Summing sines and cosines with trig.

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In summary, the conversation discusses using trigonometric formulas to write two terms as a single harmonic. The problem is equivalent to the expression sin(2x) + sin(2(x+(pi/3))), which can be expanded using the given trig formulas. The final solution is (1/2)sin2x - (sqrt(3)/2)cos2x, and it is suggested to use the identity A sin(2x) + B cos(2x) = √(A^2 + B^2) sin(2x + phi) to simplify it further.
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res3210
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Hey guys,

I just wanted to come here and see if anyone could help me out. I am getting stuck using the trig formulas for summing sines and cosines. I won't beat around the bush too much, so let's get right into it.

Homework Statement


Use a trig formula to write the two terms as a single harmonic.
sin(2x) + sin(2(x+(pi/3)))

Homework Equations


sin(A+B) = sinA*cosB + sinB*cosA
cos(A+B) = cosA*cosB - sinA*sinB
sin^2(A) + cos^2(A) = 1

The Attempt at a Solution


So I think that I might be over-thinking this, or maybe I am just not recognizing some pattern which would lead me to the solution. I'll walk through my steps:

1. I assume that to solve this, I should expand these two functions with the trig formulas I mentioned above. I am hoping that in doing so, I can combine like terms and find a single sine or cosine function which gives me all the information I need.

2. The given problem is equivalent to the expression: sin2x + sin(2x + 2pi/3)

3. From that follows:
2*sinx*cosx + sin2x*cos(2*pi/3) + cos2x*sin(2*pi/3)

4. From that, I get:
2*sinx*cosx - (1/2)sin2x - (sqrt(3)/2)cos2x

5. Following that:
2*sinx*cosx - sinx*cosx - (sqrt(3)/2)(cos^2(x) - sin^2(x))

6. Then I get:
sinx*cosx - (sqrt(3)/2)(cos^2(x) - sin^2(x))

7. Which ultimately leads me to:
(1/2)sin2x - (sqrt(3)/2)cos2x

Step seven is where I get stuck. I honestly have no clue where to go from here. Any tips or hints would be greatly appreciated!
 
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  • #3
Thanks! Would it be 2(x + phi) or 2x + phi?
 
  • #4
res3210 said:
Thanks! Would it be 2(x + phi) or 2x + phi?

2x+phi. Just substitute 2x for x into the formula. That doesn't change anything does it?
 

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