4
$\begingroup$

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!

$\endgroup$
11
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Ohh.... I see now. Well thanks! But how exactly am I supposed to think of that? $\endgroup$ – Gummy bears Sep 11 '14 at 12:36
  • 1
    $\begingroup$ @Gummybears, Sorry for the typo $\endgroup$ – lab bhattacharjee Sep 11 '14 at 12:44
  • $\begingroup$ How did you get the value of $\sin \alpha \sin \beta$? $\endgroup$ – Gummy bears Sep 11 '14 at 12:51
  • $\begingroup$ @Gummybears, Please find en.wikipedia.org/wiki/Vieta's_formulas $\endgroup$ – lab bhattacharjee Sep 11 '14 at 13:12
6
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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.

enter image description here

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, $$\cos(\theta+\phi) = 2\cos^2\frac{\theta+\phi}{2} - 1 = \frac{2a^2-d^2}{d^2} = \frac{2a^2-(a^2+b^2)}{a^2+b^2} = \frac{a^2-b^2}{a^2+b^2}$$ $$\sin(\theta+\phi) = 2 \sin\frac{\theta+\phi}{2}\cos\frac{\theta+\phi}{2} = \frac{2ab}{d^2} = \frac{2ab}{a^2+b^2} $$

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

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

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

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.