====== Question 13, Exercise 8.1 ======
Solutions of Question 13 of Exercise 8.1 of Unit 08: Fundamental of Trigonometry. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. 


===== Question 13(i)=====
Express in the form of $r \sin (\theta+\phi)$:  $12 \sin \theta-5 \cos \theta$

** Solution. **

Let $12=r\cos \varphi $ and $-5=r\sin \varphi$.\\
Squaring and adding 
\begin{align*}
& (12)^2+(-5)^2=r^2 \cos^2\varphi+r^2 \sin^2 \varphi \\
\implies & 144+25={{r}^{2}}\left( {{\cos }^{2}}\varphi +{{\sin }^{2}}\varphi  \right) \\
\implies &  169={{r}^{2}}\left( 1 \right) \\
\implies &  r=\sqrt{169}=13
\end{align*}
Also
\begin{align*}
& \frac{-5}{12}=\frac{r\sin \varphi }{r\cos \varphi } \\
\implies & \frac{-5}{12}=\tan \varphi \\
\implies & \varphi =\tan^{-1}\left(-\frac{5}{12}\right)
\end{align*}

Now 
\begin{align*} & 12\sin \theta +5\cos \theta \\
=& r\cos \varphi \sin \theta +r\sin \varphi \cos \theta \\
=& r\left( \cos \varphi \sin \theta +\sin \varphi \cos \theta  \right) \\
=& r\sin \left( \theta +\varphi  \right),
\end{align*}
where $r=13$ and $\varphi =\tan^{-1}\left(-\frac{5}{12}\right)$. GOOD


===== Question 13(ii)=====
Express in the form of $r \sin (\theta+\phi)$: $3 \sin \theta+4 \cos \theta$

** Solution. **

Let \( 3 = r \cos \varphi \) and \( 4 = r \sin \varphi \).\\
Squaring and adding:
\begin{align*}
& (3)^2 + (4)^2 = r^2 \cos^2 \varphi + r^2 \sin^2 \varphi \\
\implies & 9 + 16 = r^2 \left( \cos^2 \varphi + \sin^2 \varphi \right) \\
\implies & 25 = r^2 \left( 1 \right) \\
\implies & r = \sqrt{25} = 5.
\end{align*}
Also
\begin{align*}
& \frac{4}{3} = \frac{r \sin \varphi}{r \cos \varphi} \\
\implies & \frac{4}{3} = \tan \varphi \\
\implies & \varphi = \tan^{-1} \left( \frac{4}{3} \right).
\end{align*}
Now,
\begin{align*}
& 3 \sin \theta + 4 \cos \theta \\
=&  r \cos \varphi \sin \theta + r \sin \varphi \cos \theta \\
=&  r \left( \cos \varphi \sin \theta + \sin \varphi \cos \theta \right) \\
=&  r \sin \left( \theta + \varphi \right),
\end{align*}
where \( r = 5 \) and \( \varphi = \tan^{-1} \left( \frac{4}{3} \right) \).
	

===== Question 13(iii)=====
Express in the form of $r \sin (\theta+\phi)$: $\sin \theta-\cos \theta$

** Solution. **

Let \( 1 = r \cos \varphi \) and \( -1 = r \sin \varphi \).\\
Squaring and adding:
\begin{align*}
& (1)^2 + (-1)^2 = r^2 \cos^2 \varphi + r^2 \sin^2 \varphi \\
\implies & 1 + 1 = r^2 \left( \cos^2 \varphi + \sin^2 \varphi \right) \\
\implies & 2 = r^2 \left( 1 \right) \\
\implies & r = \sqrt{2}.
\end{align*}
Also
\begin{align*}
& \frac{-1}{1} = \frac{r \sin \varphi}{r \cos \varphi} \\
\implies & \frac{-1}{1} = \tan \varphi \\
\implies & \varphi = \tan^{-1}(-1) = -\frac{\pi}{4}.
\end{align*}
Now
\begin{align*}
& \sin \theta - \cos \theta \\
=&  r \cos \varphi \sin \theta + r \sin \varphi \cos \theta \\
=&  r \left( \cos \varphi \sin \theta + \sin \varphi \cos \theta \right) \\
=&  r \sin \left( \theta + \varphi \right),
\end{align*}
where \( r = \sqrt{2} \) and \( \varphi = -\frac{\pi}{4} \).





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