<div style='float: left; width:100%;'> <h3 id="success-trigonometric-functions-br-lecture---7success"><Success> Trigonometric functions </br> lecture - 7</Success></h3> </div>

#### <Success>General solution of trigonometric equations </Success> <div style='float: left; width:99%; left-align;font-size:30px;'> <p>$\sin x = \sin y $</p> <p>$x \in \{ n\pi + (-1)^n | n \in \Z\}$</p> <p>$\cos x = \cos y$</p> <p>$x \in \{ 2n \pi \pm y | n \in \Z\}$</p> <p>$\tan x = \tan y$</p> <p>$x \in \{ n \pi + y | n\in \Z \}$</p> </div>

### <Success>Problem</Success> <div style='float: left; width:99%;left-align;font-size:30px;'> <p>$ \text{Find the general solution to}$ $ \cosec x = \cot x + √3 $</p> </div>

### <Success>Solution</Success> <div style='float: left; width:99%;left-align;font-size:30px;'> <p>$\frac{1}{\sin x} = \frac{\cos x}{\sin x} + \sqrt{3}$</p> <p>$\text{We \ know \ that} \ 1-\cos x=2\sin^2 \frac{x}{2}$</p> <p>$1 = \cos x + \sqrt{3} \sin x$</p> <p>$\quad =2 \biggl(\frac{1}{2}\cos x + \frac{\sqrt{3}}{2}\sin x \biggl)$</p> </div>

### <Success>Solution</Success> <div style='float: left; width:99%;left-align;font-size:30px;'> <p>$=2 \biggl( \sin \frac{\pi}{6}\cos x + \cos \frac{\pi}{6} \sin x \biggl)$</p> <p>$=2 \sin \biggl(x + \frac{\pi}{6} \biggl)$</p> <p>$\Rightarrow 2 \sin \biggl(x + \frac{\pi}{6} \biggl) = 1$</p> <p>$\sin \biggl(x + \frac{\pi}{6} \biggl) = \frac{1}{2} = \sin \frac{\pi}{6}$</p> </div>

### <Success>Solution</Success> <div style='float: left; width:99%;left-align;font-size:30px;'> <p>$x + \frac{\pi}{6} \in \biggl\{ n \pi + (-1)^n \frac{\pi}{6} | n \in \Z \biggl\}$</p> <p>$x \in \biggl\{ n \pi + (-1)^n \frac{\pi}{6} - \frac{\pi}{6} | n \in \Z \biggl\}$</p> </div>

### <Success>Problem</Success> <div style='float: left; width:99%;left-align;font-size:30px;'> <p>$\text{Find all solutions to}$ $\cosec \theta + \sec \theta =1 $</p> </div>

### <Success>Solution</Success> <div style='float: left; width:99%;left-align;font-size:30px;'> <p>$\frac{1}{\sin \theta} +\frac{1}{\cos \theta} =1$</p> <p>$\cos \theta + \sin \theta = \sin \theta \cos \theta $</p> <p>Let, $t\triangleq \sin \theta + \cos \theta, \\ \Rightarrow t^2 = \sin^2\theta + \cos^2 \theta + 2\sin \theta \cos \theta \\ \Rightarrow t^2 =1+2\sin \theta \cos \theta $</p> <p>$\sin \theta \cos \theta = \frac{(t^2-1)}{2}$</p> <p>$t = \frac{(t^2-1)}{2}$</p> </div>

### <Success>Solution</Success> <div style='float: left; width:99%;left-align;font-size:30px;'> <p>$t^2-1 = 2t$</p> <p>$t^2-2t-1=0$</p> <p>$t = \frac{2 \pm \sqrt{8}}{2} = (1\pm\sqrt{2})$</p> <p>$t = \sin \theta + \cos \theta = \sqrt{2} \biggl( \sin \theta \frac{1}{\sqrt{2}} + \cos \theta \frac{1}{\sqrt{2}}\biggl)$</p> <p>$= \sqrt{2} \biggl( \sin \theta \cos\frac{\pi}{4} + \cos \theta \sin\frac{\pi}{4}\biggl)$</p> <p>$= \sqrt{2} \sin \biggl(\theta + \frac{\pi}{4} \biggl)$</p> </div>

### <Success>Solution</Success> <div style='float: left; width:99%;left-align;font-size:30px;'> <p>$|t| \leq \sqrt{2}$</p> <p>$t = \sqrt{2} \sin(\theta + \frac{\pi}{4} ) = (1-\sqrt{2})$</p> <p>$\sqrt{2} \sin \biggl(\theta + \frac{\pi}{4} \biggl) = \biggl(1-\sqrt{2}\biggl)$</p> <p>$\sin\biggl(\theta + \frac{\pi}{4} \biggl) = \biggl(\frac{1}{\sqrt{2}}-1 \biggl) = \sin \phi$</p> <p>$\Rightarrow \theta+\frac{\pi}{4} \in\biggl\{n \pi+(-1)^n \phi \mid n \in \Z \biggl\} $</p> <p>$\Rightarrow \theta \in\biggl\{n \pi+(-1)^n \phi-\frac{\pi}{4} \mid n \in \Z \biggl\}$</p> </div>

