If $p$ & $q$ are the solutions of $$a \cos x + b \sin x = c$$
Then how do I prove that, $$\cos (p + q) = \dfrac{a^2-b^2}{a^2 + b^2} $$
I tried all the adjustments I could think of, like dividing by $ \cos x $ and extracting $a$ and $b$ from the 2 equations. Also tried adding/subtracting and all the basics I know to no avail.
Any help is appreciated, thank you :)
On
Y'all no worries! I figured it out – (by a fluke?), I just want someone to confirm this is right :)
Converting to it to original question in my textbook. Let $p= \alpha$ and $q = \beta$
$\alpha$ and $\beta$ are solutions of $a\cos\theta+b\sin\theta=c$
$$a\cos\alpha+b\sin\alpha=c=a\cos\beta+b\sin\beta $$
$$b(\sin\alpha-\sin\beta)=-a(\cos\alpha-\cos\beta)$$
$$ 2b\cos\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right)=2a\sin\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right)$$
$$\frac{b}{a}=\frac{\sin\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right)}{\cos\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right)}=\tan\left(\frac{\alpha+\beta}{2}\right) $$
$$ \tan^2\left(\frac{\alpha+\beta}{2}\right)=\frac{b^2}{a^2}$$
$$\cos(\alpha+\beta)=\frac{\cos^2\left(\frac{\alpha+\beta}{2}\right)-\sin^2\left(\frac{\alpha+\beta}{2}\right)}{\cos^2\left(\frac{\alpha+\beta}{2}\right)+\sin^2\left(\frac{\alpha+\beta}{2}\right)}$$
Dividing the numerator and denominator by $\cos^2\left(\frac{\alpha+\beta}{2}\right)$ , we get,
$$\cos(\alpha+\beta)=\frac{1-\tan^2\left(\frac{\alpha+\beta}{2}\right)}{1+\tan^2\left(\frac{\alpha+\beta}{2}\right)}=\frac{1-\frac{b^2}{a^2}}{1+\frac{b^2}{a^2}}$$
$$\Rightarrow\qquad \cos(\alpha+\beta)=\frac{a^2-b^2}{a^2+b^2}$$
Thanks to everyone for their efforts :-)
On
We have
$$c-a\cos x=b\sin x$$ and by squaring and rewriting,
$$c^2-2ac\cos x+a^2\cos^2x=b^2(1-\cos^2x),$$ $$(a^2+b^2)\cos^2x-2ac\cos x+c^2-b^2=0.$$
Using the Vieta's formulas, the product of the roots is
$$\cos p\cos q=\frac{c^2-b^2}{a^2+b^2}.$$ Repeating the same reasoning symmetrically,
$$\sin p\sin q=\frac{c^2-a^2}{a^2+b^2}.$$
Now by subtraction,
$$\cos(p+q)=\frac{a^2-b^2}{a^2+b^2}.$$
Given $$ a\cos x + b\sin x = c $$ put $$ A = \sqrt {a^{\,2} + b^{\,2} } \quad \phi = \arctan \left( {{b \over a}} \right) $$ to get $$ \eqalign{ & c = a\cos x + b\sin x = A\cos \phi \cos x + A\sin \phi \sin x = \cr & = A\cos \left( {x - \phi } \right) = A\cos \left( {p - \phi } \right) = A\cos \left( {q - \phi } \right) \cr} $$ from which you obtain $$ A\cos \left( {p - \phi } \right) = A\cos \left( {q - \phi } \right)\quad \Rightarrow \quad p - \phi = - \left( {q - \phi } \right)\quad \Rightarrow \quad p + q = 2\phi $$ (apart from multiples of $2\pi$).
The rest you should do easily by yourself, I suppose.