Exercise (11.5) - Solution

$ \displaystyle \begin{array}{l}\text{1}\text{.}\ \ \ \ \ \text{Solve the following equations for 0}{}^\circ \le \theta \le 360{}^\circ .\\\\\ \ \ \ \ \ \ \text{(a)}\ \ \ \ 3\text{cos}\theta -\sin \theta =2\end{array}$

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$ \displaystyle \begin{array}{l}\text{(a)}\ \ \ \ 3\text{cos}\theta -\sin \theta =2\\\\\ \ \ \ \ \ \ \ \text{Comparing with}\ a\ \text{cos}\theta -b\sin \theta =c,\\\\\ \ \ \ \ \ \ \ a=3,b=1,c=2\\\\\therefore \,\ \ \ \ \ \ \sqrt{{{{a}^{2}}+{{b}^{2}}}}=\sqrt{{{{3}^{2}}+{{1}^{2}}}}=\sqrt{{10}}\\\\\ \ \ \ \ \ \ \ \text{Let}\ \tan \alpha =\displaystyle \frac{b}{a}\Rightarrow \tan \alpha =\displaystyle \frac{1}{3}\\\\\therefore \ \ \ \ \ \ \alpha =18{}^\circ 2{6}'\\\\\therefore \ \ \ \ \ \ 3\text{cos}\theta -\sin \theta =2\Rightarrow \sqrt{{10}}\cos (\theta +18{}^\circ 2{6}')=2\\\\\therefore \ \ \ \ \ \ \cos (\theta +18{}^\circ 2{6}')=0.6325\\\\\therefore \ \ \ \ \ \ \theta +18{}^\circ 2{6}'=50{}^\circ 4{6}'\ \ (\text{or)}\ \theta +18{}^\circ 2{6}'=360{}^\circ -50{}^\circ 4{6}'\\\\\therefore \ \ \ \ \ \ \theta =32{}^\circ 2{0}'\ \ (\text{or)}\ \theta =290{}^\circ 4{8}'\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{array}$

$ \displaystyle \ \ \ \ \ \ \ \text{(b)}\ \ \ \ 2\text{sin}\theta -3\cos \theta =3$

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$ \displaystyle \begin{array}{l}\text{(b)}\ \ \ \ 2\text{sin}\theta -3\cos \theta =3\\\\\ \ \ \ \ \ \ \ \text{Comparing with}\ a\ \text{sin}\theta -b\text{cos}\theta =c,\\\\\ \ \ \ \ \ \ \ a=2,b=3,c=3\\\\\therefore \,\ \ \ \ \ \ \sqrt{{{{a}^{2}}+{{b}^{2}}}}=\sqrt{{{{2}^{2}}+{{3}^{2}}}}=\sqrt{{13}}\\\\\ \ \ \ \ \ \ \ \text{Let}\ \tan \alpha =\displaystyle \frac{b}{a}\Rightarrow \tan \alpha =\displaystyle \frac{3}{2}\\\\\therefore \ \ \ \ \ \ \alpha =56{}^\circ 1{9}'\\\\\therefore \ \ \ \ \ \ 2\text{sin}\theta -3\cos \theta =3\Rightarrow \sqrt{{13}}\sin (\theta -56{}^\circ 1{9}')=3\\\\\therefore \ \ \ \ \ \ \sin (\theta -56{}^\circ 1{9}')=0.8321\\\\\therefore \ \ \ \ \ \ \theta -56{}^\circ 1{9}'=56{}^\circ 1{9}'\ \ (\text{or)}\ \theta -56{}^\circ 1{9}'=180{}^\circ -56{}^\circ 1{9}'\\\\\therefore \ \ \ \ \ \ \theta =112{}^\circ 3{8}'\ \ (\text{or)}\ \theta =180{}^\circ \\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{array}$

