If $\alpha$ and $\beta$ are the solutions of $a\cos \theta+ b\sin \theta= c$ show that

Question

If $\alpha$ and $\beta$ are the solutions of $a\cos \theta+ b\sin \theta= c$ show that

1) $\sin \alpha + \sin \beta = \dfrac{2bc}{a^2 + b^2}$

2) $\sin \alpha \sin \beta = \dfrac{c^2-a^2}{a^2+b^2}$

I couldn't even start the problem, and I generally have a lot of difficulty in compound angles so please help me with this.

Thanks!

Answer 1

$$a\cos\theta = c - b\sin\theta \implies a^2\cos^2\theta = c^2-2bc\sin\theta+b^2\sin^2\theta \\ \implies (a^2-a^2\sin^2\theta) = c^2-2bc\sin\theta+b^2\sin^2\theta $$

$$(a^2+b^2)\sin^2\theta - (2bc)\sin\theta+(c^2-a^2)=0 \equiv Ax^2 + Bx +C = 0$$

As $\alpha$ and $\beta$ are the solutions of the given equation, $\sin\alpha \equiv x_1$ and $\sin\beta \equiv x_2$ satisfy the above equation.

If $x_1$ and $x_2$ are the solutions of $Ax^2 + Bx +C = 0$, then ,

$$x_1+x_2 = -\frac{B}{A} \implies \sin\alpha + \sin\beta = - \frac{-2bc}{a^2+b^2} = \frac{2bc}{a^2+b^2}$$

and

$$x_1 \cdot x_2 = \frac{C}{A} \implies \sin\alpha\cdot\sin\beta=\frac{c^2-a^2}{a^2+b^2}$$