$\text{Find the value of} \sqrt{3} \cosec 20^{\circ} - \sec 20^{\circ}$

Solution: $\sqrt{3} \cosec 20^{\circ} - \sec 20^{\circ}$

$= \frac{\sqrt{3}}{\sin 20^{\circ}} - \frac{1}{\cos 20^{\circ}}$

$= \frac{\sqrt{3} \cos20^{\circ} - \sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}}$

$\sin 2A = 2 \sin A \cos A$

$= \frac{\sqrt{3} \cos 20^{\circ} - \sin 20^{\circ}}{\frac{1}{2} \sin 40^{ \circ}}$

$\sqrt{3} \cos 20^{\circ}- \sin 20^{\circ} = (a \cos 20^{\circ} - b\sin 20^{\circ}) \times c$

$\cos(A+B) = \cos A \cos B - \sin A \sin B$

$\cos(A+20^{\circ}) = \cos A \cos 20^{\circ} - \sin A \sin 20^{\circ}$

$a^2 + b^2 = 1$

$ac = \sqrt3 \ \text{and} \ bc=1$

$(ac)^2 + (bc)^2 =4$

$c^2 (a^2 + b^2) = 4 \\ \quad c=2$

$\sqrt{3} \cos 20^{\circ} - \sin 20^{\circ}$

$= (a \cos20^{\circ} - b \sin 20^{\circ})c$

$c=2, a = \frac{\sqrt{3}}{2}, b = \frac{1}{2}$

$=2\Biggl(\frac{\sqrt{3}}{2} \cos 20^{\circ} - \frac{1}{2} \sin 20^{\circ}\Biggl)$

$\cos A = \frac{\sqrt{3}}{2}, \sin A = \frac{1}{2}$

$A=30^{\circ}$

$\sqrt{3} \cos 20^{\circ} - \sin 20^{\circ} = 2(\cos (30^{\circ} +20^{\circ}))= 2 \cos 50^{\circ}$

$\sqrt{3} \cosec 20^{\circ} - \sec 20^{\circ}$

$= \frac{\sqrt{3} \cos 20^{\circ} - \sin 20^{\circ}}{\frac{1}{2} \sin 40^{ \circ}}$

$=\frac{2 \cos50^{\circ}}{\frac{1}{2}\sin40^{\circ}}$

$=4$

Show that $\tan6^{\circ} \tan42^{\circ} \tan66^{\circ} \tan78^{\circ} =1$

Solution: $\tan x = \frac{\sin x}{\cos x}$

$\frac{\sin 6^{\circ} \sin 42^{\circ} \sin 66^{\circ} \sin 78^{\circ}}{ \cos 6^{\circ} \cos 42^{\circ} \cos 66^{\circ} \cos 78^{\circ}}$

$\sin 6^{\circ} \sin 66^{\circ} = \frac{1}{2}\Biggl(\frac{1}{2} - \cos 72^{\circ}\Biggl)$

$\{\because 2 \sin A \sin B = \cos(A-B) - \cos(A+B)\}$

$\sin 42^{\circ} \sin 78^{\circ} = \frac{1}{2}\biggl(\cos 36^{\circ}+\frac{1}{2}\biggl)$

$\cos (90^{\circ} + x) = - \sin x$

$\cos (120^{\circ}) = -\sin 30^{\circ} = \frac{-1}{2}$

$\Rightarrow \sin 6^{\circ} \sin 66^{\circ} \sin 42^{\circ} \sin 78^{\circ}$

$= \frac{1}{4}\biggl(\frac{1}{2} - \cos 72^{\circ}\biggl)\biggl(\frac{1}{2} + \cos 36^{\circ}\biggl)$

$\cos 6^{\circ} \cos 66^{\circ} = \frac{1}{2} \biggl(\cos 72^{\circ} + \frac{1}{2}\biggl)$

$\{ 2 \cos A \cos B = cos(A+B) +cos(A-B)\}$

$\cos 42^{\circ} \cos 78^{\circ} = \frac{1}{2}\biggl(\frac{-1}{2} + \cos 36^{\circ} \biggl)$

