# 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