$ \displaystyle \ \ \ \ \ \ \ \text{(c)}\ \ \ \ \cos \theta +2\sin \theta =2$

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$ \displaystyle \begin{array}{l}\text{(c)}\ \ \ \ \cos \theta +2\sin \theta =2\\\\\ \ \ \ \ \ \ \ \text{Comparing with}\ a\cos \theta +b\sin \theta =c,\\\\\ \ \ \ \ \ \ \ a=1,b=2,c=2\\\\\therefore \,\ \ \ \ \ \ \sqrt{{{{a}^{2}}+{{b}^{2}}}}=\sqrt{{{{1}^{2}}+{{2}^{2}}}}=\sqrt{5}\\\\\ \ \ \ \ \ \ \ \text{Let}\ \tan \alpha =\displaystyle \frac{b}{a}\Rightarrow \tan \alpha =2\\\\\therefore \ \ \ \ \ \ \alpha =63{}^\circ 2{6}'\\\\\therefore \ \ \ \ \ \ \cos \theta +2\sin \theta =2\Rightarrow \sqrt{5}\cos (\theta -63{}^\circ 2{6}')=2\\\\\therefore \ \ \ \ \ \ \cos (\theta -63{}^\circ 2{6}')=0.8944\\\\\therefore \ \ \ \ \ \ \theta -63{}^\circ 2{6}'=26{}^\circ 3{4}'\ \ (\text{or)}\ \theta -63{}^\circ 2{6}'=360{}^\circ -26{}^\circ 3{4}'\ \\\\\therefore \ \ \ \ \ \ \theta =90{}^\circ \ \ (\text{or)}\ \theta =396{}^\circ 5{2}'\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{array}$

$ \displaystyle \ \ \ \ \ \ \ \text{(d)}\ \ \ \ 8\text{sin}\theta +6\cos \theta =5$

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$ \displaystyle \begin{array}{l}\text{(d)}\ \ \ \ 8\sin \theta +6\cos \theta =5\\\\\ \ \ \ \ \ \ \ \text{Comparing with}\ a\sin \theta +b\cos \theta =c,\\\\\ \ \ \ \ \ \ \ a=8,b=6,c=5\\\\\therefore \,\ \ \ \ \ \ \sqrt{{{{a}^{2}}+{{b}^{2}}}}=\sqrt{{{{8}^{2}}+{{6}^{2}}}}=10\\\\\ \ \ \ \ \ \ \ \text{Let}\ \tan \alpha =\displaystyle \frac{b}{a}\Rightarrow \tan \alpha =\displaystyle \frac{3}{4}\\\\\therefore \ \ \ \ \ \ \alpha =36{}^\circ 5{2}'\\\\\therefore \ \ \ \ \ \ 8\sin \theta +6\cos \theta =5\Rightarrow 10\sin (\theta +36{}^\circ 5{2}')=5\\\\\therefore \ \ \ \ \ \ \sin (\theta +36{}^\circ 5{2}')=0.5\\\\\therefore \ \ \ \ \ \ \theta +36{}^\circ 5{2}'=30{}^\circ \ \ (\text{or)}\ \theta +36{}^\circ 5{2}'=150{}^\circ \ (\text{or)}\ \theta +36{}^\circ 5{2}'=390{}^\circ \\\\\therefore \ \ \ \ \ \ \theta =-6{}^\circ 5{2}'\ \ (\text{or)}\ \theta =113{}^\circ {8}'\ (\text{or)}\ \theta =353{}^\circ {8}'\\\\\ \ \ \ \ \ \ \text{Since}\ 0{}^\circ \le \theta \le 360{}^\circ ,\theta =-6{}^\circ 5{2}'\ \text{is impossible}\text{.}\\\\\therefore \ \ \ \ \ \ \theta =113{}^\circ {8}'\ (\text{or)}\ \theta =353{}^\circ {8}'\ \ \ \ \ \ \ \ \ \ \ \ \end{array}$

$ \displaystyle \ \ \ \ \ \ \ \text{(e)}\ \ \ \ \sqrt{3}\text{cos}\theta +\sin \theta =\sqrt{2}$

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$ \displaystyle \begin{array}{l}\text{(e)}\ \ \ \ \sqrt{3}\cos \theta +\sin \theta =\sqrt{2}\\\\\ \ \ \ \ \ \ \ \text{Comparing with}\ a\cos \theta +b\sin \theta =c,\\\\\ \ \ \ \ \ \ \ a=\sqrt{3},b=1,c=\sqrt{2}\\\\\therefore \,\ \ \ \ \ \ \sqrt{{{{a}^{2}}+{{b}^{2}}}}=\sqrt{{3+{{1}^{2}}}}=2\\\\\ \ \ \ \ \ \ \ \text{Let}\ \tan \alpha =\displaystyle \frac{b}{a}\Rightarrow \tan \alpha =\displaystyle \frac{1}{{\sqrt{3}}}\\\\\therefore \ \ \ \ \ \ \alpha =30{}^\circ \\\\\therefore \ \ \ \ \ \ \sqrt{3}\cos \theta +\sin \theta \Rightarrow 2\cos (\theta -30{}^\circ )=\sqrt{2}\\\\\therefore \ \ \ \ \ \ \cos (\theta -30{}^\circ )=\displaystyle \frac{{\sqrt{2}}}{2}\\\\\therefore \ \ \ \ \ \ \theta -30{}^\circ =45{}^\circ \ \ (\text{or)}\ \theta -30{}^\circ =315{}^\circ \ \\\\\therefore \ \ \ \ \ \ \theta =75{}^\circ \ \ (\text{or)}\ \theta =345{}^\circ \ \ \ \ \ \end{array}$