$\cos 6^{\circ} \cos 42^{\circ} \cos 66^{\circ} \cos 78^{\circ}$

$=\frac{1}{4}\biggl(\frac{1}{2}+ \cos 72^{\circ}\biggl) \biggl(\cos 36^{\circ} - \frac{1}{2} \biggl)$

L.H.S. $=\frac{\frac{1}{4}\biggl(\frac{1}{2} - \cos 72^{\circ}\biggl) \biggl(\frac{1}{2} + \cos 36^{\circ}\biggl)}{\frac{1}{4}\biggl(\frac{1}{2} + \cos 72^{\circ}\biggl) \biggl(\cos 36^{\circ} -\frac{1}{2}\biggl)}$

$= \frac{\frac{1}{4} + \frac{1}{2} \cos 36^{\circ} - \frac{1}{2} \cos 72^{\circ} - \cos 36^{\circ} \cos 72^{\circ}}{\frac{1}{2}\cos 36^{\circ}-\frac{1}{4}+ \cos36^{\circ} \cos72^{\circ} -\frac{1}{2} \cos 72^{\circ}}$

$\frac{1}{4} - \cos 36^{\circ} \cos 72^{\circ} = \cos 36^{\circ} \cos 72^{\circ} - \frac{1}{4}$

$2 \cos 36^{\circ} \cos 72^{\circ} = \frac{1}{2}$

$\cos 36^{\circ} \cos 72^{\circ} = \frac{1}{4}$

$\Rightarrow$ L.H.S.$= \frac{\frac{1}{4} + \frac{1}{2} \cos 36^{\circ} - \frac{1}{2} \cos 72^{\circ} - \frac{1}{4}}{\frac{1}{2}\cos 36^{\circ}-\frac{1}{4}+ \frac{1}{4} -\frac{1}{2} \cos 72^{\circ}}$

$=\frac{\frac{1}{2} \cos 36^{\circ} - \frac{1}{2} \cos 72^{\circ}}{\frac{1}{2} \cos 36^{\circ} - \frac{1}{2} \cos 72^{\circ}}=1$

$\cos 36^{\circ} \cos 72^{\circ}= \sin 54^{\circ} \sin 18^{\circ}$

$(\because \cos 36^{\circ}=\sin 54^{\circ} ; \cos 72^{\circ} = \sin 18^{\circ})$

$\sin 18^{\circ} =\frac{(\sqrt{5}-1)}{4}$

$\sin 3x = 3 \sin x - \sin^3 x$

$\sin 54^{\circ} = \frac{(\sqrt{5}+1)}{4}$

$\sin 54^{\circ} \sin 18^{\circ} = \frac{(\sqrt{5}-1)}{4}\frac{(\sqrt{5}+1)}{4}$

$= \frac{(5-1)}{16}= \frac{1}{4}$

Show that $\sin^4 \frac{\pi}{8}+ \sin^4 \frac{3\pi}{8}+ \sin^4 \frac{5\pi}{8}+ \sin^4 \frac{7\pi}{8}=\frac{3}{2}$

Solution: $\frac{5\pi}{8} = \frac{\pi}{8} + \frac{\pi}{2}$

$\sin^4 \biggl(\frac{5 \pi}{8}\biggl) = \sin^4 \biggl(\frac{\pi}{2} + \frac{\pi}{8} \biggl)$

$\quad = \cos^4 \biggl(\frac{\pi}{8} \biggl)$

$\frac{7 \pi}{8} =\frac{\pi}{2} + \frac{3\pi}{8}$

$\sin^4 \frac{\pi}{8}+ \cos^4\frac{\pi}{8} + \sin^4\frac{3\pi}{8}+ \cos^4\frac{3\pi}{8}$

$\sin^4 \frac{\pi}{8}+ \cos^4\frac{\pi}{8}= \biggl(\sin^2 \frac{\pi}{8}+ \cos^2\frac{\pi}{8}\biggl)^2 -2 \sin^2\frac{\pi}{8}\cos^2\frac{\pi}{8},\\ \quad \quad \{\because a^4+b^4 = (a^2+b^2)^2 - 2a^2b^2\}$

