The maximum value of 3 cos x + 4 sin x + 5 is

Solution:

$3 \cos x+4 \sin x+5$

first express $3 \cos x+4 \sin x$ as $a \cos (x+A)$

i. e. $3 \cos x+4 \sin x=a[\cos x \cos A-\sin x \sin A]$

Now, equate the coefficients of $\sin x$ and $\cos x$

We get,

$a \cos A=3$

$-a \sin A=4$

i. e $\frac{a \sin A}{a \cos A}=\frac{-4}{3}$ i. e $\tan A=-\frac{4}{3}$

i. e $A=\tan ^{-1}\left(\frac{-4}{3}\right)$

also $(a \cos A)^{2}+(a \sin A)^{2}=9+16=25$

$a^{2}=25$

$a=5$

$\therefore 3 \cos x+4 \sin x=5 \cos \left(x-\tan ^{-1}\left(-\frac{4}{3}\right)\right)$

i. e. $3 \cos x+4 \sin x+5=5 \cos \left(x-\tan ^{-1}\left(-\frac{4}{3}\right)\right)+5$

Since minimum value of $\cos \theta=-1$

$\Rightarrow(3 \cos x+4 \sin x+5)_{\min }=5(-1)+5$

$=-5+5=0$

i. e minimum value of $3 \cos x+4 \sin x+5=0$