Question

If $a\sin x + b\cos x = c$ is satisfied by A and B then find (a) sin( A + B) (b) tan(A + B)

Answer 1

(a) $\displaystyle \sin(A+B)=\frac{2ab}{a^2+b^2}$

(b) $\displaystyle \tan(A+B)=-\frac{2ab}{a^2-b^2}$

Answer 2

We start with the equation

$$a\sin x + b\cos x = c.$$

A standard technique is to write the left‐side in the form

$$R\sin(x+\delta),$$

where

$$R=\sqrt{a^2+b^2},\quad \cos\delta=\frac{a}{R},\quad \sin\delta=\frac{b}{R}.$$

Thus the equation becomes

$$R\sin(x+\delta)=c\quad\Longrightarrow\quad \sin(x+\delta)=\frac{c}{R}.$$

For the two distinct solutions $x=A$ and $x=B$ (choosing the two values provided by the inverse sine function with the appropriate shifts), we can take

$$\begin{cases} A+\delta=\sin^{-1}\left(\frac{c}{R}\right),\\[1mm] B+\delta=\pi-\sin^{-1}\left(\frac{c}{R}\right). \end{cases}$$

Adding these we have

$$A+B+2\delta=\pi.$$

Thus,

$$A+B=\pi-2\delta.$$

Now, we compute part (a):

(a) Find $\sin(A+B)$:

$$\sin(A+B)=\sin(\pi-2\delta)=\sin 2\delta.$$

Using the double-angle formula

$$\sin2\delta=2\sin\delta\cos\delta,$$

and substituting $\sin\delta=\dfrac{b}{R}$ and $\cos\delta=\dfrac{a}{R}$ (with $R=\sqrt{a^2+b^2}$), we obtain

$$\sin(A+B)=2\left(\frac{b}{R}\right)\left(\frac{a}{R}\right)=\frac{2ab}{a^2+b^2}.$$

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(b) Find $\tan(A+B)$:

Since

$$A+B=\pi-2\delta,$$

we have

$$\tan(A+B)=\tan(\pi-2\delta)=-\tan2\delta.$$

Now use the double-angle formula for tangent:

$$\tan2\delta=\frac{2\tan\delta}{1-\tan^2\delta}.$$

But $\tan\delta=\frac{\sin\delta}{\cos\delta}=\frac{b}{a}$; hence

$$\tan2\delta=\frac{2(b/a)}{1-(b/a)^2}=\frac{2b/a}{\frac{a^2-b^2}{a^2}}=\frac{2ab}{a^2-b^2}.$$

Then

$$\tan(A+B)=-\tan2\delta=-\frac{2ab}{a^2-b^2}.$$