A sin θ ± B cos θ = C and Maximum and Minimum Range of Sine Function

(a)    Express $ \displaystyle {5\sin x+12\cos x}$ in the form of $ \displaystyle R\sin (x+\theta )$, where R > 0 and 0° < θ < 90°.
(b)    Hence or otherwise find the maximum value of $ \displaystyle f(x)$ where $ \displaystyle f(x)=\frac{{30}}{{5\sin x+12\cos x+17}}$ .
         State the values of x, in the range 0° < x < 360°, at which they occur.

Solution
 
        Let $ \displaystyle 5\sin x+12\cos x=R\sin (x+\theta ),$ where $ \displaystyle R\cos \theta =5$ and $ \displaystyle R\sin \theta =12.$
        
        $ \displaystyle \therefore {{(R\cos \theta )}^{2}}+{{(R\sin \theta )}^{2}}={{5}^{2}}+{{12}^{2}}$

        $ \displaystyle \therefore {{R}^{2}}{{\cos }^{2}}\theta +{{R}^{2}}{{\sin }^{2}}\theta =169$

        $ \displaystyle \therefore {{R}^{2}}({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )=169$

        $ \displaystyle \therefore {{R}^{2}}(1)=169$

        $ \displaystyle \therefore R=\sqrt{{169}}=13$

        And $ \displaystyle \frac{{R\sin \theta }}{{R\cos \theta }}=\frac{{12}}{5}\Rightarrow \tan \theta =2.4\Rightarrow \theta =67{}^\circ 2{3}'\text{ }$

        $ \displaystyle \therefore 5\sin x+12\cos x=13\sin (x+67{}^\circ 2{3}')$

        $ \displaystyle f(x)=\frac{{30}}{{5\sin x+12\cos x+17}}$ (given)

        $ \displaystyle \ \ \ \ \ \ \ =\frac{{30}}{{13\sin (x+67{}^\circ 2{3}')+17}}$

        $ \displaystyle \ \ \ \ \ \ \ =\frac{{30}}{{g(x)+17}}$ where $ \displaystyle g(x)=13\sin (x+67{}^\circ 2{3}')$
        
        Therefore $ \displaystyle f(x)$ is maximum when $ \displaystyle g(x)$ is minimum and vice versa.

        Since $ \displaystyle -1\le \sin (x+67{}^\circ 2{3}')\le 1$,

        $ \displaystyle -13\le 13\sin (x+67{}^\circ 2{3}')\le 13$ 

        $ \displaystyle -13\le g(x)\le 13$

        Hence the minimum value of $ \displaystyle g(x)=-13$

        $ \displaystyle \therefore 13\sin (x+67{}^\circ 2{3}')=-13$ 

        $ \displaystyle \ \ \ \sin (x+67{}^\circ 2{3}')=-1$

        $ \displaystyle \ \ \ x+67{}^\circ 2{3}'=270{}^\circ $

        $ \displaystyle \therefore x=202{}^\circ 3{7}'$  

        $ \displaystyle \therefore x=202{}^\circ 3{7}'$

        Therefore the maximum value of $ \displaystyle f(x)$ is $ \displaystyle \frac{{30}}{{-13+17}}=\frac{{15}}{2}$ when $ \displaystyle x=202{}^\circ 3{7}'.$ 

        The maximum value of $ \displaystyle g(x)=13.$ 

        $ \displaystyle \therefore 13\sin (x+67{}^\circ 2{3}')=13$ 

        $ \displaystyle \ \ \ \sin (x+67{}^\circ 2{3}')=1$ 

        $ \displaystyle \ \ \ x+67{}^\circ 2{3}'=90{}^\circ $ 

        $ \displaystyle \therefore x=22{}^\circ 3{7}'$ 

        Therefore the minimum value of $ \displaystyle f(x)$ is $ \displaystyle \frac{{30}}{{13+17}}=1$ when $ \displaystyle x=22{}^\circ 3{7}'.$

Illustration