# Products, Sums, Linear Combinations, and Applications

Rewriting A*sinx + B*cosx using cosine sum and difference

Tip: This isn't the place to ask a question because the teacher can't reply.

## Key Questions

As below.

#### Explanation:

color(blue)("Sum to Product Formulas"

color(green)("Product to Sum Formulas"

• Main approach to solve a trig equation : Use Trig Transformation Identities to transform it to a product of a few basic trig equations. Solving a trig equation finally results in solving a few basic trig equations.
Transformation Trig Identities that convert Sums to Products .
1. cos a + cos b = 2cos (a +b)/2cos (a - b)/2
2. cos a - cos b = -2sin (a + b)/2sin (a - b)/2
3. sin a + sin b = 2sin (a + b)/2cos (a - b)/2
4. sin a - sin b = 2cos (a + b)/2sin (a - b)/2
5. tan a + tan b = sin (a + b)/cos acos b.
6. tan a - tan b = sin (a - b)/cos a
cos b
Example 1 . Transform f(x) = sin a + cos a to a product.
Solution. Use Identity (3) to transform f(x) = sin a + sin (Pi/2 - a) = 2sin (Pi/4)sin (a + Pi/4)
Example 2 . Transform f(x) = sin x + sin 3x + sin 2x to a product. Use Identity (3) to transform the sum (sin x + sin 3x), then put in common factor.
f(x) = (2sin 2acos a) + 2sin acos a = 2cos a(2sin 3a/2*cos a/2)

Substitute

$a \cos x + b \sin x = \sqrt{{a}^{2} + {b}^{2}} \cos \left(x - \textrm{\arctan 2} \left(b / , a\right)\right)$

where arctan2 is the two parameter, four quadrant inverse tangent.

#### Explanation:

I've been answering all these old math questions. It's hard to know if anyone reads the answers.

The linear combination of a cosine and a sine of the same angle corresponds to a scaling and a phase shift. Let's explain how that works.

The linear combination of a cosine and a sine of the same angle is an expression of the form:

$a \cos x + b \sin x$

That looks very much like the sum angle formula for sine or the difference angle formula for cosine:

$\sin \left(\alpha + \beta\right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$

$\cos \left(\alpha - \beta\right) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$

Indeed, we can take our linear combination and transform it into a scaled version of either of these. Pro Tip: When given a choice, prefer cosine to sine.

Let's multiply by a scale factor $r$ and set $\alpha = x , \beta = \theta .$

$r \cos \left(x - \theta\right) = r \cos \theta \cos x + r \sin \theta \sin x$

Matching to our linear combination leads us to want to solve for $r$ and $\theta$:

$a = r \cos \theta$

$b = r \sin \theta$

We've seen this before. It's how we turn turn the polar coordinates $P \left(r , \theta\right)$ to rectangular coordinates, here $\left(a , b\right) .$ So our task is to turn $\left(a , b\right)$ to polar coordinates.

Let's remind ourselves how to do that, perhaps in a bit more detail than usually seen.

${a}^{2} + {b}^{2} = {\left(r \cos \theta\right)}^{2} + {\left(r \sin \theta\right)}^{2} = {r}^{2} \left({\cos}^{2} \theta + {\sin}^{2} \theta\right) = {r}^{2}$

$r = \sqrt{{a}^{2} + {b}^{2}}$

$\sin \theta = \frac{b}{r}$

$\cos \theta = \frac{a}{r}$

$\tan \theta = \frac{b}{a}$

$\theta = \textrm{\arctan 2} \left(b / , a\right)$

Usually this is written as the regular arctangent, but that's not really right. The regular arctangent only covers two quadrants, and doesn't work on the y axis. This is the two parameter, four quadrant inverse tangent, which returns a valid theta for all real input pairs. The $/ ,$ is deliberate, reminding us there are two separate parameters, and which is which.

So now we're assured $a = r \cos \theta$ and $b = r \sin \theta$ so

$a \cos x + b \sin x = r \cos \theta \cos x + r \sin \theta \sin x = r \cos \left(x - \theta\right)$

$a \cos x + b \sin x = \sqrt{{a}^{2} + {b}^{2}} \cos \left(x - \textrm{\arctan 2} \left(b / , a\right)\right)$

When at $\left({71.06047}^{\circ} W , {43.08350}^{\circ} N\right)$ a distant cell tower is at heading ${131}^{\circ} \left(S E\right)$, and at $\left({71.06137}^{\circ} W , {43.08007}^{\circ} N\right)$ it's at heading ${99}^{\circ} \left(E\right)$. Where is the tower?

#### Explanation:

Someone from San Antonio requested an answer three years ago! Hope they've figured it out by now.

Here some real life trig I've been meaning to do. I wanted to know where that cell phone tower I can see from my house is. It's on a hill in the distance.

From one spot near my house, maybe the one in this picture, I pointed my phone at the tower with my GPS app on and got this:

The relevant information is:

$\left({71.06047}^{\circ} W , {43.08350}^{\circ} N\right)$ heading ${131}^{\circ} \left(S E\right)$

At another spot I got

$\left({71.06137}^{\circ} W , {43.08007}^{\circ} N\right)$ heading ${99}^{\circ} \left(E\right)$

The heading is relative to due north. Here's a figure.

We treat west as negative. I translated the origin to (-71.06, 43,08) and multiplied the coordinates by 100,000 so our new problem is:

$\left(- 047 , 350\right)$ heading ${131}^{\circ} \left(S E\right)$

$\left(- 137 , 007\right)$ heading ${99}^{\circ} \left(E\right)$

Find Tower coordinates $\left(x , y\right)$

To solve we write the equations of the two lines and find the meet.

When measured relative to the y axis like that, the heading angle $\theta$ converts to a slope as $m = \cot \theta$.

So our two lines

$y - 350 = \left(x + 47\right) \cot 131$

$\left(y - 7\right) = \left(x + 137\right) \cot 99$

meet when

 350 + (x+47) cot 131 = 7+ (x+137) cot 99 

$x = \frac{7 + 137 \cot 99 - 350 - 47 \cot 131}{\cot 131 - \cot 99}$ x

 xâ‰ˆ 455.537

y â‰ˆ-86.849 #

That means my tower is at GPS coordinates

$\left(- 71.06 + .00456 , 43.08 - .000868\right) = \left(- 71.0554 , 43.07132\right)$

Let's check Google Maps. Pretty good, off by around 400 feet.

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