### <Success>Problem</Success> <div style='float: left; width:99%;left-align;font-size:30px;'> <p>$ \text{Find the general solution to}$ $(1+ \sec 2 \theta)(1+ \sec 4\theta) = \cot \theta$</p> </div>

### <Success>Solution</Success> <div style='float: left; width:99%;left-align;font-size:30px;'> <p>$\biggl(1+ \frac{1}{ \cos2 \theta}\biggl)\biggl(1+ \frac{1}{ \cos4 \theta}\biggl) = \frac{ \cos \theta}{ \sin \theta}$</p> <p>$\text{Multipy L.H.S. and R.H.S. by}$ $\cos 2\theta \cos4\theta \sin\theta$</p> <p>$(1+ \cos2 \theta)(1+ \cos4 \theta) \sin \theta = \cos \theta \cos2 \theta \cos4\theta$</p> <p>$2 \cos^2 \theta \ 2 \cos^2 2\theta \sin\theta = \cos \theta \cos2 \theta \cos4 \theta$</p> <p>$4 \sin \theta \cos^2\theta \cos^2 2 \theta - cos \theta \cos2 \theta \cos4 \theta =0$</p> <p>$ \cos \theta \cos2 \theta(4 \sin \theta \cos\theta \cos 2 \theta - \cos 4 \theta) =0$</p> </div>

### <Success>Solution</Success> <div style='float: left; width:99%;left-align;font-size:30px;'> <p>$ \cos \theta \cos2 \theta(2 \sin2 \theta \cos2 \theta - \cos 4 \theta) =0$</p> <p>$ \cos \theta \cos2 \theta ( \sin4 \theta - \cos4 \theta) = 0$</p> <p>$ \cos \theta =0, or \cos2 \theta =0, or \sin 4 \theta - \cos4 \theta =0$</p> <p>$ \cos \theta =0 \Leftrightarrow \theta \in \biggl\{ (2n \pm 1) \frac{\pi}{2} | n \in \Z \biggl\}$</p> <p>$ \cos2 \theta =0 \Leftrightarrow \theta \in \biggl\{ (2n \pm 1) \frac{\pi}{4} | n \in \Z \biggl\}$</p> <p>$ \sin 4 \theta - \cos 4 \theta =0$</p> </div>

### <Success>Solution</Success> <div style='float: left; width:99%;left-align;font-size:30px;'> <p>$\Leftrightarrow \sin4 \theta \ \frac{1}{\sqrt{2}} - \cos4 \theta \ \frac{1}{\sqrt{2}}=0$</p> <p>$\Leftrightarrow \sin4 \theta \cos\frac{\pi}{4} - \cos4 \theta \sin\frac{\pi}{4} =0$</p> <p>$\Leftrightarrow \sin (4 \theta - \frac{\pi}{4}) = 0 $</p> <p>$\Leftrightarrow 4 \theta - \frac{\pi}{4} \in \{ n \pi | n \in \Z \}$</p> <p>$\theta \in \{ \frac{n \pi}{4} + \frac{\pi}{16} | n \in \Z \}$</p> </div>

### <Success>Solution</Success> <div style='float: left; width:99%;left-align;font-size:30px;'> <p>$(1 + \sec 2 \theta)(1 + \sec 4 \theta) = \cot \theta$</p> <p>$\text{General solution}$</p> <p>$ \{ (2n \pm 1) \frac{\pi}{2} | n \in \Z \} \cup \{ (2n \pm 1) \frac{\pi}{4} | n \in \Z \} \cup \{ \frac{n \pi}{4} + \frac{\pi}{16} | n \in \Z\}$</p> </div>

### <Success>Problem</Success> <div style='float: left; width:99%;left-align;font-size:30px;'> <p>$ \text{Find the smallest positive value of x such that }$</p> <p>$ \tan(x + 100^{\circ}) = \tan(x + 50^{\circ}) \tan(x) \tan(x - 50^{\circ})$</p> </div>

