If 6cosx + 8sinx = 9, Find the Value of sin(α+β)

Question

If \( \alpha \) and \( \beta \) are two different values of \(x\) lying between \[ 0 \text{ and } 2\pi \] which satisfy:

\[ 6\cos x+8\sin x=9 \]

find the value of:

\[ \sin(\alpha+\beta) \]

Solution

\[ 6\cos x+8\sin x=9 \]

\[ \frac{6}{10}\cos x+\frac{8}{10}\sin x=\frac{9}{10} \]

\[ \frac35\cos x+\frac45\sin x=\frac9{10} \]

Let

\[ \cos\theta=\frac35, \qquad \sin\theta=\frac45 \]

Then,

\[ \cos(x-\theta)=\frac9{10} \]

Therefore,

\[ x-\theta=\cos^{-1}\frac9{10} \]

or

\[ x-\theta=2\pi-\cos^{-1}\frac9{10} \]

Hence,

\[ \alpha=\theta+\cos^{-1}\frac9{10} \]

\[ \beta=\theta+2\pi-\cos^{-1}\frac9{10} \]

Adding,

\[ \alpha+\beta = 2\theta+2\pi \]

\[ \sin(\alpha+\beta) = \sin2\theta \]

\[ = 2\sin\theta\cos\theta \]

\[ = 2\cdot\frac45\cdot\frac35 \]

\[ = \frac{24}{25} \]

Therefore,

\[ \boxed{\sin(\alpha+\beta)=\frac{24}{25}} \]