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.



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

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

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 ,

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


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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.