### <Success>Trigonometric functions <br> lecture - 6</Success> <div style='float: left; width:100%;'> </div>

#### <Success>$\text{General solution of} \cos x= \cos y$</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <ol> <li>If $\cos x= \cos y$, then</li> </ol> <p>$ x = 2n\pi + y$ or</p> <p>$x = 2n\pi - y$</p> <p>for some $n \in \Z $</p> <ol start="2"> <li>For any $n \in \Z$</li> </ol> <p>$\cos (2n\pi+y) = \cos(2n\pi-y) = \cos y$</p> </div>

#### <Success>$\text{Theorem}$</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$\text{Prove} \cos (2n\pi + y) = \cos (2n\pi - y) = \cos y$</p> <p><strong>$\text{Proof:}$</strong></p> <p>$\because \cos (A+B) = \cos A \cos B - \sin A \sin B$</p> <p>$\cos (2n \pi + y) = \cos 2n \pi \cos y - \sin 2n\pi \sin y=\cos y$</p> <p>$\cos (2n \pi - y) = \cos 2n \pi \cos y + \sin 2n\pi \sin y=\cos y$ ,$n \in \Z$</p> <p>$\Rightarrow \cos (2n\pi + y) = \cos (2n\pi - y) = \cos y$</p> </div>

#### <Success>$\text{General solution of} \cos x= \cos y$</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$\cos x = \cos y$</p> <p>$\Rightarrow \cos x - \cos y = 0$</p> <p>$ \cos A - \cos B = 2\sin \frac{(B-A)}{2} \sin \frac{(B+A)}{2}$</p> <p>$\Rightarrow 2 \sin\frac{(y-x)}{2} \sin\frac{(y+x)}{2}=0$</p> <p>either $\sin \frac{(y-x)}{2}=0,$</p> <p>or $\sin\frac{(y+x)}{2}=0 $</p> </div>

#### <Success>$\text{General solution of} \cos x= \cos y$</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$ \sin (\frac {y-x}{2})=0$</p> <p>$ \Rightarrow \sin (\frac {x-y}{2})=0$</p> <p>($ \sin \theta=0$ $ \Rightarrow \theta=n \pi$ $,n \in \Z$)</p> <p>$ \frac{x-y}{2}=n\pi $ for some $n \in \Z$</p> <p>$ x=2n\pi-y $ for some $n \in \Z$</p> </div> ====== #### <Success>$\text{General solution of} \cos x= \cos y$</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$ \sin (\frac {x+y}{2})=0$</p> <p>$ \frac{x+y}{2}=n\pi $ for some $n \in \Z$</p> <p>$ x=2n\pi-y $ for some $n \in \Z$</p> <p>As $\cos x = \cos y$, when</p> <p>either $\sin \frac{(y-x)}{2}=0, \Rightarrow x=2n\pi+y$</p> <p>or $\sin \frac{(y+x)}{2}=0, \Rightarrow x=2n\pi-y$</p> </div> <div style='float: right;text-align:left; width:0%;font-size:30px;'> </div>

### <Success>$\text{Example}$</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$\cos x = -\frac{1}{2}$</p> <p>$=\cos (\frac{\pi}{2}+\frac{\pi}{6})$</p> <p>$\cos x = \cos\frac{2\pi}{3}$</p> <table> <thead> <tr> <th>n</th> <th>$2n\pi \pm \frac{2 \pi}{3}$</th> </tr> </thead> <tbody> <tr> <td>0</td> <td>$\frac{2 \pi}{3}, \frac{-2 \pi}{3} $</td> </tr> <tr> <td>1</td> <td>$2 \pi + \frac{2 \pi}{3}, 2 \pi - \frac{2 \pi}{3}$</td> </tr> <tr> <td>-1</td> <td>$-2 \pi + \frac{2\pi}{3}, -2\pi -\frac{2 \pi}{3}$</td> </tr> </tbody> </table> </div>

#### <Success>$\text{General solution of} \tan x= \tan y$</Success> <div style='float: left;text-align:left; width:90%;font-size:30px;'> <ul> <li> <p>If $\tan x = \tan y$, then $\\x = n\pi + y$ for some $n \in \Z$</p> </li> <li> <p>For any integer $n \in \Z \\ \tan (n \pi +y) = \tan (y)$</p> </li> </ul> </div> <div style='float: right; width:10%;'> </div>