$ \displaystyle \ \ \ \ \ \ \ \text{(f)}\ \ \ \ \text{sin}\theta -\cos \theta =\sqrt{2}$

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$ \displaystyle \begin{array}{l}\text{(f)}\ \ \ \ \sin \theta -\cos \theta =\sqrt{2}\\\\\ \ \ \ \ \ \ \ \text{Comparing with}\ a\sin \theta -b\cos \theta =c,\\\\\ \ \ \ \ \ \ \ a=1,b=1,c=\sqrt{2}\\\\\therefore \,\ \ \ \ \ \ \sqrt{{{{a}^{2}}+{{b}^{2}}}}=\sqrt{{{{1}^{2}}+{{1}^{2}}}}=\sqrt{2}\\\\\ \ \ \ \ \ \ \ \text{Let}\ \tan \alpha =\displaystyle \frac{b}{a}\Rightarrow \tan \alpha =1\\\\\therefore \ \ \ \ \ \ \alpha =45{}^\circ \\\\\therefore \ \ \ \ \ \ \sin \theta -\cos \theta =\sqrt{2}\Rightarrow \sqrt{2}\sin (\theta -45{}^\circ )=\sqrt{2}\\\\\therefore \ \ \ \ \ \ \sin (\theta -45{}^\circ )=1\\\\\therefore \ \ \ \ \ \ \theta -45{}^\circ =90{}^\circ \ \ \\\\\therefore \ \ \ \ \ \ \theta =135{}^\circ \end{array}$

$ \displaystyle \ \ \ \ \ \ \ \text{(g)}\ \ \ \ 4\text{sin}\theta +3\cos \theta =0$

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$ \displaystyle \begin{array}{l}\text{(g)}\ \ \ \ 4\sin \theta +3\cos \theta =0\\\\\ \ \ \ \ \ \ \ \text{Comparing with}\ a\sin \theta +b\cos \theta =c,\\\\\ \ \ \ \ \ \ \ a=4,b=3,c=2\\\\\therefore \,\ \ \ \ \ \ \sqrt{{{{a}^{2}}+{{b}^{2}}}}=\sqrt{{{{4}^{2}}+{{3}^{2}}}}=5\\\\\ \ \ \ \ \ \ \ \text{Let}\ \tan \alpha =\displaystyle \frac{b}{a}\Rightarrow \tan \alpha =\displaystyle \frac{3}{4}\\\\\therefore \ \ \ \ \ \ \alpha =36{}^\circ 5{2}'\\\\\therefore \ \ \ \ \ \ 4\sin \theta +3\cos \theta =0\Rightarrow 5\sin (\theta +36{}^\circ 5{2}')=0\\\\\therefore \ \ \ \ \ \ \sin (\theta +36{}^\circ 5{2}')=0\\\\\therefore \ \ \ \ \ \ \theta +36{}^\circ 5{2}'=0{}^\circ \ \ \text{(or)}\ \ \ \theta +36{}^\circ 5{2}'=180{}^\circ \ \text{(or)}\ \ \ \theta +36{}^\circ 5{2}'=360{}^\circ \\\\\therefore \ \ \ \ \ \ \theta =-36{}^\circ 5{2}'\ \ \text{(or)}\ \ \ \theta =143{}^\circ {8}'\ \text{(or)}\ \ \ \theta =323{}^\circ {8}'\\\\\ \ \ \ \ \ \ \ \text{Since }0{}^\circ \le \theta \le 360{}^\circ ,\theta =-36{}^\circ 5{2}'\ \text{is impossible}\text{.}\\\\\therefore \ \ \ \ \ \ \theta =143{}^\circ {8}'\ \text{(or)}\ \ \ \theta =323{}^\circ {8}'\end{array}$

a cos θ ± b sin θ = c နှင့် a sin θ ± b cos θ = c ညီမျှခြင်း ပုံသေနည်းများ ဖြစ်ပေါ်လာပုံ