### <Success>Solution</Success> <div style='float: left; width:99%;left-align;font-size:30px;'> <p>$\frac{\sin(x + 100^{\circ})}{\cos(x + 100^{\circ})} = \frac{\sin(x + 50^{\circ}) \sin(x) \sin(x - 50^{\circ})}{\cos(x + 50^{\circ}) \cos(x) \cos(x - 50^{\circ})}$</p> <p>Multiply L.H.S. and R.H.S. by $\cos(x+100) \cos(x+50) \cos(x) \cos(x-50)$</p> <p>$\Rightarrow 4 \sin(x+100) \cos(x +50) \cos(x) \cos(x-50) = 4 \sin(x+50) \sin(x) \sin(x-50) \cos(x+100)$</p> </div>

### <Success>Solution</Success> <div style='float: left; width:99%;left-align;font-size:30px;'> <p><strong>L.H.S.</strong></p> <p>$(2 \sin(x+100) \cos x)(2 \cos(x+50) \cos(x-50))$</p> <p>$= (\sin(2x+100) + \sin 100) \times (\cos 2x + \cos100)$</p> <p>$= \sin(2x +100) \cos2x + \sin(2x +100) \cos100 + \sin100 \cos2x + \sin100 \cos100 $</p> <p><strong>R.H.S.</strong></p> <p>$4 \sin(x+50) \sin x \sin(x-50) \cos(x+100)$</p> <p>$=(2 \sin(x+50) \sin(x-50) \times (2 \sin x \cos(x+100))$</p> <p>$=(\cos100 - \cos2x) \times(\sin(2x+100) -\sin100)$</p> </div>

### <Success>Solution</Success> <div style='float: left; width:99%;left-align;font-size:30px;'> <p>$=\sin(2x+100)\cos100 - \sin100 \cos100 -\cos2x \sin(2x+100) + \cos2x \sin100$</p> <p>$\because$ L.H.S. = R.H.S.</p> <p>$\Rightarrow \sin(2x+100) \cos2x + \sin(2x+100) \cos100 + \sin100 \cos2x + \sin100 \cos100 \quad =\sin(2x+100)\cos100 - \sin100 \cos100 -\cos2x \sin(2x+100) + \cos2x \sin100 $</p> <p>$2 \sin(2x+100) \cos2x + 2 \sin100 \cos100 = 0$</p> <p>$2 \sin(2x + 100) \cos2x + 2 \sin100 \cos100 = 0$</p> <p>$\sin(4x + 100) + \sin100 + \sin200 =0$</p> </div>

### <Success>Solution</Success> <div style='float: left; width:99%;left-align;font-size:30px;'> <p>$\sin(4x + 100) + 2 \sin150 \cos50 = 0$ ($\because \sin 150 = 1/2)$</p> <p>$\sin(4x + 100) = -\cos50 = -\sin40$</p> <p>$4x + 100 \in \{n \pi+(-1)^n[(-40) \times \frac{\pi}{180}] |n \in \Z \}$</p> <p>$4x \in \{ n \pi + (-1)^n [\frac{-2 \pi}{9}] \frac{-5 \pi}{9} | n \in \Z\}$</p> <p>$x \in \{ \frac{n\pi}{4} +(-1)^n[\frac{-\pi}{18}] -\frac{5\pi}{36} | n \in \Z\}$</p> </div>

### <Success>Solution</Success> <div style='float: left; width:99%;left-align;font-size:30px;'> <p>$n=1, \quad \frac{\pi}{4} - \biggl( \frac{-\pi}{18} \biggl) - \frac{5\pi}{36}$</p> <p>$=\frac{\pi}{4} + \frac{\pi}{18} - \frac{5\pi}{36}$</p> <p>$=45^{\circ} + 10^{\circ} - 25^{\circ}$</p> <p>$=30^{\circ}$</p> <p>$=\frac{\pi}{6}$</p> </div>

### <Success>Problem</Success> <div style='float: left; width:99%;left-align;font-size:30px;'> <p>$\text {For all}$ $ x \in \biggl[0, \frac{\pi}{2}\biggl]$, $\text{ show that}$ $\cos (\sin x) \geq \sin (\cos x)$</p> </div>

### <Success>Solution</Success> <div style='float: left; width:99%;left-align;font-size:30px;'> <p>$\sin \biggl(\frac{\pi}{2} - \sin x\biggl) \geq \sin (\cos x)$</p> <p>Let , $A =\biggl(\frac{\pi}{2} - \sin x\biggl) \quad\quad B=(\cos x), x\in \biggl[{0,\frac{\pi}{2}}\biggl]$</p> <p>$\frac{\pi}{2} -1 \leq \frac{\pi}{2} - \sin x \leq \frac{\pi}{2}$</p> <p>$0 \leq \cos x \leq 1$</p> <p>$\biggl[\frac{\pi}{2} -1, \frac{\pi}{2}\biggl] \subset \biggl[0,\frac{\pi}{2} \biggl]$</p> <p>$[0,1] \subset \biggl[0, \frac{\pi}{2}\biggl]$ $\Rightarrow A,B \in \biggl[0, \frac{\pi}{2}\biggl]$</p> </div>