### <Success>$\text{Theorem}$</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$\text{Prove} \tan (n \pi +y) = \tan y$ for all $n \in \Z$</p> <p>$\text{Proof:}$</p> <p>$\tan (A+B) = \frac{\tan A +\tan B}{1- \tan A \tan B}$</p> <p>$\tan (n \pi + y) = \frac{ \tan n\pi + \tan y}{1- \tan n\pi \ \tan y}$</p> <p>For any $n \in \Z $</p> <p>$ \tan n \pi= \frac{ \sin n\pi}{ \cos n\pi}$ = 0 ($\because \sin n\pi = 0)$</p> <p>$\Rightarrow \tan (n \pi + y) = \frac{ \tan n\pi + \tan y}{1- \tan n\pi \ \tan y}= \tan y$</p> </div> <div style='float: right; width:10%;'> </div>

#### <Success>$\text{General solution of} \tan x= \tan y$</Success> <div style='float: left;text-align:left; width:50%;font-size:30px;'> <p>$\tan x = \tan y$</p> <p>$ \Rightarrow \tan x- \tan y =0$</p> <p>$ \Rightarrow \frac{\sin x}{\cos x} - \frac{\sin y}{\cos y} =0$</p> <p>$ \Rightarrow \frac{\sin x \cos y - \cos x \sin y}{ \cos x \cos y} =0$</p> <p>$ \Rightarrow \sin x \cos y - \cos x \sin y=0$</p> <p>$ \Rightarrow \sin (x-y) = 0$</p> <p>$ \Rightarrow (x-y) = n\pi$</p> <p>$ \Rightarrow x = n\pi + y \quad$ for some $n \in \Z $</p> </div>

### <Success>Problem</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Find the general solution to $a\cos \theta +b \sin \theta = c$</p> <p>(where, a, b, c are real).</p> </div> ====== ### <Success>Solution</Success> <div style='float: left;text-align:left; width:50%;font-size:30px;'> <p>$\frac{a}{\sqrt{a^2 + b^2}} \cos \theta + \frac{b}{\sqrt{a^2 + b^2}} \sin \theta = \frac{c}{\sqrt{a^2 + b^2}}$</p> <p>$\frac{a}{\sqrt{a^2 + b^2}} = \cos \phi$</p> <p>$\frac{b}{\sqrt{a^2 + b^2}} = \sin \phi$</p> </div> <div style='float: right; width:50%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/5696b93b-9792-452b-9ab2-062819b02400/public"></p> </div> <div style='float: right; width:1%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/61ad725a-311d-42e2-2c6c-7e7d41877200/public"></p> </div>

### <Success>Solution</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$\cos \phi \cos\theta + \sin \phi \sin \theta = \frac{c}{\sqrt{a^2+b^2}}$</p> <p>$ \cos (\theta- \phi) = \frac{c}{\sqrt{a^2+b^2}} = \cos y$</p> <p>$\Biggl \vert \frac{c}{\sqrt{a^2+b^2}} \Biggl \vert \leq 1 $</p> <p>We know that for any $n \in \Z$</p> <p>$\cos (2n\pi+y) = \cos (2n\pi-y) = cos y$</p> <p>$\Rightarrow \theta - \phi = 2n \pi \pm y$ for all $n \in \Z $</p> <p>$ \theta \in \{\phi + 2n \pi \pm y \vert \ n \in \Z\} $</p> </div>

### <Success>Problem</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>The solution set of the system of equations</p> <p>$x+y = \frac{2 \pi}{3}$</p> <p>$\cos x + \cos y = \frac{3}{2}$</p> <p>where $x, y$ are real is __?</p> </div> ====== ### <Success>Solution</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$y = \frac{2 \pi}{3} -x$</p> <p>$\cos x + \cos \biggl(\frac{2 \pi}{3} -x\biggl) = \frac{3}{2}$</p> <p>$\cos x + \cos \frac{2\pi}{3} \cos x + \sin \frac{2\pi}{3} \sin x = \frac{3}{2}$</p> <p>$\cos x - \frac{1}{2} \cos x + \frac{\sqrt{3}}{2} \sin x = \frac{3}{2}$</p> <p>$\frac{1}{2} \cos x + \frac{\sqrt{3}}{2} \sin x = \frac{3}{2}$</p> <p>$\cos \frac{\pi}{3} \cos x + \sin \frac{\pi}{3} \sin x = \frac{3}{2}$</p> <p>$\cos \biggl(x-\frac{\pi}{3} \biggl) = \frac{3}{2} \rightarrow$ no solution</p> </div>

