cos \left( {\alpha + \beta } \right) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \\ \sin \left( {\alpha + \beta } \right) = \cos \alpha \sin \beta + \sin \alpha \cos \beta \\ \end{array}$

The formulas for the difference are easily derived by changing beta into -beta. The formula for tan follows from sin/cos.
• Mar 11th 2006, 03:15 PM
ThePerfectHacker
Let me just state something useful. If you know any one of these formulas you can easily derive the other three know that sine and cosine are co-functions.
• Mar 12th 2006, 12:26 PM
c_323_h
Quote:

Originally Posted by ThePerfectHacker
Let me just state something useful. If you know any one of these formulas you can easily derive the other three know that sine and cosine are co-functions.

i would appreciate it if you could show me? i don't quite understand how this would apply.

$\displaystyle \sin\theta=\cos(\frac{\pi}{2}-\theta)$
• Mar 12th 2006, 07:20 PM
c_323_h
Quote:

Originally Posted by ThePerfectHacker
Let me just state something useful. If you know any one of these formulas you can easily derive the other three know that sine and cosine are co-functions.

ahh i see now.