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.

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 $\varphi =\tan^{-1}\left(-\frac{5}{12}\right)$.

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

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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