Help with a proof (1 Viewer)

[Intilegience]

Hi, I was struggling with this question in the Cambridge textbook so I stumbled upon a solution online which seems to work, however I dont understand how the person thought of this and why this works. I also dont understand what the part about the pi -pi stuff is about either

If someone who understands this could explain it in a more easier way I would very much appreciate this, and if anyone has a different more methodical way pls could you share, even if it is longer as long as it is more "inside the box" I would be very grateful.

Thanks.

 

[WeiWeiMan]

Hi, I was struggling with this question in the Cambridge textbook so I stumbled upon a solution online which seems to work, however I dont understand how the person thought of this and why this works. I also dont understand what the part about the pi -pi stuff is about either

If someone who understands this could explain it in a more easier way I would very much appreciate this, and if anyone has a different more methodical way pls could you share, even if it is longer as long as it is more "inside the box" I would be very grateful.

Thanks.

since 0<a1<a2<π/2, w1 and w2 are in Q1. furthermore, since they both have the same modulus of 1, you can form a rhombus with diagonals w1+w2 and w1-w2

arg(w1-w2)=arg(w1+w2)-π/2 {rhombus diagonals are perpendicular}
w1+w2 (D) bisects the angle W2OW1 so arg(w1+w2)=1/2(a2-a1)+a1=1/2(a1+a2)
therefore arg(w1-w2)=1/2(a1+a2)-π/2=1/2(a1+a2-π)

 

[scaryshark09]

 

[kendricklamarlover101]

Hi, I was struggling with this question in the Cambridge textbook so I stumbled upon a solution online which seems to work, however I dont understand how the person thought of this and why this works. I also dont understand what the part about the pi -pi stuff is about either

If someone who understands this could explain it in a more easier way I would very much appreciate this, and if anyone has a different more methodical way pls could you share, even if it is longer as long as it is more "inside the box" I would be very grateful.

Thanks.

Another way to do it algebraically is to represent the complex numbers as cis(a1) and cis(a2) then use the sum to product trigonometric identities.

 

[Luukas.2]

Bear in mind that the question asks for Arg(w₁ - w₂), with a capital A - not just for arg(w₁ - w₂).

Thus, you seek not just the argument of the complex number, but specifically its principal argument.

Now, as and are acute, and so the found argument is the principal argument, but something should be noted in the answer on this point, IMO.