Question

if sinθ and cosθ are the roots of equation ax² -bx+c=0 then ?

Answer 1

Answer on Question #60865 – Math – Trigonometry

Question

If $\sin\theta$ and $\cos\theta$ are the roots of equation $ax^2 - bx + c = 0$ then?

Solution

At first, factorize the polynomial $ax^2 - bx + c$:

$$a x ^ {2} - b x + c = a (x - \sin \theta) (x - \cos \theta), \quad a \neq 0$$

Simplifying the right-hand side obtain

$$a x ^ {2} - b x + c = a x ^ {2} - a (\sin \theta + \cos \theta) x + a \cdot \sin \theta \cos \theta$$

Coefficients of the terms $x, x^0 = 1$ in the left-hand and right-hand sides must be equal, hence by Vieta's formulas obtain

$$\left\{ \begin{array}{l} a \cdot \sin \theta \cos \theta = c \\ a (\sin \theta + \cos \theta) = b \end{array} \right.$$

Let's derive relation between $a, b$ and $c$:

$$b ^ {2} = a ^ {2} (\sin \theta + \cos \theta) ^ {2} = a ^ {2} (\sin^ {2} \theta + \cos^ {2} \theta + 2 \sin \theta \cos \theta) = a ^ {2} (1 + 2 \sin \theta \cos \theta) = a ^ {2} + 2 a c.$$

That is,

$$b ^ {2} = a ^ {2} + 2 a c,$$

hence

$$c = \frac {b ^ {2} - a ^ {2}}{2 a}.$$

Let's find $\theta$:

$$\sin \theta + \cos \theta = \sqrt {2} \cdot \cos \left(\theta - \frac {\pi}{4}\right) = \frac {b}{a}$$ $$\cos \left(\theta - \frac {\pi}{4}\right) = \frac {b}{\sqrt {2} a}$$ $$\theta - \frac {\pi}{4} = \pm \arccos \frac {b}{\sqrt {2} a} + 2 \pi n, \quad n \in \mathbb {Z}, \text{ where arccos is the inverse of the cosine.}$$

Finally, there are two possible families of solutions:

$$\left\{ \begin{array}{l} \theta = \frac {\pi}{4} + \arccos \frac {b}{\sqrt {2} a} + 2 \pi n, \\ \theta = \frac {\pi}{4} - \arccos \frac {b}{\sqrt {2} a} + 2 \pi n, \end{array} \right. \quad n \in \mathbb {Z}$$

Answer: $\theta = \frac{\pi}{4} + \arccos \frac{b}{\sqrt{2}a} + 2\pi n$ or $\theta = \frac{\pi}{4} - \arccos \frac{b}{\sqrt{2}a} + 2\pi n, n \in \mathbb{Z}$.

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