### <Success>Trigonometric functions <br> lecture - 5</Success> <div style='float: left; width:50%;font-size:30px;'> </div>

### <Success>Problem 1</Success> <div style='float: left; font-size: 30px; width:100%;'> <p>$\text{Find the value of} \sqrt{3} \cosec 20^{\circ} - \sec 20^{\circ}$</p> <p><strong>Solution:</strong> $\sqrt{3} \cosec 20^{\circ} - \sec 20^{\circ}$</p> <p>$= \frac{\sqrt{3}}{\sin 20^{\circ}} - \frac{1}{\cos 20^{\circ}}$</p> <p>$= \frac{\sqrt{3} \cos20^{\circ} - \sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}}$</p> <p>$\sin 2A = 2 \sin A \cos A$</p> <p>$= \frac{\sqrt{3} \cos 20^{\circ} - \sin 20^{\circ}}{\frac{1}{2} \sin 40^{ \circ}}$</p> </div>

### <Success>Solution</Success> <div style='float: left; font-size: 30px; width:100%;'> <p>$\sqrt{3} \cos 20^{\circ}- \sin 20^{\circ} = (a \cos 20^{\circ} - b\sin 20^{\circ}) \times c$</p> <p>$\cos(A+B) = \cos A \cos B - \sin A \sin B$</p> <p>$\cos(A+20^{\circ}) = \cos A \cos 20^{\circ} - \sin A \sin 20^{\circ}$</p> <p>$a^2 + b^2 = 1$</p> <p>$ac = \sqrt3 \ \text{and} \ bc=1$</p> <p>$ (ac)^2 + (bc)^2 =4 $</p> <p>$c^2 (a^2 + b^2) = 4 \\ \quad c=2$</p> </div>

### <Success>Solution</Success> <div style='float: left; font-size: 30px; width:100%;'> <p>$\sqrt{3} \cos 20^{\circ} - \sin 20^{\circ}$</p> <p>$ = (a \cos20^{\circ} - b \sin 20^{\circ})c$</p> <p>$c=2, a = \frac{\sqrt{3}}{2}, b = \frac{1}{2}$</p> <p>$=2\Biggl(\frac{\sqrt{3}}{2} \cos 20^{\circ} - \frac{1}{2} \sin 20^{\circ}\Biggl)$</p> <p>$\cos A = \frac{\sqrt{3}}{2}, \sin A = \frac{1}{2}$</p> <p>$A=30^{\circ}$</p> <p>$\sqrt{3} \cos 20^{\circ} - \sin 20^{\circ} = 2(\cos (30^{\circ} +20^{\circ}))= 2 \cos 50^{\circ}$</p> </div>

### <Success>Solution</Success> <div style='float: left; font-size: 30px; width:100%;'> <p>$\sqrt{3} \cosec 20^{\circ} - \sec 20^{\circ} $</p> <p>$= \frac{\sqrt{3} \cos 20^{\circ} - \sin 20^{\circ}}{\frac{1}{2} \sin 40^{ \circ}}$</p> <p>$=\frac{2 \cos50^{\circ}}{\frac{1}{2}\sin40^{\circ}}$</p> <p>$=4$</p> </div>

### <Success>Problem 2</Success> <div style='float: left; font-size: 30px; width:100%;'> <p>Show that $\tan6^{\circ} \tan42^{\circ} \tan66^{\circ} \tan78^{\circ} =1$</p> <p><strong>Solution:</strong> $ \tan x = \frac{\sin x}{\cos x}$</p> <p>$ \frac{\sin 6^{\circ} \sin 42^{\circ} \sin 66^{\circ} \sin 78^{\circ}}{ \cos 6^{\circ} \cos 42^{\circ} \cos 66^{\circ} \cos 78^{\circ}}$</p> <p>$ \sin 6^{\circ} \sin 66^{\circ} = \frac{1}{2}\Biggl(\frac{1}{2} - \cos 72^{\circ}\Biggl)$</p> <p>$\{\because 2 \sin A \sin B = \cos(A-B) - \cos(A+B)\}$</p> </div>

### <Success>Solution</Success> <div style='float: left; font-size: 30px; width:100%;'> <p>$ \sin 42^{\circ} \sin 78^{\circ} = \frac{1}{2}\biggl(\cos 36^{\circ}+\frac{1}{2}\biggl)$</p> <p>$\cos (90^{\circ} + x) = - \sin x$</p> <p>$\cos (120^{\circ}) = -\sin 30^{\circ} = \frac{-1}{2}$</p> <p>$\Rightarrow \sin 6^{\circ} \sin 66^{\circ} \sin 42^{\circ} \sin 78^{\circ}$</p> <p>$= \frac{1}{4}\biggl(\frac{1}{2} - \cos 72^{\circ}\biggl)\biggl(\frac{1}{2} + \cos 36^{\circ}\biggl)$</p> </div>

