Well!This is a nice problem

If sin(A+B)=sinA+sinB\sin(A+B)=\sin A+\sin B Then how many distinct values of A,BA,B are there?

Note by Sumukh Bansal
3 years, 6 months ago

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Infinitely many.

Both A/πA/\pi and B/πB / \pi can be any integer, and the equation will be fulfilled.

Pi Han Goh - 3 years, 6 months ago

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^^ This

I'm not sure if this is a complete solution set, but I found that

A=2πkA = 2\pi k for some integer kk will be a solution for all BB

A+B=2πkA+B = 2\pi k for some integer kk will be a solution

Brandon Monsen - 3 years, 6 months ago

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@Brandon Monsen I could go into more detail by expanding sin(A+B) = sinA cosB + cosA sinB, but that's not needed. The question only asked for "How many solutions are there?"

Pi Han Goh - 3 years, 6 months ago

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@Pi Han Goh Oh when I said "not sure if this is a complete solution set" I was referring to myself. Yours is perfect for what the problem asks, just figured I'd give more since I'd done it anyways and only took a few seconds to type up

Brandon Monsen - 3 years, 6 months ago

@Pi Han Goh What if there is a condition that A,B360A,B\le360^\circ

Sumukh Bansal - 3 years, 6 months ago

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@Sumukh Bansal There's still infinitely many. The condition that A+B=2πA+B=2\pi means that AA can take any real number between 00 and 2π2\pi and there will be a value of BB which satisfies the equations

Brandon Monsen - 3 years, 6 months ago

@Brandon Monsen Yes, after doing the expansion and manipulation I find that either

(i) cos(A)=cos(B)A+B=2πk\cos(A) = \cos(B) \Longrightarrow A + B = 2\pi k or

(ii) either cos(A)=1(A,B)=(2πk,B)\cos(A) = 1 \Longrightarrow (A,B) = (2\pi k, B) or cos(B)=1(A,B)=(A,2πk)\cos(B) = 1 \Longrightarrow (A,B) = (A,2\pi k),

provides the complete set of solutions.

Brian Charlesworth - 3 years, 6 months ago
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