Question

Y=a sinx + b cosx prove that y^2 + (dy/dy)^2 = a^2 + b^2

Answer 1

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Question :

Y=asinx+bcosx .prove that y²+(dy/dx)²=a²+b²

Given :

Y=asinx+bcosx

To prove:

$\rm{y^2+\bigg(\dfrac{dy}{dx}\bigg)^{2} = a^2+b^2}$

Solution:

$\large \green{\bold{ \underline{ \underline{step \: by \: step \: explaination}}}}$

$\\ \implies \large \sf \: y \: = asinx \: + b \: cos \: x$

$\\ \color{pink} \: \: \: \: \bigstar\large \red{ \bf{ \underline{squaring \: both \: sides}}}$

$\\ \implies \large \sf \: {y}^{2} = {(asinx + bcosx)}^{2}$

$\\ \implies \large \sf \: {y}^{2} = {a}^{2} {sin}^{2} x + {b}^{2} {cos}^{2} x + 2asinxbcosx$

$\\ \implies \large \sf \: {y}^{2} = {a}^{2} {sin}^{2} x + {b}^{2} {cos}^{2} x + 2asinx bcosx \: \: ... \:. .......equation.(1)$

______________________________________

$\\ \implies \large \sf \: y = asinx \: + bcosx$

$\\ \color{blue} \bigstar \pink{\bf{ \underline{ differentiating \: both \: side \: with \: respect \: to \: x}}}$

$\\ \implies \large \sf \: \frac{dy}{dx} = a \frac{dy}{dx} sinx + b \frac{dy}{dx}$

$\\ \implies \large \sf \: \frac{dy}{dx} = acosx - bsinx \: \:$

$\color{green} \large \bigstar \blue{ \bf{ \underline{squaring \: both \: sides}}}$

$\\ \implies \large \sf \: \bigg( { \frac{dy}{dx} } \bigg)^{2} = {(acosx - bsinx)}^{2}$

$\\ \implies \large \sf \: \bigg({ \frac{dy}{dx} }^{2} \bigg) = {a}^{2} {cos}^{2} x + {b}^{2} {sin}^{2} x - 2acosxbsinx \: \: .......equation(2)$

$\color{cyan} \large \bigstar \red{ \bf{ \underline{adding \: both \: equations \: (1) \: and \: (2)}}}$

$\\ \implies \large \sf \: y + \bigg({ \frac{dy}{dx} } \bigg)^{2} = {a}^{2} {sin}^{2} x + {b}^{2} {cos}^{2} x + 2asinxcosx + {a}^{2} {cos}^{2} x + {b}^{2} {sin}^{2} x + 2absinxcosx - 2abcosxsin$

$\\ \implies \large \sf \: y + \bigg({ \frac{dy}{dx} } \bigg)^{2} = {a}^{2} ( {sin}^{2} x + {cos}^{2}x) + {b}^{2} ( {sin}^{2} x + {cos}^{2} x) + \cancel{2absinxcosx} - \cancel{2basinxcosx}$

$\\ \implies \large \sf \: y + \bigg({ \frac{dy}{dx} } \bigg)^{2} = {a}^{2} + {b}^{2}$

$\: \therefore\boxed{ \bf{ \underline{ y + \bigg( { \frac{dy}{dx} } \bigg)^{2} = {a}^{2} + {b}^{2} }}}$

hence proved