$ \displaystyle a$ နဲ့ $ \displaystyle b$ ဟာ အပေါင်းကိန်း နှစ်ခုဖြစ်တယ် ဆိုပါစို့။ ၎င်း တို့ဟာ မျဉ်းပြတ်နှစ်ခုရဲ့ အလျားများ ဖြစ်တယ်လို့ သတ်မှတ်လိုက်မယ်။

အလျား $ \displaystyle a$ ယူနစ် ရှိတဲ့ မျဉ်းပြတ် နဲ့ အလျား $ \displaystyle b$ ယူနစ်ရှိတဲ့ မျဉ်းပြတ် တို့ဟာထောင့်မှန်တြိဂံ တစ်ခုရဲ့ ထောင့်မှန်ဆောင်အနားများဖြစ်ပြီး ထောင့်မှန်ခံအနားရဲ့ အလျားကတော့ $ \displaystyle R$ ယူနစ် ဖြစ်မယ်ဆိုရင် Pythagoras’ Theorem အရ…

$ \displaystyle \ \ \ \ \ \ R^2=a^2+b^2$

$ \displaystyle \therefore \ \ R= \sqrt{a^2+b^2}$ ဖြစ်ပါတယ်။

ထောင့်မှန်ခံအနား(hypotenuse) နဲ့ ထောင့်မှန်ဆောင်အနား(leg) တစ်ဖက် ကြားမှာရှိတဲ့ ထောင့်က $ \displaystyle \alpha$ ဖြစ်တယ်လို့ သတ်မှတ်ပါမယ်။ ဒါဆိုရင် …

$ \displaystyle \ \ \ \ \ \ a=R\cos\alpha$

$ \displaystyle \ \ \ \ \ \ b= R\sin\alpha$

$ \displaystyle \therefore \ \ \ \frac{b}{a}=\frac{{R\sin \alpha }}{{R\cos \alpha }}\ \Rightarrow \ \tan \alpha =\frac{b}{a}$ ဖြစ်ပါတယ်။

$ \displaystyle a$ နဲ့ $ \displaystyle b$ ဟာ အပေါင်းကိန်းများလို့ သတ်မှတ်ထားလို့ $ \displaystyle \alpha$ က ထောင့်ကျဉ်း (acute angle) အဖြစ်သာ ယူပါမယ်။

$ \displaystyle a\cos \theta +b\sin \theta =c$ ဆိုတဲ့ ညီမျှခြင်းရဲ့ $ \displaystyle a$ နဲ့ $ \displaystyle b$ နေရာမှာ အထက်မှာ သတ်မှတ်ခဲ့တဲ့ တန်ဖိုးတွေ အစားသွင်းလိုက်ရင် ...

$ \displaystyle \begin{array}{l}\ \ \ \ \ R\cos \theta \cos \alpha +R\sin \theta \sin \alpha =c\\\\\therefore \ \ \ R\left( {\cos \theta \cos \alpha +\sin \theta \sin \alpha } \right)=c\\\\\therefore \ \ \ R\cos \left( {\theta -\alpha } \right)=c\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{array}{l}\begin{array}{|r||l|l|l|c|c|c|c|c|c|} \hline \therefore \ \ \ \sqrt{{{{a}^{2}}+{{b}^{2}}}}\cos \left( {\theta -\alpha } \right)=c \\ \hline \end{array}\end{array}\end{array}$

နောက်ထပ် ညီမျှခြင်းတွေကို ဆက်ပြီး အစားသွင်းကြည့်မယ်...။

$ \displaystyle \begin{array}{l}\ \ \ \ \ a\cos \theta -b\sin \theta =c\\\\\ \ \ \ \ R\cos \theta \cos \alpha -R\sin \theta \sin \alpha =c\\\\\therefore \ \ \ R\left( {\cos \theta \cos \alpha -\sin \theta \sin \alpha } \right)=c\\\\\therefore \ \ \ R\cos \left( {\theta +\alpha } \right)=c\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{array}{l}\begin{array}{|r||l|l|l|c|c|c|c|c|c|} \hline \therefore \ \ \ \sqrt{{{{a}^{2}}+{{b}^{2}}}}\cos \left( {\theta +\alpha } \right)=c \\ \hline \end{array}\end{array}\end{array}$

