Trig problem using formuals to transform the equation

In summary, the conversation discusses solving a trigonometric equation involving the identity sin(2x) - cos(2x) = sqrt(2)sin(2x + Aπ). The solution is found by equating coefficients and using the expansion for sin(A+B). The final answer is A = 7/4.
  • #1
agentc0re
2
0

Homework Statement



sin(2x)-cos(2x)=sqrt(2)sin(2x+Api), then the number 0 < A = ____ < 2

Homework Equations



http://bit.ly/9njiUW <-- trig reference sheet with formulas and identites

The Attempt at a Solution



So I start with the left side of the equation and attempt to make it look like the right in order to figure out what A is. Using some identities i started out:

2sin(x)cos(x) - cos^2(x) + sin^2(x)
2sin(x)cos(x) - ( 1 - sin^2(x) ) + sin^2(x)
2sin(x)cos(x) - 1 + sin^2(x) +sin^2(x)
2sin(x)cos(x) - 1 + 2sin^2(x)
2( 1/2[ sin(x + x) + sin(x - x) ] ) - 1 + 2sin^2(x)
sin(x + x) + sin(x - x) - 1 + 2sin^2(x)
sin(2x) + sin(0) - 1 + 2sin^2(x)
sin(2x) + 0 - 1 + 2sin^2(x)
sin(2x) -1 + 2sin^2(x)

Then I realized I made a big loop with the "sin(2x)" part and am not sure how to go from there.
 
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  • #2
The only identity you need is the expansion for sin(A+B). When you do that you'll have

[tex]sin2x-cos2x=\sqrt{2}sin2x.cosA\pi+\sqrt{2}cos2x.sinA\pi[/tex]

And notice that the right side is of the form [tex]C_1sin2x-C_2cos2x[/tex] where [tex]C_1,C_2[/tex] are just some constants.

This means you can equate both sides, so [tex]1=C_1[/tex] and [tex]1=C_2[/tex]
 
  • #3
I still don't understand how you've mentioned to transform the right side equals the left side.
1. I don't understand how i would transform anything on the right side to have a [tex]\sqrt{2}[/tex]
2. I do know the answer but i don't know how to get there. I used wolframalpha and the answer is "A=[tex]\frac{7}{4}[/tex]"

Maybe I didn't explain myself well enough? I dunno... Hrmm... Well here's an example of another problem that I was able to work out that is like the one I've asked help for.

[tex]sin(x-\pi) = Asin(x)[/tex]

[tex]sin(x)cos(\pi)-cos(x)sin(\pi)[/tex]
[tex]sin(x)cos(\pi)-cos(x)(0)[/tex]
[tex]sin(x)cos(\pi)[/tex]
[tex]sin(x)(-1)[/tex]
[tex]-1sin(x)[/tex]
[tex]A= -1[/tex]

So that's how I've attempted to try and solve this one but with no luck. :/
 
  • #4
It's exactly the same process, but in a little more complicated fashion.

In that other example you ended up with [tex]-sinx=Asinx[/tex] and then since sinx is on both sides of the equation, you can equate the coefficients which are -1 and A, so then -1=A.

Another way to look at it is to take them to one side, [tex]-sinx-Asinx=0[/tex]

and then [tex]sinx(-1-A)=0[/tex] and since sinx can be all values from -1 to 1 and not just 0, we have to consider when the other factor -1-A=0, thus A=-1

The same deal applies for your question...

Equate the coefficients of both sides (thus equate the coefficients of sin2x and cos2x) or equivalently you can take everything to one side and factor out sin2x and cos2x which will give you the same answer.

[tex]
sin2x-cos2x=\sqrt{2}sin2x.cosA\pi+\sqrt{2}cos2x.sinA\pi
[/tex]

[tex]
sin2x-\sqrt{2}sin2x.cosA\pi-cos2x-\sqrt{2}cos2x.sinA\pi=0
[/tex]

[tex]sin2x(1-\sqrt{2}cosA\pi)+cos2x(-1-\sqrt{2}sinA\pi)=0[/tex]

Now simply make each factor equal to 0.It's pretty much the same as when you equate coefficients of different variables, say if [tex]ax^2+bx+c\equiv 2x^2+3x+4[/tex] then a=2,b=3,c=4.
 

Related to Trig problem using formuals to transform the equation

1. What is the formula for transforming a trigonometric equation?

The formula for transforming a trigonometric equation is: f(x) = a * sin(bx + c) + d, where a is the amplitude, b is the frequency, c is the phase shift, and d is the vertical shift.

2. How do I determine the amplitude of a trigonometric equation?

The amplitude of a trigonometric equation is equal to a, the coefficient of the sine or cosine function. It represents the maximum vertical distance from the midline of the graph.

3. What does the frequency in a trigonometric equation represent?

The frequency, b, in a trigonometric equation determines how many cycles of the function occur within a given interval. It is equal to divided by the period of the function.

4. How do I find the phase shift of a trigonometric equation?

The phase shift, c, of a trigonometric equation is equal to -c/b, where b is the frequency. It represents the horizontal shift of the graph.

5. Can I use trigonometric equations to solve real-world problems?

Yes, trigonometric equations can be used to solve a variety of real-world problems involving periodic motion, such as finding the height of a roller coaster at a given time or determining the optimal angle for a ramp.

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