Trigonometry - Simplification of Trigonometric Expressions

Without using tables, find $\;$ $\tan \left(\alpha - \dfrac{\pi}{4}\right)$, if $\cos \alpha = \dfrac{-9}{41}$, $\pi < \alpha < \dfrac{3 \pi}{2}$

Given: $\;$ $\pi < \alpha < \dfrac{3 \pi}{2}$

$\implies$ $\alpha$ lies in the third quadrant.

$\implies$ Both $\sin \alpha$, $\cos \alpha$ are negative and $\tan \alpha$ is positive.

Given: $\;$ $\cos \alpha = \dfrac{-9}{41}$

$\therefore \;$ $\sin \alpha = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - \left(\dfrac{-9}{41}\right)^2} = \dfrac{-40}{41}$

$\therefore \;$ $\tan \alpha = \dfrac{\sin \alpha}{\cos \alpha} = \dfrac{-40 / 41}{-9 / 41} = \dfrac{40}{9}$

Now, $\;$ $\tan \left(\alpha - \dfrac{\pi}{4}\right) = \dfrac{\tan \alpha - \tan \dfrac{\pi}{4}}{1 - \tan \alpha \tan \dfrac{\pi}{4}} = \dfrac{\dfrac{40}{9} - 1}{1 + \dfrac{40}{9}} = \dfrac{31}{49}$

Trigonometry - Simplification of Trigonometric Expressions

Without using tables, calculate $\;$ $\sin 2 \alpha$, $\;$ if $\;$ $\sin \left(\dfrac{\alpha}{2}\right) + \cos \left(\dfrac{\alpha}{2}\right) = \dfrac{-1}{2}$ $\;$ and $\alpha$ belongs to the fourth quadrant.

$\because \;$ $\alpha$ is in the fourth quadrant $\implies$ $\sin \alpha$ is negative and $\cos \alpha$ is positive

Given: $\;$ $\sin \left(\dfrac{\alpha}{2}\right) + \cos \left(\dfrac{\alpha}{2}\right) = \dfrac{-1}{2}$

i.e. $\;$ $\left[\sin \left(\dfrac{\alpha}{2}\right) + \cos \left(\dfrac{\alpha}{2}\right)\right]^2 = \left(\dfrac{-1}{2}\right)^2$

i.e. $\;$ $\sin^2 \left(\dfrac{\alpha}{2}\right) + \cos^2 \left(\dfrac{\alpha}{2}\right) + 2 \sin \left(\dfrac{\alpha}{2}\right) \cos \left(\dfrac{\alpha}{2}\right) = \dfrac{1}{4}$

i.e. $\;$ $1 + \sin \left(2 \times \dfrac{\alpha}{2}\right) = \dfrac{1}{4}$

i.e. $\;$ $\sin \alpha = \dfrac{1}{4} - 1 = \dfrac{-3}{4}$

$\therefore \;$ $\cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - \left(\dfrac{-3}{4}\right)^2} = \dfrac{\pm\sqrt{7}}{4}$

$\because \;$ $\alpha$ is in the fourth quadrant, $\cos \alpha = \dfrac{\sqrt{7}}{4}$

Now, $\;$ $\sin 2 \alpha = 2 \sin \alpha \cos \alpha = 2 \times \left(\dfrac{-3}{4}\right) \times \dfrac{\sqrt{7}}{4} = \dfrac{-3 \sqrt{7}}{8}$

Trigonometry - Simplification of Trigonometric Expressions

Without using tables, find: $\;\;\;$ $\cot \left(\dfrac{\alpha}{2}\right)$, $\;$ given $\;$ $20 \sin^2 \alpha + 21 \cos \alpha - 24 = 0$, $\;$ $\dfrac{7 \pi}{4} < \alpha < 2 \pi$

