Question

If $\alpha$ and $\beta$ are two different values $\theta$ lying between $0$ and $2\pi$ which satisfy the equation $6\cos \theta +8\sin \theta =9$. Find $\sin \left(\alpha +\beta \right)$.

Answer 1

$6\cos \theta +8\sin \theta =9$

$6\cos \theta =9-8\sin \theta$

$squaring.$

$36{\cos }^2\theta =81+64{\sin }^2\theta -144\sin \theta$

$36\left(1-{\sin }^2\theta \right)=81-144\sin \theta +64{\sin }^2\theta$

$100{\sin }^2\theta -144\sin \theta +45=0$

$\alpha \&\beta arediffValuesof\theta$

$thenproductofroots=\sin \alpha \sin \beta =\frac{45}{100}$

$\frac{9}{20}$

$Again8\sin \theta =9-6\cos \theta$

$64{\sin }^2\theta =81+36{\cos }^2\theta -108\cos \theta$

$100{\cos }^2\theta -108\cos \theta +17=0$

$productofroots=\cos \alpha \cos \beta =\frac{17}{100}$

$Now\cos \alpha \cos \beta -\sin \alpha \sin \beta =\frac{17}{100}-\frac{45}{100}$

$\cos \left(\alpha +\beta \right)=\frac{-7}{25}$

$\sin \left(\alpha +\beta \right)=\frac{24}{25}$