Question

if mSin theta +n cos theta=p and mCos theta - n Sin theta = q, then prove that m^2+n^2=p^2+q^2

Answer 1

$(m \sin( \alpha ) + n \cos( \alpha ) ) = p \\ sq \: both \: sides \\ {m}^{2} {sin}^{2} \alpha + {n}^{2} {cos}^{2} \alpha + mnsin2 \alpha \\ = {p}^{2} .....(1) \\ \\ \\ also \\ \\ m \cos( \alpha ) - n \sin( \alpha ) = q \\ sq \: both \: sides \: \\ m {}^{2} {cos}^{2} \alpha + {n}^{2} {sin}^{2} \alpha - mn \sin(2 \alpha ) \\ = {q}^{2} .........(2) \\ \\ add(1) \: and \: (2) \\ \\ {m}^{2} ( {sin}^{2} \alpha + {cos}^{2} \alpha ) + {n}^{2} ( {sin}^{2} \alpha + {cos}^{2} \alpha ) \\ = {p}^{2} + {q}^{2} \\ m {}^{2} + {n}^{2} = p {}^{2} + {q}^{2}$

hence, proved...


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