Products, Sums, Linear Combinations, and Applications

Add yours
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

  • Answer:

    As below.

    Explanation:

  • 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)

  • 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) ) #

  • 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.

    enter image source here

    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:

    enter image source here

    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.

    enter image source here

    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.

    enter image source here

Questions

  • Eddie answered · 1 year ago
  • dk_ch answered · 1 year ago