Given $\;$ $20 \sin^2 \alpha + 21 \cos \alpha - 24 = 0$

i.e. $\;$ $20 \left(1 - \cos^2 \alpha\right) + 21 \cos \alpha - 24 = 0$

i.e. $\;$ $- 20 \cos^2 \alpha + 21 \cos \alpha - 4 = 0$

i.e. $\;$ $20 \cos^2 \alpha - 21 \cos \alpha + 4 = 0$

i.e. $\;$ $20 \cos^2 \alpha - 16 \cos \alpha - 5 \cos \alpha + 4 = 0$

i.e. $\;$ $5 \cos \alpha \left(4 \cos \alpha - 1\right) - 4 \left(4 \cos \alpha - 1\right) = 0$

i.e. $\;$ $\left(5 \cos \alpha - 4\right) \left(4 \cos \alpha - 1\right) = 0$

i.e. $\;$ $5 \cos \alpha - 4 = 0$ $\;$ or $\;$ $4 \cos \alpha - 1 = 0$

i.e. $\;$ $\cos \alpha = \dfrac{4}{5}$ $\;$ or $\;$ $\cos \alpha = \dfrac{1}{4}$

Since $\;$ $\dfrac{7 \pi}{4} < \alpha < 2 \pi$

$\implies$ $\alpha$ is in the fourth quadrant.

$\implies$ $\cos \alpha$ is positive and $\sin \alpha$ is negative.

Now, when $\;$ $\cos \alpha = \dfrac{4}{5}$, $\;$ $\sin \alpha = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - \dfrac{16}{25}} = \pm \dfrac{3}{5}$

When $\;$ $\cos \alpha = \dfrac{1}{4}$, $\;$ $\sin \alpha = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - \dfrac{1}{16}} = \pm \dfrac{\sqrt{15}}{4}$

$\because \;$ $\alpha$ is in the fourth quadrant, $\;$ $\sin \alpha = \dfrac{-3}{5}$ $\;$ or $\;$ $\sin \alpha = \dfrac{- \sqrt{15}}{4}$

Further, since $\;$ $\dfrac{7 \pi}{4} < \alpha < 2 \pi$

$\implies$ $\dfrac{7 \pi}{8} < \dfrac{\alpha}{2} < \pi$

$\implies$ $\dfrac{\alpha}{2}$ lies in the second quadrant.

$\implies$ $\sin \left(\dfrac{\alpha}{2}\right)$ is positive and $\cos \left(\dfrac{\alpha}{2}\right)$ is negative.

Now, $\;$ $\cos \alpha = 2 \cos^2 \left(\dfrac{\alpha}{2}\right) - 1$

$\implies$ $\cos^2 \left(\dfrac{\alpha}{2}\right) = \dfrac{1 + \cos \alpha}{2} = \dfrac{1 + \dfrac{4}{5}}{2} = \dfrac{9}{10}$ $\;$ or $\;$ $\cos^2 \left(\dfrac{\alpha}{2}\right) = \dfrac{1 + \dfrac{1}{4}}{2} = \dfrac{5}{8}$

$\implies$ $\cos \left(\dfrac{\alpha}{2}\right) = \dfrac{\pm 3}{\sqrt{10}}$ $\;$ or $\;$ $\cos \left(\dfrac{\alpha}{2}\right) = \dfrac{\pm \sqrt{5}}{2 \sqrt{2}}$

$\because \;$ $\dfrac{\alpha}{2}$ is in the second quadrant, $\;$ $\cos \left(\dfrac{\alpha}{2}\right) = \dfrac{-3}{\sqrt{10}}$ $\;$ or $\;$ $\cos \left(\dfrac{\alpha}{2}\right) = \dfrac{- \sqrt{5}}{2 \sqrt{2}}$

Now, $\;$ $\sin \alpha = 2 \sin \left(\dfrac{\alpha}{2}\right) \cos \left(\dfrac{\alpha}{2}\right)$ $\implies$ $\sin \left(\dfrac{\alpha}{2}\right) = \dfrac{\sin \alpha}{2 \cos \left(\dfrac{\alpha}{2}\right)}$

When $\;$ $\sin \alpha = \dfrac{-3}{5}$, $\cos \left(\dfrac{\alpha}{2}\right) = \dfrac{-3}{\sqrt{10}}$, $\;$ $\sin \left(\dfrac{\alpha}{2}\right) = \dfrac{\dfrac{-3}{5}}{-2 \times \dfrac{3}{\sqrt{10}}} = \dfrac{1}{\sqrt{10}}$

