# How do we Find the maximum and minimum vlaues of 1/(3sinx-4cosx+7)?

Mar 20, 2018

Let

$y = \frac{1}{3 \sin x - 4 \cos x + 7}$

=>y=1/((5(3/5sinx-4/5cosx)+7)

Considering
$\frac{3}{5} = \cos \alpha$

So $\frac{4}{5} = \sin \alpha$

And $\alpha = {\tan}^{-} 1 \left(\frac{4}{3}\right)$

=>y=1/((5(cosalphasinx-sinalphacosx)+7)

$\implies y = \frac{1}{\left(5 \sin \left(x - \alpha\right) + 7\right)}$

We know $- 1 \le \sin \left(x - \alpha\right) \le + 1$

Hence $y$ will be maximum when $\sin \left(x + \alpha\right) = - 1$

So $= {y}_{\text{max}} = \frac{1}{5 \left(- 1\right) + 7} = \frac{1}{2}$

And $y$ will be minimum when $\sin \left(x + \alpha\right) = 1$

So $= {y}_{\text{min}} = \frac{1}{5 \cdot 1 + 7} = \frac{1}{12}$