$ \displaystyle \begin{array}{l}\ \ \ \ \ a\sin \theta +b\cos \theta =c\\\\\ \ \ \ \ R\sin \theta \cos \alpha +R\cos \theta \sin \alpha =c\\\\\therefore \ \ \ R\left( {\sin \theta \cos \alpha +\cos \theta \sin \alpha } \right)=c\\\\ \therefore \ \ \ R\sin \left( {\theta +\alpha } \right)=c\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{array}{l}\begin{array}{|r||l|l|l|c|c|c|c|c|c|} \hline \therefore \ \ \ \sqrt{{{{a}^{2}}+{{b}^{2}}}}\sin \left( {\theta +\alpha } \right)=c \\ \hline \end{array}\end{array}\end{array}$

$ \displaystyle \begin{array}{l}\ \ \ \ \ a\sin \theta -b\cos \theta =c\\\\\ \ \ \ \ R\sin \theta \cos \alpha -R\cos \theta \sin \alpha =c\\\\\therefore \ \ \ R\left( {\sin \theta \cos \alpha -\cos \theta \sin \alpha } \right)=c\\\\\therefore \ \ \ R\sin \left( {\theta -\alpha } \right)=c\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{array}{l}\begin{array}{|r||l|l|l|c|c|c|c|c|c|} \hline \therefore \ \ \ \sqrt{{{{a}^{2}}+{{b}^{2}}}}\sin \left( {\theta -\alpha } \right)=c \\ \hline \end{array}\end{array}\end{array}$

အချုပ်ဆိုရသော် ...

$ \displaystyle \text{Original Expression}$	$ \displaystyle \text{Combined Expression}$	$ \displaystyle \tan\alpha$
$ \displaystyle a\cos\theta +b\sin \theta =c$	$ \displaystyle \sqrt{{{{a}^{2}}+{{b}^{2}}}}\cos \left( {\theta -\alpha } \right)=c$	$ \displaystyle \tan \alpha =\frac{b}{a}$
$ \displaystyle a\cos\theta -b\sin \theta =c$	$ \displaystyle \sqrt{{{{a}^{2}}+{{b}^{2}}}}\cos \left( {\theta +\alpha } \right)=c$	$ \displaystyle \tan \alpha =\frac{b}{a}$
$ \displaystyle a\sin\theta +b\cos \theta =c$	$ \displaystyle \sqrt{{{{a}^{2}}+{{b}^{2}}}}\sin \left( {\theta +\alpha } \right)=c$	$ \displaystyle \tan \alpha =\frac{b}{a}$
$ \displaystyle a\sin\theta -b\cos \theta =c$	$ \displaystyle \sqrt{{{{a}^{2}}+{{b}^{2}}}}\sin \left( {\theta -\alpha } \right)=c$	$ \displaystyle \tan \alpha =\frac{b}{a}$

Problem Study : Trigonometric Equations

 

Solve the equation

(i) $ \displaystyle 3\sin x-5\cos x=0$ for $ \displaystyle 0{}^\circ < x < 360{}^\circ$.

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Let $ \displaystyle R\cos \theta =3$ and $ \displaystyle R\sin \theta =5$.

$ \displaystyle \therefore R=\sqrt{{{{3}^{2}}+{{5}^{2}}}}=\sqrt{{34}}$

$ \displaystyle \ \ \ \tan \theta =\frac{5}{3}=1.6667$

$ \displaystyle \therefore \theta =59{}^\circ {2}'$

$ \displaystyle \therefore 3\sin x-5\cos x=\sqrt{{34}}\sin (x-59{}^\circ {2}')$

$ \displaystyle \therefore \sqrt{{34}}\sin (x-59{}^\circ {2}')=0$

$ \displaystyle \therefore \sin ((x-59{}^\circ {2}')=0$

$ \displaystyle \therefore x-59{}^\circ {2}'=0{}^\circ $ (or) $ \displaystyle x-59{}^\circ {2}'=180{}^\circ $

$ \displaystyle \therefore x=59{}^\circ {2}'$ (or) $ \displaystyle x=239{}^\circ {2}'$

(ii) $ \displaystyle 5{{\sin }^{2}}y+9\cos y-3=0\ \text{for }$ $ \displaystyle 0{}^\circ < y < 360{}^\circ$.