### <Success>Solution</Success> <div style='float: left; font-size: 30px; width:100%;'> <p>$ \cos 6^{\circ} \cos 66^{\circ} = \frac{1}{2} \biggl(\cos 72^{\circ} + \frac{1}{2}\biggl)$</p> <p>$ \{ 2 \cos A \cos B = cos(A+B) +cos(A-B)\}$</p> <p>$ \cos 42^{\circ} \cos 78^{\circ} = \frac{1}{2}\biggl(\frac{-1}{2} + \cos 36^{\circ} \biggl)$</p> <p>$\cos 6^{\circ} \cos 42^{\circ} \cos 66^{\circ} \cos 78^{\circ}$</p> <p>$ =\frac{1}{4}\biggl(\frac{1}{2}+ \cos 72^{\circ}\biggl) \biggl(\cos 36^{\circ} - \frac{1}{2} \biggl)$</p> </div>

### <Success>Solution</Success> <div style='float: left; font-size: 30px; width:100%;'> <p>L.H.S. $=\frac{\frac{1}{4}\biggl(\frac{1}{2} - \cos 72^{\circ}\biggl) \biggl(\frac{1}{2} + \cos 36^{\circ}\biggl)}{\frac{1}{4}\biggl(\frac{1}{2} + \cos 72^{\circ}\biggl) \biggl(\cos 36^{\circ} -\frac{1}{2}\biggl)}$</p> <p>$= \frac{\frac{1}{4} + \frac{1}{2} \cos 36^{\circ} - \frac{1}{2} \cos 72^{\circ} - \cos 36^{\circ} \cos 72^{\circ}}{\frac{1}{2}\cos 36^{\circ}-\frac{1}{4}+ \cos36^{\circ} \cos72^{\circ} -\frac{1}{2} \cos 72^{\circ}}$</p> <p>$\frac{1}{4} - \cos 36^{\circ} \cos 72^{\circ} = \cos 36^{\circ} \cos 72^{\circ} - \frac{1}{4}$</p> <p>$2 \cos 36^{\circ} \cos 72^{\circ} = \frac{1}{2}$</p> </div>

### <Success>Solution</Success> <div style='float: left; font-size: 30px; width:100%;'> <p>$ \cos 36^{\circ} \cos 72^{\circ} = \frac{1}{4}$</p> <p>$ \Rightarrow$ L.H.S.$= \frac{\frac{1}{4} + \frac{1}{2} \cos 36^{\circ} - \frac{1}{2} \cos 72^{\circ} - \frac{1}{4}}{\frac{1}{2}\cos 36^{\circ}-\frac{1}{4}+ \frac{1}{4} -\frac{1}{2} \cos 72^{\circ}}$</p> <p>$=\frac{\frac{1}{2} \cos 36^{\circ} - \frac{1}{2} \cos 72^{\circ}}{\frac{1}{2} \cos 36^{\circ} - \frac{1}{2} \cos 72^{\circ}}=1$</p> </div> ====== ### <Success>Solution</Success> <div style='float: left; font-size: 30px; width:100%;'> <p>$ \cos 36^{\circ} \cos 72^{\circ}= \sin 54^{\circ} \sin 18^{\circ} $</p> <p>$(\because \cos 36^{\circ}=\sin 54^{\circ} ; \cos 72^{\circ} = \sin 18^{\circ})$</p> <p>$ \sin 18^{\circ} =\frac{(\sqrt{5}-1)}{4} $</p> <p>$ \sin 3x = 3 \sin x - \sin^3 x$</p> <p>$\sin 54^{\circ} = \frac{(\sqrt{5}+1)}{4} $</p> <p>$ \sin 54^{\circ} \sin 18^{\circ} = \frac{(\sqrt{5}-1)}{4}\frac{(\sqrt{5}+1)}{4}$</p> <p>$ = \frac{(5-1)}{16}= \frac{1}{4}$</p> </div>