$=1- \frac{1}{2}\biggl( 2\sin \frac{\pi}{8} \cos\frac{\pi}{8} \biggl)^2$

$=1-\frac{1}{2} \sin^2\frac{\pi}{4} = \frac{3}{4}$

$\sin^4\frac{3 \pi}{8}+\cos^4\frac{3 \pi}{8} = \cos^4\frac{\pi}{8}+\sin^4\frac{\pi}{8} = \frac{3}{4}$

$[\sin^4\frac{3 \pi}{8} = \sin^4(\frac{\pi}{2}-\frac{\pi}{8}) = \cos^4\frac{\pi}{8}$

$\cos^4\frac{3 \pi}{8}=\sin^4\frac{\pi}{8}]$

$\Rightarrow \sin^4 \frac{\pi}{8}+ \sin^4 \frac{3\pi}{8}+ \sin^4 \frac{5\pi}{8}+ \sin^4 \frac{7\pi}{8}=\frac{3}{4} + \frac{3}{4}=\frac{3}{2}$

Equations involving trigonometric functions of a variable

e.g. $\sin x + \tan x = 2$

No

Example: $\sin x = 2$ has no solution

$\sin (x) = \sin (x+2\pi)$

$\cos (x) = \cos(x+2\pi)$

$\tan (x) = \tan (x+x\pi)$

Let, $\sin x + \tan x = 2$

Put, $x=\theta \quad\quad \sin \theta + \tan \theta=2$

$\sin (\theta + 2\pi) + \tan(\theta+2\pi)$

$= \sin \theta+ \tan\theta=2$

Solutions which lie in the interval $[0,2\pi]$,are Principal solutions.

Here, Red line representing the line $\sin x = \frac{1}{2}$

So, $x=\frac{\pi}{6} , x=\frac{5\pi}{6}$ are Principal solutions.

$\sin x=0 \Leftrightarrow x=n \pi , \quad n\in \Z$

$\cos x=0 \Leftrightarrow x=\biggl(n+\frac{1}{2} \biggl) \pi , \quad n\in \Z$

1.If $\sin x = \sin y$ then

$x=n \pi +(-1)^ny, \quad$ for some $\ n \in \Z$

2.For any $n \in \Z,$

$\sin (n \pi + (-1)^ny) =\sin (y)$

$\sin (n \pi + (-1)^n \frac{\pi}{6}) = \sin \frac{\pi}{6} = \frac{1}{2}$

$\sin x = \sin \frac{\pi}{6}$

$\sin (n \pi +(-1)^ny) \\ (\because\sin(A+B) = \sin A \cos B+ \cos A \sin B)$

$=(\sin n \pi \cos(-1)^ny)+ \cos n\pi \sin(-1)^ny$

$= \cos n\pi \sin (-1)^ny$

$\biggl( n \ \text{even} \Rightarrow \cos n\pi = 1 , \\ \quad n \ \text{odd} \Rightarrow \cos n\pi = -1 \biggl)$

$(\cos n\pi = (-1)^n)$

$\Rightarrow \sin (n \pi +(-1)^ny)=(-1)^n \sin(-1)^ny$

$\biggl( n \ \text{even} \Rightarrow \sin y , \quad n \ \text{odd} \Rightarrow -\sin (-y)\biggl)$

$=\sin y$

$\Rightarrow \sin (n \pi +(-1)^ny) = \sin y$

$\sin x = \sin y$

$\Rightarrow \sin x - \sin y = 0$

$\Rightarrow 2 \cos \biggl(\frac{x+y}{2}\biggl) \sin \biggl(\frac{x-y}{2}\biggl) =0$

either $\quad$ $\cos \biggl( \frac{x+y}{2} \biggl) = 0$

$\frac{x+y}{2} = \biggl(m+\frac{1}{2} \biggl) \pi$ for some $\ m \in \Z$

$x+y = (2m+1) \pi$

$x = (2m+1) \pi -y$

$x= (2m+1) \pi + (-1)^{2m+1}y$

or, $\sin \biggl ( \frac{x-y}{2} \biggl ) = 0$

$\sin \theta = 0 , \theta =m \pi, m\in \Z)$

$\frac{x-y}{2} = m \pi, m \in \Z$

$x= 2m \pi + y$ for some $\ m \in \Z$

$x = 2m\pi +(-1)^{2m}y$

$x = n\pi + (-1)^ny$ for some $\in \Z$

Thank you