Question

If x=a*sinQ + b*cosQ and y=a*cosQ + b*sinQ

Prove that (x^2+y^2)(a^+y^2) - 4abxy = (a^2 - b^2)^2

Answer 1

Answer on Question #62146 – Math – Trigonometry

Question

If

$$\left\{ \begin{array}{l} x = a \sin Q + b \cos Q, \\ y = a \cos Q + b \sin Q. \end{array} \right.$$

Prove that $(x^{2} + y^{2})(a^{2} + b^{2}) - 4abxy = (a^{2} - b^{2})^{2}$

Solution

$$\begin{array}{l} x ^ {2} + y ^ {2} = (a \sin Q + b \cos Q) ^ {2} + (a \cos Q + b \sin Q) ^ {2} = \\ = a ^ {2} (\sin Q) ^ {2} + 2 a b \sin Q \cos Q + b ^ {2} (\cos Q) ^ {2} + 2 a b \sin Q \cos Q + \\ + b ^ {2} (\sin Q) ^ {2} + 2 a b \sin Q \cos Q + a ^ {2} (\cos Q) ^ {2} = a ^ {2} + b ^ {2} + 4 a b \sin Q \cos Q; \\ \end{array}$$ $$\begin{array}{l} x y = a ^ {2} \sin Q \cos Q + a b (\sin Q) ^ {2} + a b (\cos Q) ^ {2} + b ^ {2} \sin Q \cos Q = \\ = a b + \left(a ^ {2} + b ^ {2}\right) \sin Q \cos Q; \\ \end{array}$$ $$\begin{array}{l} (x ^ {2} + y ^ {2}) (a ^ {2} + b ^ {2}) - 4 a b x y = \\ = (a ^ {2} + b ^ {2} + 4 a b \sin Q \cos Q) (a ^ {2} + b ^ {2}) - 4 a b (a b + (a ^ {2} + b ^ {2}) \sin Q \cos Q) = \\ = (a ^ {2} + b ^ {2}) ^ {2} - 4 a ^ {2} b ^ {2} + (4 a b \sin Q \cos Q - 4 a b \sin Q \cos Q) (a ^ {2} + b ^ {2}) = \\ = a ^ {4} + 2 a ^ {2} b ^ {2} + b ^ {4} - 4 a ^ {2} b ^ {2} = a ^ {4} - 2 a ^ {2} b ^ {2} + b ^ {4} = (a ^ {2} - b ^ {2}) ^ {2}, \\ \end{array}$$

which proves that

$$(x ^ {2} + y ^ {2}) (a ^ {2} + b ^ {2}) - 4 a b x y = (a ^ {2} - b ^ {2}) ^ {2}.$$

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