If $2cos\theta-sin\theta=x$ and $cos\theta-3sin\theta=y$. Prove that $2x^2+y^2-2xy=5$.

Given: $2cos\theta-sin\theta=x$ and $cos\theta-3sin\theta=y$. 

To do: To prove that $2x^2+y^2-2xy=5$.

Solution:

As given,

$( 2 cos\theta - sin\theta)=x$   and $( cos\theta - 3 sin\theta)=y$

Put the values of $x$ and $y$ in the equation  

LHS$=2x^2 + y^2 - 2xy$

$=2( 2cos\theta - sin\theta)2 + (cos\theta - 3 sin\theta)2 - 2(2 cos\theta - sin\theta)(cos\theta - 3 sin\theta)$

$=2(4cos^2\theta - 4cos\theta sin\theta + sin^2\theta) + (cos^2\theta - 6cos\theta sin\theta + 9sin^2\theta) - 2(2cos^2\theta - 7cos\theta sin\theta + 3sin^2\theta)$

$=8cos^2\theta - 8cos\theta sin\theta + 2sin^2\theta + cos^2\theta - 6cos\theta sin\theta + 9sin^2\theta - 4cos^2\theta + 14cos\theta sin\theta - 6sin^2\theta$

$=5cos^2\theta + 5sin^2\theta$

$=5( cos2\theta + sin2\theta)$

$=5( 1)=5$        $(\therefore  cos^2\theta + sin^2\theta = 1)$

$=$RHS 

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Updated on: 10-Oct-2022

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