Products, Sums, Linear Combinations, and Applications

Key Questions

How do you use the transformation formulas to go from product to sum and sum to product?

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 acos 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)

Nghi N. · 3 · Apr 12 2015
How do you use linear combinations to solve trigonometric equations?

Answer:

Substitute

$a cos x + b sin x = sqrt{a^2 + b^2} cos(x -text{arctan2}( b //, a) )$

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:

$sin(alpha + beta ) = sin alpha cos beta + cos alpha sin beta$

$cos(alpha - beta) = cos alpha cos beta + sin alpha sin beta$

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 $alpha = x, beta=theta.$

$r cos(x - theta) = r cos theta cos x + r sin theta sin x$

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 $P(r, theta)$ to rectangular coordinates, here $(a,b).$ So our task is to turn $(a,b)$ to polar coordinates.

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

Squaring and adding we get

$a^2 + b^2 = (r cos theta)^2 + (r sin theta)^2 = r^2(cos^2 theta + sin^2 theta) = r^2$

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

$sin theta = b/r$

$cos theta = a/r$

$tan theta = b/a$

$theta = text{arctan2}( b //, a)$

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

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

$a cos x + b sin x = sqrt{a^2 + b^2} cos(x -text{arctan2}( b //, a) )$

Dean R. · 1 · Apr 27 2018
How do you apply trigonometric equations to solve real life problems?

Answer:

When at $(71.06047^circ W, 43.08350^circ N)$ a distant cell tower is at heading $131^circ (SE)$ , and at $(71.06137^circ W, 43.08007^circ N)$ it's at heading $99^circ (E)$ . 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:

$(71.06047^circ W, 43.08350^circ N)$ heading $131^circ (SE)$

At another spot I got

$(71.06137^circ W, 43.08007^circ N)$ heading $99^circ (E)$

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:

$(-047, 350)$ heading $131^circ (SE)$

$(-137, 007)$ heading $99^circ (E)$

Find Tower coordinates $(x,y)$

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

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

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

meet when

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

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

$x≈ 455.537$

$y ≈-86.849$

That means my tower is at GPS coordinates

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

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

Dean R. · 2 · Apr 27 2018
What is the sum to product formulas?

Answer:

As below.

Explanation:

$color(blue)("Sum to Product Formulas"$

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$color(green)("Product to Sum Formulas"$

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sankarankalyanam · 1 · Mar 19 2018