Question

2cosa-sina=x,cosa-2sina=y,then prove that 2xsquare+y square-2xy=5

Answer 1

Answer on Question #60533 – Math – Trigonometry

Question

2cosa-sina=x,cosa-2sina=y, then prove that 2xsquare+y square-2xy=5

Solution

$$\left\{ \begin{array}{l} 2 \cos a - \sin a = x \\ \cos a - 2 \sin a = y \end{array} \right.$$ $$x^2 = 4 \cos^2 \alpha + \sin^2 \alpha - 4 \cos \alpha \cdot \sin \alpha$$ $$y^2 = \cos^2 \alpha + 4 \sin^2 \alpha - 4 \cos \alpha \cdot \sin \alpha$$ $$\begin{array}{l} xy = (2 \cos \alpha - \sin \alpha) (\cos \alpha - 2 \sin \alpha) = 2 \cos^2 \alpha - 4 \cos \alpha \cdot \sin \alpha - \cos \alpha \cdot \sin \alpha + 2 \sin^2 \alpha \\ = 2 - 5 \cos \alpha \cdot \sin \alpha \end{array}$$ $$\begin{array}{l} 2 x^2 + y^2 - 2 x y = 2 (4 \cos^2 \alpha + \sin^2 \alpha - 4 \cos \alpha \cdot \sin \alpha) + \\ + \left(\cos^2 \alpha + 4 \sin^2 \alpha - 4 \cos \alpha \cdot \sin \alpha\right) - 2 (2 - 5 \cos \alpha \cdot \sin \alpha) = \\ = 8 \cos^2 \alpha + 2 \sin^2 \alpha - 8 \cos \alpha \cdot \sin \alpha + \cos^2 \alpha + 4 \sin^2 \alpha - 4 \cos \alpha \cdot \sin \alpha - 4 + \\ + 10 \cos \alpha \cdot \sin \alpha = 9 \cos^2 \alpha + 6 \sin^2 \alpha - 2 \cos \alpha \cdot \sin \alpha = 3 \cos^2 \alpha + 6 (\cos^2 \alpha + \sin^2 \alpha) - 2 \cos \alpha \cdot \sin \alpha = \\ = 3 \cos^2 \alpha + 6 - 2 \cos \alpha \cdot \sin \alpha = 5. \end{array}$$

Hence

$$3 \cos^2 \alpha + 1 - 2 \cos \alpha \cdot \sin \alpha = 0,$$ $$3 \cos^2 \alpha + \cos^2 \alpha + \sin^2 \alpha - 2 \cos \alpha \cdot \sin \alpha = 0,$$ $$4 \cos^2 \alpha - 2 \cos \alpha \cdot \sin \alpha + \sin^2 \alpha = 0$$

If $\cos \alpha = 0$ then it follows from (1) that $\sin \alpha = 0$, which does not hold, because $\cos^2 \alpha + \sin^2 \alpha = 1$ according to the Pythagorean identity.

If $\cos \alpha \neq 0$ then dividing by $\cos^2 \alpha$ in (1) one gets

$$\tan^2 \alpha - 2 \tan \alpha + 4 = 0$$

Equation (2) does not have real solutions, because its discriminant is negative:

$$D = 2^2 - 4 \cdot 1 \cdot 4 < 0.$$

Thus, $2x^2 + y^2 - 2xy \neq 5$ when $x = 2\cos \alpha - \sin \alpha$, $y = \cos \alpha - 2\sin \alpha$.

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