Question

minimum value of 3cosx+4sinx-8 is​

Answer 1

Step-by-step explanation:

We have,

$y = 3 \cos(x) + 4 \sin(x) - 8$

$\implies \: y = 5( \frac{3}{5} \cos(x) + \frac{4}{5} \sin(x)) -8$

$\implies \: y = 5 \sin(x + \alpha ) - 8 \: \: (where \: \: \alpha = \tan ^{ - 1} ( \frac{3}{4} ) )$

Now,

$- 1 \leqslant \sin(x + \alpha ) \leqslant 1$

$\implies -5 \leqslant 5 \sin(x + \alpha ) \leqslant 5$

$\implies - 5 - 8 \leqslant 5 \sin(x + \alpha ) - 8 \leqslant 5 - 8$

$\implies - 13 \leqslant y \leqslant - 3$

so, the minimum value of 3cos(x) + 4sin(x) - 8 is -13