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Question Number 17152 by Tinkutara last updated on 01/Jul/17

If sinA = sinB and cosA = cosB, then (1) A = B + nπ, n ∈ I (2) A = B − nπ, n ∈ I (3) A = 2nπ + B, n ∈ I (4) A = nπ − B, n ∈ I

$$\mathrm{If}\:\mathrm{sin}{A}\:=\:\mathrm{sin}{B}\:\mathrm{and}\:\mathrm{cos}{A}\:=\:\mathrm{cos}{B},\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:{A}\:=\:{B}\:+\:{n}\pi,\:{n}\:\in\:{I} \\ $$$$\left(\mathrm{2}\right)\:{A}\:=\:{B}\:−\:{n}\pi,\:{n}\:\in\:{I} \\ $$$$\left(\mathrm{3}\right)\:{A}\:=\:\mathrm{2}{n}\pi\:+\:{B},\:{n}\:\in\:{I} \\ $$$$\left(\mathrm{4}\right)\:{A}\:=\:{n}\pi\:−\:{B},\:{n}\:\in\:{I} \\ $$

Answered by ajfour last updated on 01/Jul/17

(3).

$$\left(\mathrm{3}\right). \\ $$

Commented by ajfour last updated on 01/Jul/17

since sin A=sin B we have B= nπ+(−1)^n A while cos A=cos B ⇒ B=2nπ±A if n is odd both dont give the same answer, while if n is even, we need to reject B=2nπ−A . so B=A+2n𝛑 where n∈I .

$$\mathrm{since}\:\mathrm{sin}\:\mathrm{A}=\mathrm{sin}\:\mathrm{B} \\ $$$$\mathrm{we}\:\mathrm{have}\:\:\:\:\mathrm{B}=\:\mathrm{n}\pi+\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{A}\:\: \\ $$$$\mathrm{while}\:\mathrm{cos}\:\mathrm{A}=\mathrm{cos}\:\mathrm{B} \\ $$$$\Rightarrow\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{B}=\mathrm{2n}\pi\pm\mathrm{A} \\ $$$$\mathrm{if}\:\mathrm{n}\:\mathrm{is}\:\mathrm{odd}\:\mathrm{both}\:\mathrm{dont}\:\mathrm{give}\:\mathrm{the} \\ $$$$\mathrm{same}\:\mathrm{answer}, \\ $$$$\mathrm{while}\:\mathrm{if}\:\mathrm{n}\:\mathrm{is}\:\mathrm{even},\:\mathrm{we}\:\mathrm{need}\:\mathrm{to} \\ $$$$\mathrm{reject}\:\mathrm{B}=\mathrm{2n}\pi−\mathrm{A}\:. \\ $$$$\mathrm{so}\:\:\boldsymbol{\mathrm{B}}=\boldsymbol{\mathrm{A}}+\mathrm{2}\boldsymbol{\mathrm{n}\pi}\:\:\mathrm{where}\:\mathrm{n}\in{I}\:. \\ $$

Commented by Tinkutara last updated on 01/Jul/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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