### <Success>Solution</Success> <div style='float: left; width:60%;left-align;font-size:30px;'> <p>Now we know that,</p> <p>Here, A $\geq$ B $\Leftrightarrow$ $\sin A \geq \sin B$</p> <p>Showing $\frac{\pi}{2} - \sin x \geq \cos x \ in \ x \in [0, \frac{\pi}{2}]$ is equivalent to showing</p> <p>$\cos x + \sin x \leq \frac{\pi}{2} \ for \ all \ x \in [0,\frac{\pi}{2}].$</p> </div> <div style='float: right; width:1%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/12c71326-ded1-4f81-8a02-13e7bc078300/public"></p> </div> <div style='float: right; width:40%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/52dee5f2-8613-4f27-036e-60bcd0a59000/public"></p> </div>

### <Success>Solution</Success> <div style='float: left; width:99%;left-align;font-size:30px;'> <p>$\sin x + \cos x = \sqrt{2} \biggl(\frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x\biggl)$</p> <p>=$\sqrt{2}\sin \biggl(x+\frac{\pi}{4}\biggl) ,\quad\quad x \in \biggl[0, \frac{\pi}{2} \biggl]$</p> <p>$\sin \biggl(x+\frac{\pi}{4}\biggl) \leq 1$</p> </div>

### <Success>Solution</Success> <div style='float: left; width:99%;left-align;font-size:30px;'> <p>$\therefore \sin x + \cos x \leq \sqrt{2}$</p> <p>$\because \sqrt{2} < \frac{\pi}{2}$</p> <p>($\because 1.41 < 1.57$)</p> <p>$\Rightarrow \sin x + \cos x <\frac{\pi}{2} \ for \ all \ x \in \biggl[0, \frac{\pi}{2}\biggl]$</p> <p>$\Rightarrow \frac{\pi}{2} - \sin x \geq \cos x \ for \ all \ x \in \biggl[0, \frac{\pi}{2}\biggl]$</p> <p>$\Rightarrow \sin \biggl(\frac{\pi}{2} - \sin x\biggl) \geq \sin (\cos x)$ $\Rightarrow \cos (\sin x) \geq \sin (\cos x)$</p> </div>

### <Success>Problem</Success> <div style='float: left; width:99%;left-align;font-size:30px;'> <p>$ \text{Find the smallest positive number p for which the equation}$</p> <p>$\cos (p \sin x) = \sin(p \cos x)$ $\text{has a solution}$ where, $x\in [0, 2\pi]$</p> </div>

### <Success>Solution</Success> <div style='float: left; width:99%;left-align;font-size:30px;'> <p>$ \sin(\frac{\pi}{2} - p \sin x) = \sin(p \cos x)$</p> <p>$\frac{\pi}{2} - p \sin x = n \pi + (-1)^n p \cos x \ for \ some \ n \in \Z$</p> <p>For $n=0$, $\frac{\pi}{2} - p \sin x = p \cos x$</p> <p>$p( \sin x + \cos x ) = \frac{\pi}{2}$</p> <p>$p\sqrt{2} \sin (x + \frac{\pi}{4}) = \frac{\pi}{2}$</p> </div>

### <Success>Solution</Success> <div style='float: left; width:99%;left-align;font-size:30px;'> <p>$\sin (x + \frac{\pi}{4}) \leq 1$</p> <p>$\Rightarrow p \geq \frac{n}{2\sqrt{2}}$</p> <p>For $n=1$</p> <p>$\frac{\pi}{2} - p \sin x = \pi - p \cos x$</p> <p>$p\sqrt{2} . \cos (x + \frac{\pi}{4}) = \frac{\pi}{2}$</p> <p>Similarly here , $p \geq \frac{n}{2\sqrt{2}}$</p> </div>

### <Success>Solution</Success> <div style='float: left; width:99%;left-align;font-size:30px;'> <p>$\frac{\pi}{2} - p \sin x = n \pi + (-1)^n p \cos x$</p> <p>$p((-1)^n \cos x + \sin x) = \frac{-\pi}{2}(2n-1)$</p> <p>$p\sqrt{2}(\frac{(-1)^n}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x) = -\frac{\pi}{2}(2n-1)$</p> <p>$\because |a \cos x + b \sin x | \leq \sqrt(a^2+b^2)$</p> <p>$\Rightarrow |p|\sqrt{2} \geq \frac{\pi}{2}|2n-1|$</p> <p>$p \geq \frac{\pi}{2\sqrt{2}}$</p> </div>

### <Success>Thank you</Success> <div style='float: left; width:99%;left-align;font-size:30px;'> </div>