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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!

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

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

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

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

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  • $\begingroup$ If alpha and beta are solutions of the equation, then how did you take x1 and x2 as sin alpha and beta? $\endgroup$ – Aditya Jul 21 at 4:08
  • $\begingroup$ I just used it as an equivalence to the quadratic equation, $\endgroup$ – Ak19 Jul 21 at 4:52
  • $\begingroup$ Didn’t know we can do that. Could u elaborate on that? $\endgroup$ – Aditya Jul 21 at 7:49
  • $\begingroup$ If we have a quadratic equation $$Ax^2+Bx+ C = 0 \implies x = \frac{-B\pm\sqrt{B^2-4AC}}{2A}$$ Now the sum of the roots is $$x_1 + x_2 = \frac{-B+\sqrt{B^2-4AC}}{2A}+ \frac{-B-\sqrt{B^2-4AC}}{2A} = -\frac{B}{A}$$ and their product is $$x_1x_2 = \frac{-B^2-(B^2-4AC)}{4A^2} = \frac{C}{A}$$ Here replace $x = \sin\theta$, $x_1 = \sin\alpha$ and $x_2 = \sin\beta$ $\endgroup$ – Ak19 Jul 21 at 7:54
  • $\begingroup$ That’s true, but what I am still confused about is, alpha and beta are the roots of the equation, not sin alpha and sin beta. So how can we just use them interchangeably $\endgroup$ – Aditya Jul 21 at 8:14

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