Question

The minimum value of 3 cos x + 4 sin x + 8 is​

Answer 1

Answer:

13

Step-by-step explanation:

FORMULA :

The range of a trigonometric  expression , acosx+bsinx+c is given by

$[c-\sqrt{a^{2}+b^{2} },c+\sqrt{a^{2}+b^{2} } ]$. i.e., the maximum is $c+\sqrt{a^{2}+b^{2} }$ .

So maximum value of 3cosx+4sinx+8 = $8+\sqrt{3^{2}+4^{2} }$

                                                               = 8 + 5 = 13.