Trig problem using formuals to transform the equation

[agentc0re]

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.

 

[Mentallic]

The only identity you need is the expansion for sin(A+B). When you do that you'll have

sin2x-cos2x=\sqrt{2}sin2x.cosA\pi+\sqrt{2}cos2x.sinA\pi

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

This means you can equate both sides, so 1=C_1 and 1=C_2

 

[agentc0re]

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 \sqrt{2}
2. I do know the answer but i don't know how to get there. I used wolframalpha and the answer is "A=\frac{7}{4}"

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.

sin(x-\pi) = Asin(x)

sin(x)cos(\pi)-cos(x)sin(\pi)
sin(x)cos(\pi)-cos(x)(0)
sin(x)cos(\pi)
sin(x)(-1)
-1sin(x)
A= -1

So that's how I've attempted to try and solve this one but with no luck. :/

 

[Mentallic]

It's exactly the same process, but in a little more complicated fashion.

In that other example you ended up with -sinx=Asinx 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, -sinx-Asinx=0

and then sinx(-1-A)=0 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.

<br /> sin2x-cos2x=\sqrt{2}sin2x.cosA\pi+\sqrt{2}cos2x.sinA\pi<br />

<br /> sin2x-\sqrt{2}sin2x.cosA\pi-cos2x-\sqrt{2}cos2x.sinA\pi=0<br />

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

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