### <Success>Problem 3</Success> <div style='float: left; font-size: 30px; width:100%;'> <p>Show that $\sin^4 \frac{\pi}{8}+ \sin^4 \frac{3\pi}{8}+ \sin^4 \frac{5\pi}{8}+ \sin^4 \frac{7\pi}{8}=\frac{3}{2}$</p> <p><strong>Solution:</strong> $ \frac{5\pi}{8} = \frac{\pi}{8} + \frac{\pi}{2}$</p> <p>$ \sin^4 \biggl(\frac{5 \pi}{8}\biggl) = \sin^4 \biggl(\frac{\pi}{2} + \frac{\pi}{8} \biggl)$</p> <p>$\quad = \cos^4 \biggl(\frac{\pi}{8} \biggl) $</p> <p>$ \frac{7 \pi}{8} =\frac{\pi}{2} + \frac{3\pi}{8}$</p> </div>

### <Success>Solution</Success> <div style='float: left; font-size: 30px; width:100%;'> <p>$\sin^4 \frac{\pi}{8}+ \cos^4\frac{\pi}{8} + \sin^4\frac{3\pi}{8}+ \cos^4\frac{3\pi}{8}$</p> <p>$\sin^4 \frac{\pi}{8}+ \cos^4\frac{\pi}{8}= \biggl(\sin^2 \frac{\pi}{8}+ \cos^2\frac{\pi}{8}\biggl)^2 -2 \sin^2\frac{\pi}{8}\cos^2\frac{\pi}{8},\\ \quad \quad \{\because a^4+b^4 = (a^2+b^2)^2 - 2a^2b^2\}$</p> <p>$ =1- \frac{1}{2}\biggl( 2\sin \frac{\pi}{8} \cos\frac{\pi}{8} \biggl)^2$</p> <p>$ =1-\frac{1}{2} \sin^2\frac{\pi}{4} = \frac{3}{4}$</p> </div>

### <Success>Solution</Success> <div style='float: left; font-size: 30px; width:100%;'> <p>$\sin^4\frac{3 \pi}{8}+\cos^4\frac{3 \pi}{8} = \cos^4\frac{\pi}{8}+\sin^4\frac{\pi}{8} = \frac{3}{4}$</p> <p>$[\sin^4\frac{3 \pi}{8} = \sin^4(\frac{\pi}{2}-\frac{\pi}{8}) = \cos^4\frac{\pi}{8}$</p> <p>$\cos^4\frac{3 \pi}{8}=\sin^4\frac{\pi}{8}]$</p> <p>$\Rightarrow \sin^4 \frac{\pi}{8}+ \sin^4 \frac{3\pi}{8}+ \sin^4 \frac{5\pi}{8}+ \sin^4 \frac{7\pi}{8}=\frac{3}{4} + \frac{3}{4}=\frac{3}{2} $</p> </div>

### <Success>Trigonometric equations</Success> <div style='float: left; width:100%;text-allign:left;font-size:30px;'> <p>Equations involving trigonometric functions of a variable</p> <p>e.g. $\sin x + \tan x = 2$</p> </div>

##### <Success>Does solution of trigonometric equation always exist ?</Success> <div style='float: left; text-align: left; width:100%;font-size:30px;'> <p>No</p> <p><strong>Example:</strong> $\sin x = 2$ has no solution</p> </div>

### <Success> Is solution of trigonometric equations unique ?</Success> <div style='float: left; text-align: left; font-size: 30px; width:100%;'> <p>$\sin (x) = \sin (x+2\pi)$</p> <p>$\cos (x) = \cos(x+2\pi)$</p> <p>$\tan (x) = \tan (x+x\pi)$</p> <p>Let, $\sin x + \tan x = 2$</p> <p>Put, $x=\theta \quad\quad \sin \theta + \tan \theta=2$</p> <p>$\sin (\theta + 2\pi) + \tan(\theta+2\pi)$</p> <p>$ = \sin \theta+ \tan\theta=2$</p> </div>

#### <Success>Principal solutions</Success> <div style='float: left; font-size: 30px; width:99%;'> <p>Solutions which lie in the interval $[0,2\pi]$,are <span style="color:blue">Principal solutions</span>.</p> <p>Here, Red line representing the line $\sin x = \frac{1}{2}$</p> <p>So, $ x=\frac{\pi}{6} , x=\frac{5\pi}{6}$ are Principal solutions.</p> </div> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/aaa0161a-afbb-49cb-6e05-ce17ef0cfb00/public"></p> <div style='float: right; width:1%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/137e41e5-4f1e-4511-fc33-34dba5fdaf00/public"></p> </div>

#### <Success>General solution to trigonometric equation</Success> <div style='float: left; text-align: left; width:100%;font-size:30px;'> <p>$\sin x=0 \Leftrightarrow x=n \pi , \quad n\in \Z$</p> <p>$\cos x=0 \Leftrightarrow x=\biggl(n+\frac{1}{2} \biggl) \pi , \quad n\in \Z$</p> </div>

