Question

If $\alpha ,\beta$ are two different vlaues of $\theta$ lying between 0 and $2\pi$ which satisfy the equations 6 cos \theta + 8 sin\theta= 9, find the value of sin $(\alpha +\beta )$.

Answer 1

We have,
6 cos $\theta$ + 8 sin $\theta$ = 9 ... (i)
$\Rightarrow 8sin\theta =9-6cos\theta \Rightarrow (8sin\theta {)}^2=(9-6cos\theta {)}^2$
[$\because$ Squaring both sides]
$\Rightarrow 64si{n}^2\theta =81+36co{s}^2\theta -108cos\theta \Rightarrow 64si{n}^2\theta =81+36co{s}^2\theta -108cos\theta \Rightarrow 64(1-co{s}^2\theta )=81+36co{s}^2\theta -108cos\theta \Rightarrow 64-64co{s}^2\theta +64co{s}^2\theta -108cos\theta +81-64=0\Rightarrow 100co{s}^2\theta -108cos\theta +17=0$ ...(ii),
Therefore, $cos\alpha$ and $cos\beta$ are roots of equation (ii)
$\therefore cos\alpha +cos\beta =\frac{17}{100}$ ...(iii)
Again, 6 $cos\theta +8sin\theta =9$
$\Rightarrow (6cos\theta {)}^2=(9-8sin\theta {)}^2$
[$\because$ Squaring both sides]
$\Rightarrow 36co{s}^2\theta =81+64si{n}^2\theta -144sin\theta \Rightarrow 36(1-si{n}^2\theta )=81+64si{n}^2\theta -144cos\theta \Rightarrow 36-36si{n}^2\theta =81+64si{n}^2\theta -144sin\theta \Rightarrow 64si{n}^2\theta +36si{n}^2\theta -144sin\theta +81-36=0$
$\Rightarrow 100si{n}^2\theta -144sin\theta +45=0$ ...(iv)
It is given that $\alpha ,\beta$ are roots of equation (ii). So, sin$\alpha$ and $sin\beta$ are the roots of eqaution (iv)
$\therefore sin\alpha \times sin\beta =\frac{45}{100}$
Now, cos$(\alpha +\beta )=cos\alpha cos\beta -sin\alpha sin\beta$
= $\frac{17}{100}-\frac{45}{100}$
[Using equation(iii) and (v)]
= $\frac{-28}{100}=\frac{-7}{25}$
Now, $sin(\alpha +\beta )=\sqrt{1-(cos\theta {)}^2}=\sqrt{1-\frac{49}{625}}=\sqrt{\frac{625-49}{625}}=\sqrt{\frac{576}{625}}=\sqrt{\frac{24}{25}}\therefore sin(\alpha +\beta )=\frac{24}{25}$