Cosine of the sum of two solutions of trigonometric equation $a\cos \theta + b\sin \theta = c$

Question: If $\alpha$ and $\beta$ are the solutions of $a\cos \theta + b\sin \theta = c$, then show that: $$\cos (\alpha + \beta) = \frac{a^2 - b^2}{a^2 + b^2}$$

No idea how to even approach the problem. I tried taking two equations, by substituting $\alpha$ and $\beta$ in place of $\theta$ in the equation and manipulating them, but that didn't get me anywhere. Please help!

We have $\displaystyle a\cos\theta=c-b\sin\theta,$

Squaring we get, $\displaystyle(c-b\sin\theta)^2=(a\cos\theta)^2=a^2(1-\sin^2\theta)$

$\displaystyle\iff (a^2+b^2)\sin^2\theta-2bc\sin\theta+c^2-a^2=0$

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

Similarly find $\displaystyle\cos\alpha\cos\beta$ by squaring $\displaystyle b\sin\theta=c-a\cos\theta$

Finally use $\displaystyle\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$

• Ohh.... I see now. Well thanks! But how exactly am I supposed to think of that? – Gummy bears Sep 11 '14 at 12:36
• @Gummybears, Sorry for the typo – lab bhattacharjee Sep 11 '14 at 12:44
• How did you get the value of $\sin \alpha \sin \beta$? – Gummy bears Sep 11 '14 at 12:51
• @Gummybears, Please find en.wikipedia.org/wiki/Vieta's_formulas – lab bhattacharjee Sep 11 '14 at 13:12

We have $\displaystyle a\cos\alpha+b\sin\alpha=c=a\cos\beta+b\sin\beta$

$\displaystyle a(\cos\alpha-\cos\beta)=-b(\sin\alpha-\sin\beta)$

Now use Prosthaphaeresis Formulas to find $\displaystyle\tan\frac{\alpha+\beta}2$ assuming $\displaystyle\sin\frac{\alpha-\beta}2\ne0$

Then use $\displaystyle\cos2A=\frac{1-\tan^2A}{1+\tan^2A}$

Here's a picture showing angles $\theta$ (at $P$) and $\phi$ (at $Q$) such that $$a \cos\theta + b \sin\theta = c = a \cos\phi + b \sin \phi$$ The measure of $\angle PAQ$ is the sum of these angles.

Note that $P$ and $Q$ lie on the circle with diameter $\overline{AB}$, and that the diameter bisects $\angle PAQ$. From here, we have many approaches to the final relation; here's one: Clearly, $$\cos\frac{\theta+\phi}{2} = \frac{a}{d} \qquad\qquad \sin\frac{\theta+\phi}{2} = \frac{b}{d}$$ so that, by the Double-Angle Formulas, $$\cos(\theta+\phi) = 2\cos^2\frac{\theta+\phi}{2} - 1 = \frac{2a^2-d^2}{d^2} = \frac{2a^2-(a^2+b^2)}{a^2+b^2} = \frac{a^2-b^2}{a^2+b^2}$$ $$\sin(\theta+\phi) = 2 \sin\frac{\theta+\phi}{2}\cos\frac{\theta+\phi}{2} = \frac{2ab}{d^2} = \frac{2ab}{a^2+b^2}$$

Here's a geometric solution (which thus has some in common with lab's Prosthaphaeresis Formula solution); it amounts just to interpreting the given equation as for the intersection of a line and a circle, and then using symmetry and doing some easy algebra.

By construction, the solutions $\alpha$ and $\beta$ are the angles between the positive $x$-axis and the points of intersection of the unit circle $C$ with the line $L$ defined by $a x + b y = c$ (since the original question says solutions, we'll assume there are two points of intersection). We may as well assume too that the points are not endpoints of a diameter (equivalently that $c \neq 0$), as the claim is a trivial calculation in that case.

Now, by symmetry, (1) the line $L'$ through $0$ and the point on $L$ closest to $0$ makes an angle $$\theta := \frac{1}{2} (\alpha + \beta)$$ with the positive $x$-axis, and (2) $L' \perp L$, so $L$ has equation $b x - a y = 0$. This line intersects $C$ at two points and substituting gives that the intersections satisfy $$x^2 = \frac{a^2}{a^2 + b^2}$$ and so $$y^2 = \frac{b^2}{a^2 + b^2}.$$ By definition these quantities are respectively $\cos^2 \theta$ and $\sin^2 \theta$, and so the cosine double angle identity gives $$\cos(\alpha + \beta) = \cos(2 \theta) = \cos^2 \theta - \sin^2 \theta = \frac{a^2 - b^2}{a^2 + b^2}$$ as desired.

Use Weierstrass Substitution to form a Quadratic Equation in $\tan\dfrac\theta2$

Now using Vieta's formula, we can find $\displaystyle\tan\dfrac\alpha2+\tan\dfrac\beta2,\tan\dfrac\alpha2\tan\dfrac\beta2$

Then $\displaystyle\tan\left(\dfrac\alpha2+\dfrac\beta2\right)=\dfrac{\tan\dfrac\alpha2+\tan\dfrac\beta2}{1-\tan\dfrac\alpha2\tan\dfrac\beta2}$

• @Gummybears, Here is my third method – lab bhattacharjee Sep 11 '14 at 16:49