### <Success>Problem</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Find the general value of $\theta$ satisfying</p> <p>$2 \sin^2 \theta - 3 \sin \theta -2 = 0$</p> </div> ====== ### <Success>Solution</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$z = \sin \theta$</p> <p>$2z^2 - 3z-2 = 0$</p> <p>$z = \frac {3 \pm\sqrt{9+16}}{4}$</p> <p>$z = \frac{3 \pm 5}{4}, \quad z = \frac{3-5}{4} = \frac{-1}{2}$</p> <p>$\sin \theta = \frac{-1}{2}$</p> </div>

### <Success>Solution</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$\sin \theta =-\frac{1}{2}$</p> <p>$\because \sin (\pi+x) = - \sin x \Rightarrow x=\frac{\pi}{6}$</p> <p>$\sin \frac{7\pi}{6}=-\sin \frac{\pi}{6} =-\frac{1}{2}$</p> <p>$\therefore \sin \theta = \sin \frac{7\pi}{6}$</p> <p>$\sin x = \sin y$</p> <p>$x \in \{ n\pi + (-1)^ny | n \in \Z \}$</p> <p>Therefore General solution is , $\{ n\pi + (-1)^n \frac{7\pi}{6} | n \in \Z \}$</p> </div>

### <Success>Problem</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Find the general value of $\theta$ satisfying</p> <p>$\tan^2 \theta + \sec2\theta=1$</p> </div> ====== ### <Success>Solution</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> $\tan^2 \theta +\frac{1}{\cos^2 \theta - \sin^2 \theta} =1$ <p>$\tan^2 \theta + \frac{\cos^2 \theta + \sin^2 \theta}{\cos^2 \theta - \sin^2 \theta} =1$</p> <p>$\tan^2 \theta + \frac{1+ \tan^2 \theta}{1 - \tan^2 \theta}=1$</p> </div>

### <Success>Solution</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$\tan^2 \theta (1-\tan^2 \theta) +1 +\tan^2 \theta = 1-\tan^2 \theta$</p> <p>$2 \tan^2 \theta +1- \tan^4 \theta = 1-\tan^2 \theta$</p> <p>$\tan^4 \theta - 3\tan^2 \theta = 0$</p> <p>$\tan^2 \theta( \tan^2 \theta -3) = 0$</p> <p>$\tan \theta=0 = \tan 0 \Rightarrow \theta=n \pi$</p> <p>$\tan^2 \theta -3 =0$</p> <p>$\Rightarrow \tan \theta=+\sqrt{3},\tan \theta=-\sqrt{3}$</p> </div> </div>

### <Success>Solution</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$\tan \theta=\sqrt{3}=\tan \frac{\pi}{3}$</p> <p>$\tan x=\tan y$</p> <p>$x \in \{ n\pi +y | n \in \Z \}$</p> <p>$ \theta \in \{ n\pi +\frac{\pi}{3} | n \in \Z \}$</p> <p>$ \tan \theta = - \sqrt{3} = \tan (-\frac{\pi}{3})$</p> <p>$\theta \in \{ n\pi -\frac{\pi}{3} | n \in \Z \}$</p> </div> ====== ### <Success>Solution</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$\tan ^2 \theta+\sec^2 \theta=1 \ $</p> <p>$\Rightarrow \tan ^2 \theta (\tan ^2 \theta-3 )=0 \ $</p> <p>$\Rightarrow \tan \theta=0 \ or \ \tan \theta=\sqrt{3} \ or \ \tan \theta=-\sqrt{3}\ $</p> <p>Therefore General solution is ,$\{ n \pi | n \in \Z\} \cup \{n \pi+\frac{\pi}{3} \mid n \in \Z \} \cup \{n \pi-\frac{\pi}{3} \mid n \in \Z \}$</p> </div>