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$ \displaystyle 5(1-{{\cos }^{2}}y)+9\cos y-3=0$

$ \displaystyle 5-5{{\cos }^{2}}y+9\cos y-3=0$

$ \displaystyle 5{{\cos }^{2}}y-9\cos y-2=0$

$ \displaystyle (5\cos y+1)(\cos y-2)=0$

$ \displaystyle \therefore \cos y=-\frac{1}{5}\ $ or $ \displaystyle \cos y=2$

Since $\displaystyle -1\le \cos y\le 1$, $ \displaystyle \cos y=2$ is impossible.

$ \displaystyle \therefore \cos y=-\frac{1}{5}=-0.2$

$ \displaystyle \therefore \text{basic acute angle = }78{}^\circ 2{8}'$

Since $ \displaystyle \cos y < 0$, $ \displaystyle y$ lies in the second or third quadrant.

$ \displaystyle \therefore y=180{}^\circ -78{}^\circ 2{8}'\ \text{or }y=180{}^\circ +78{}^\circ 2{8}'$

$ \displaystyle \therefore y=101{}^\circ 3{2}'\ \text{or }y=258{}^\circ 2{8}'$

(iii) $ \displaystyle 6{{\sin }^{2}}x=5+\cos x\ \text{for }0{}^\circ < x < 180{}^\circ .$

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$\displaystyle 6{{\sin }^{2}}x=5+\cos x$

$ \displaystyle 6(1-{{\cos }^{2}}x)=5+\cos x$

$ \displaystyle 6-6{{\cos }^{2}}x=5+\cos x$

$ \displaystyle 6{{\cos }^{2}}x+\cos x-1=0$

$ \displaystyle (3\cos x-1)(2\cos x+1)=0$

$ \displaystyle \cos x=\frac{1}{3}\ \ \text{or}\ \cos x=-\frac{1}{2}$

$ \displaystyle \therefore x=70{}^\circ 3{1}'\ \text{or}\ x=120{}^\circ $

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

Equation of the Type : a sin θ ± b cos θ = c

If a and b are positive,                           $ \displaystyle a\sin \theta \pm b\cos \theta $ can be written in the form $ \displaystyle R\sin (\theta \pm \alpha ),$                           $ \displaystyle a\cos \theta \pm b\sin \theta $ can be written in the form $\displaystyle R\sin (\theta \mp \alpha ),$   where $ R = \sqrt{a^2 + b^2}, R \cos \alpha = a , R \sin \alpha = b$ and $ \displaystyle \tan \alpha =\frac{b}{a}$ with $ \displaystyle {{0}^{{}^\circ }}<\alpha <{{90}^{{}^\circ }}.$

Example (1)      Solve the equation $ \displaystyle 8\sin \theta +6\cos \theta =5$ for $ \displaystyle {{0}^{{}^\circ }}\le \theta \le {{360}^{{}^\circ }}$.

Solution
         
              $ \displaystyle 8\sin \theta +6\cos \theta =5$

              Let $ \displaystyle R\cos \alpha =8$ and $ \displaystyle R\sin \alpha =6$.

              $ \displaystyle \therefore R=\sqrt{{{{8}^{2}}+{{6}^{2}}}}=\sqrt{{100}}=10$ and $ \displaystyle \tan \alpha =\frac{6}{8}\Rightarrow \alpha =36{}^\circ 5{2}'$

              Since $ \displaystyle 8\sin \theta +6\cos \theta =R\sin (\theta +\alpha )$,

             $ \displaystyle R\sin (\theta +\alpha )=5\Rightarrow 10\sin (\theta +36{}^\circ 5{2}')=5\Rightarrow \sin (\theta +36{}^\circ 5{2}')=\frac{1}{2}$

             $ \displaystyle \begin{array}{l}\therefore \theta +36{}^\circ 5{2}'=30{}^\circ \\\end{array}$ (1^st quadrant) or

                 $ \displaystyle \theta +36{}^\circ 5{2}'=150{}^\circ $(2^nd quadrant) or

                 $ \displaystyle \theta +36{}^\circ 5{2}'=390{}^\circ $(1^st quadrant)

             $ \displaystyle \therefore \theta =-6{}^\circ 5{2}'$ or $ \displaystyle \ \theta =113{}^\circ {8}'$ or $ \displaystyle \theta =353{}^\circ {8}'$

             Since $ \displaystyle {{0}^{{}^\circ }}\le \theta \le {{360}^{{}^\circ }}$, $ \displaystyle \theta =-6{}^\circ 5{2}'\ $is impossible.
             
             $ \displaystyle \therefore \theta =113{}^\circ {8}'$ or $ \displaystyle \theta =353{}^\circ {8}'$.