Post a New Question

Precalculs

posted by on .

I have no idea how to do these type of problems.

-------Problem--------
Solve each equation on the interval 0 less than or equal to theta less than 2 pi

42. SQRT(3) sin theta + cos theta = 1
----------------------

There is an example prior to the excersises that attempts to explain how to do these types of problems. I do not understand it...

-------EXAMPLE---------
Solve the equation on the interval 0 less than or equal to theta less than 2 pi

a sin theta + b cos theta = c (2)

where a, b, and c are constants and either a does not equal 0 or b does not equal 0.

We divide each side each side of equation (2) by SQRT(a^2 + b^2). Then

a sin theta /SQRT(a^2 + b^2) + b cos theta/SQRT(a^2 + b^2) = c/SQRT(a^2 + b^2) (3)

There is a unique angle phi, 0 less than or equal to phi less than 2 pi, for which

cos phi = a/SQRT(a^2 + b^2) and sin phi = b/SQRT(a^2 + b^2) (4)

Figure 36
h t t p : / / i m g 2 6 9 . i m a g e s h a c k . u s / i m g 2 6 9 / 3 0 0 2 / c a p t u r e n m k . j p g

See Figure 36. Equation (3) may be written as sin theta cos phi + cos theta sin phi = c/SQRT(a^2 + b^2)

or, equivalentley,

sin(theta + phi) = c/SQRT(a^2 + b^2) (5)

where phi satisfies equation (4).

If |c| > SQRT(a^2 + b^2), then sin(theta + phi) > 1 or sin(theta + phi) < -1, and equation (5) has no solution.

If |c| less than or equal to SQRT(a^2 + b^2), then the solutions of equation (5) are

theta + phi =sin^-1 c/SQRT(a^2 + b^2) or theta + phi = pi - sin^-1 c/SQRT(a^2 + b^2)

Because the angle phi is determined by equations (4), these are the solutions to equation (2).
------------------------

Well I didn't understand but here is my attempt at solving the PROBLEM which is clearly wrong

-----ATEMPT AT A SOLUTION-------

Solve each equation on the interval 0 less than or equal to theta less than 2 pi

42. SQRT(3) sin theta + cos theta = 1

I just plugged and chugged into what the book claims to be the solutions into the formula

theta + phi =sin^-1 c/SQRT(a^2 + b^2) or theta + phi = pi - sin^-1 c/SQRT(a^2 + b^2)

theta + phi = sin^-1 1/SQRT(3 +1) = sin^-1 1/2 = pi/6

or

theta + phi = pi - pi/6 = (5 pi)/6

-----------------------

Well I'm lost so if you could guide me in some way to solve these type of problems that would be extremely appreciated... TAHNKS!!!!

  • Precalculs - ,

    wow!
    Let me show you how I would do this question.
    (I will use x instead of theta for easier typing)

    √3sinx + cosx = 1
    √3cosx = 1-cosx
    square both sides
    3cos^2 x = 1 - 2cosx + cos^2 x
    3(1-sin^2 x) = 1 - 2cosx + cos^2 x
    this reduces easily to
    2cos^2 x - cosx - 1 = 0
    (2cosx + 1)(cosx - 1) = 0
    cosx = -1/2 or cosx = 1
    x or theta = 120º , 240º or 0 , 360º
    x = 0, 2pi/3, 4pi/3, 2pi

    since we squared we must test each answer in the original equation
    4pi/3 does not work, so
    x = 0, 2pi/3, 2pi

  • Here is what they want - Precalculs - ,

    In a rather complicated way they are trying to say that
    any linear combination of sine and cosine can be expressed as a single sine or cosine function of the form a(sin(x+ß)

    let a sin(x+ß) = √3sinx + cosx
    = a(sinxcosß + cosxsinß)

    This is true if
    a sinxcosß = √3sinx AND a cosxsinß = cosx
    then
    cosß = √3/a and sinß = 1/a

    now sinß/cosß = (1/a)/(√3/a) = 1/√3
    tanß = 1/√3
    ß = 30º or pi/6
    then sin 30º = 1/a
    1/2 = 1/a ---> a = 2

    so our original √3 sin x + cos x = 1 becomes
    2sin(x+30º) = 1
    sin(x+30º) = 1/2
    x+30 = 0 or x+30 = 150º
    x = 0º or 120º
    the period of sin(x+30) is 360º so we can add 360 to any answer as long as that keeps us in our domain
    so x = 0, 120 , or 360

    just like I had before

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question