### <Success>Problem</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Find the values of $x \in(-\pi, \pi)$ which satisfy the equation</p> <p>$2^{1+|\cos x|+ |\cos ^2 x |+ \cdots }=4$</p> </div> ====== ### <Success>Solution</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$m \in \Z, \quad |\cos ^m x |=(|\cos x|)^m$</p> <p>$1+|\cos x|+ |\cos ^2 x |+\cdots$</p> <p>$= 1+|\cos x|+|\cos x|^2 +|\cos x|^3+ \cdots \cdots $</p> <p>Put, $ c = |\cos x| $</p> <p>$= 1+c+c^2+c^3+\cdots $</p> <p>$= \frac{1}{1-c}$ iff $|c|<1$</p> </div>

### <Success>Solution</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$x \neq 0$ and $x \in(-\pi, \pi)$</p> <p>$ \frac{1}{2^{1-|\cos x|}}=4$</p> <p>$ \Rightarrow 1-|\cos x|=\frac{1}{2}$</p> <p>$ \Rightarrow |\cos x|=\frac{1}{2}$</p> <p>either $\cos x=\frac{1}{2}$ or $\cos x=-\frac{1}{2}$</p> <p>$x \in \{2n\pi \pm \frac{\pi}{3} | n \in \Z \} \cup \{2n\pi \pm \frac{2\pi}{3}| n \in \Z\}$</p> </div>

### <Success>Solution</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>So, values of x∈(−π,π) which satisfy the given equation are ,</p> <p>$\frac{- \pi}{3}, \frac{\pi}{3}, \frac{2\pi}{3},\frac{-2\pi}{3}$</p> <p>[$2\pi - \frac{2 \pi}{3} =\frac{4\pi}{3} ∉ (−π,π)$]</p> </div>

### <Success>Problem</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Let m be an odd integer.</p> <p>If $\sin mx = \sum_{p=0}^m b_p \sin^px$ for every $x$, then find the value of $b_0$ and $b_1.$</p> </div> ====== ### <Success>Solution</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$\sin mx = b_0 + b_1\sin x + b_2 \sin^2x + …. +b_m\sin^m x$</p> <p>Put x= 0 ,</p> <p>$b_0 =0$</p> </div>

### <Success>Solution</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$\lim_{x \to 0} \frac{\sin mx}{\sin x} = \lim_{x \to 0}(b_1 + b_2 \sin x + b_3 \sin^2 x + …+b_m \sin^{m-1}x) $</p> <p>$m = b_1$</p> <p>$\therefore b_1=m_1, \quad b_0=0$</p> </div>

### <Success>Problem</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Find all the solution of</p> <p>$ 4\cos^2 x \sin x - 2\sin^2 x = 3 \sin x$</p> </div> ====== ### <Success>Solution</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> $4(1-\sin^2 x) \sin x -2\sin^2 x = 3 \sin x$ <p>$ \sin x[4(1-\sin^2 x) - 2\sin x - 3] = 0$</p> <p>$ \sin x[1- 2\sin x - 4 \sin^2 x] =0 $</p> <p>For $ \sin x = 0 ,x \in \{n\pi | n\in z \} $</p> </div>

### <Success>Solution</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>For $4 \sin^2 x + 2\sin x -1 =0$</p> <p>$\sin x = \frac{-2 \pm \sqrt{4+16}}{8}$</p> <p>$=\biggl\{ \frac{-1 -\sqrt{5}}{4}, \frac{\sqrt{5} -1 }{4} \biggl\}$</p> <p>$=\biggl\{ \sin\frac{-3\pi}{10}, \sin\frac{\pi}{10} \biggl\}$</p> </div>

### <Success>Solution</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$\sin x = \sin\frac{\pi}{10}$</p> <p>$\Rightarrow x \in \{ n\pi + (-1)^n \frac{\pi}{10} | n \in \Z \}$</p> <p>$\sin x = \sin \biggl(\frac{-3 \pi}{10}\biggl)$</p> <p>$\Rightarrow x \in \{ n\pi + (-1)^n\frac{-3\pi}{10} \biggl\vert n \in \Z \}$</p> <p>$\Rightarrow$ General solution is $\{n\pi | n\in \Z \} \cup \{n\pi + (-1)^n\frac{\pi}{10} | n\in \Z \}\cup \{n\pi + (-1)^n\frac{-3\pi}{10} | n\in \Z \}$</p> </div>

<div> <h2><Success>Thank you</Success></h2> </div>