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

Solution 1:

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

Solution 2:

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

Solution 3:

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