Question

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

Answer 1

Given,

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

Now,

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

On squaring both sides, we get

${\left(6\cos \theta \right)}^2={(9-8\sin \theta )}^2$

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

Since,

$\alpha and\beta$ are different roots.

Product of roots

$\sin \alpha .\sin \beta =\frac{45}{100}$ ………. (1)

Taking again,

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

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

On squaring both sides, we get

$64(1-{\cos }^2\theta )=81+36{\cos }^2\theta -108$

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

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

Product of roots $\cos \alpha .\cos \beta =\frac{17}{100}$ ……… (2)

From equation (1) and (2), we have

$\cos \alpha \cos \beta -\sin \alpha \sin \beta =\frac{17}{100}-\frac{45}{100}=-\frac{-7}{100}$

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

Hence, this is required answer.