#### <Success>$\text{General solution for} \sin x = \sin y$</Success> <div style='float: left; font-size: 30px; width:100%;'> <p>1.If $\sin x = \sin y$ then</p> <p>$x=n \pi +(-1)^ny, \quad$ for some $\ n \in \Z$</p> <p>2.For any $n \in \Z,$</p> <p>$\sin (n \pi + (-1)^ny) =\sin (y)$</p> <p>$\sin (n \pi + (-1)^n \frac{\pi}{6}) = \sin \frac{\pi}{6} = \frac{1}{2}$</p> </div>

<Success>$\text{General solution for} \sin x = \sin{\pi}/{6} $</Success> <div style='float: left; font-size: 30px;text-allign:left;width:99%;'> <p>$ \sin x = \sin \frac{\pi}{6} $</p> <table> <thead> <tr> <th>$n$</th> <th>$ n \pi + (-1)^n \frac{\pi}{6}, n \in \Z $</th> </tr> </thead> <tbody> <tr> <td>$ 0$</td> <td>$ \frac{\pi}{6}$</td> </tr> <tr> <td>$ 1$</td> <td>$ \pi + (-\frac{\pi}{6}) = \frac{5\pi}{6}$</td> </tr> <tr> <td>$ 2$</td> <td>$ 2 \pi + \frac{\pi}{6}$</td> </tr> <tr> <td>$ 3$</td> <td>$ 3 \pi - \frac{\pi}{6}$</td> </tr> </tbody> </table> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/690f66ef-aff2-459b-4cf5-833b30a05200/public"></p> </div> <div style='float: right; width:1%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/1361bcac-68a9-4fe3-a6a1-8117af376d00/public"></p> </div>

#### <Success>$\text{Proof of}\ \sin (n \pi +(-1)^ny) = \sin y$</Success> <div style='float: left; font-size: 30px; width:100%;'> <p>$\sin (n \pi +(-1)^ny) \\ (\because\sin(A+B) = \sin A \cos B+ \cos A \sin B)$</p> <p>$=(\sin n \pi \cos(-1)^ny)+ \cos n\pi \sin(-1)^ny$</p> <p>$ = \cos n\pi \sin (-1)^ny$</p> <p>$ \biggl( n \ \text{even} \Rightarrow \cos n\pi = 1 , \\ \quad n \ \text{odd} \Rightarrow \cos n\pi = -1 \biggl)$</p> <p>$(\cos n\pi = (-1)^n)$</p> </div> ====== #### <Success>$\text{Proof of}\ \sin (n \pi +(-1)^ny) = \sin y$</Success> <div style='float: left; font-size: 30px; width:100%;'> <p>$\Rightarrow \sin (n \pi +(-1)^ny)=(-1)^n \sin(-1)^ny$</p> <p>$\biggl( n \ \text{even} \Rightarrow \sin y , \quad n \ \text{odd} \Rightarrow -\sin (-y)\biggl)$</p> <p>$=\sin y$</p> <p>$\Rightarrow \sin (n \pi +(-1)^ny) = \sin y$</p> </div>

#### <Success>$\text {General solution for} \sin x = \sin y$</Success> <div style='float: left; font-size: 30px; width:100%;'> <p>$\sin x = \sin y$</p> <p>$\Rightarrow \sin x - \sin y = 0$</p> <p>$\Rightarrow 2 \cos \biggl(\frac{x+y}{2}\biggl) \sin \biggl(\frac{x-y}{2}\biggl) =0$</p> <p>either $\quad$ $\cos \biggl( \frac{x+y}{2} \biggl) = 0 $</p> <p>$\frac{x+y}{2} = \biggl(m+\frac{1}{2} \biggl) \pi$ for some $\ m \in \Z$</p> <p>$x+y = (2m+1) \pi$</p> </div>

#### <Success>$\text{General solution for} \sin x = \sin y$</Success> <div style='float: left; font-size: 30px; width:100%;'> <p>$x = (2m+1) \pi -y$</p> <p>$x= (2m+1) \pi + (-1)^{2m+1}y$</p> <p>or, $ \sin \biggl ( \frac{x-y}{2} \biggl ) = 0$</p> <p>$\sin \theta = 0 , \theta =m \pi, m\in \Z) $</p> <p>$\frac{x-y}{2} = m \pi, m \in \Z$</p> <p>$ x= 2m \pi + y$ for some $\ m \in \Z$</p> <p>$x = 2m\pi +(-1)^{2m}y$</p> <p>$x = n\pi + (-1)^ny$ for some $\in \Z$</p> </div>

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