When $\;$ $\sin \alpha = \dfrac{-\sqrt{15}}{4}$, $\cos \left(\dfrac{\alpha}{2}\right) = \dfrac{-5}{2 \sqrt{2}}$, $\;$ $\sin \left(\dfrac{\alpha}{2}\right) = \dfrac{\dfrac{-\sqrt{15}}{4}}{-2 \times \dfrac{\sqrt{5}}{2\sqrt{2}}} = \dfrac{3}{2\sqrt{6}}$

Now, $\;$ $\cot \left(\dfrac{\alpha}{2}\right) = \dfrac{\cos \left(\dfrac{\alpha}{2}\right)}{\sin \left(\dfrac{\alpha}{2}\right)}$

When $\;$ $\cos \left(\dfrac{\alpha}{2}\right) = \dfrac{-3}{\sqrt{10}}$, $\;$ $\sin \left(\dfrac{\alpha}{2}\right) = \dfrac{1}{\sqrt{10}}$, $\;$ $\cot \left(\dfrac{\alpha}{2}\right) = \dfrac{\dfrac{-3}{\sqrt{10}}}{\dfrac{1}{\sqrt{10}}} = -3$

When $\;$ $\cos \left(\dfrac{\alpha}{2}\right) = \dfrac{-\sqrt{5}}{2\sqrt{2}}$, $\;$ $\sin \left(\dfrac{\alpha}{2}\right) = \dfrac{3}{2\sqrt{6}}$, $\;$ $\cot \left(\dfrac{\alpha}{2}\right) = \dfrac{\dfrac{- \sqrt{5}}{2 \sqrt{2}}}{\dfrac{3}{2 \sqrt{6}}} = \dfrac{-\sqrt{15}}{3}$

Trigonometry - Simplification of Trigonometric Expressions

Calculate, without using tables: $\;\;\;$ $\left(\sin 4 \alpha + 2 \sin 2 \alpha\right) \cos \alpha$ $\;\;$ if $\;\;$ $\sin \alpha = \dfrac{1}{4}$, $\;$ $\alpha$ is in the first quadrant.

Given $\;$ $\alpha$ $\;$ is in the first quadrant $\implies$ both $\sin \alpha$, $\cos \alpha \;$ are positive.

Given: $\;$ $\sin \alpha = \dfrac{1}{4}$

$\therefore \;$ $\cos \alpha = \sqrt{1 - \sin^2 \alpha}= \sqrt{1 - \dfrac{1}{16}} = \dfrac{\sqrt{15}}{4}$

Now, $\;$ $\sin 2 \alpha = 2 \sin \alpha \cos \alpha = 2 \times \dfrac{1}{4} \times \dfrac{\sqrt{15}}{4} = \dfrac{\sqrt{15}}{8}$

$\cos 2 \alpha = \cos^2 \alpha - \sin^2 \alpha = \dfrac{15}{16} - \dfrac{1}{16} = \dfrac{14}{16} = \dfrac{7}{8}$

and, $\;$ $\sin 4 \alpha = \sin \left[2 \left(2 \alpha\right)\right] = 2 \sin \left(2 \alpha\right) \cos \left(2 \alpha\right) = 2 \times \dfrac{\sqrt{15}}{8} \times \dfrac{7}{8} = \dfrac{7 \sqrt{15}}{32}$

Given expression: $\;$ $\left(\sin 4 \alpha + 2 \sin 2 \alpha\right) \cos \alpha$

$= \left(\dfrac{7 \sqrt{15}}{32} + 2 \times \dfrac{\sqrt{15}}{8}\right) \times \dfrac{\sqrt{15}}{4}$

$= \dfrac{15 \sqrt{15}}{32} \times \dfrac{\sqrt{15}}{4} = \dfrac{225}{128}$

Trigonometry - Simplification of Trigonometric Expressions

Verify the equality without using tables: $\;\;\;$ $\cos 20^\circ + 2 \sin^2 55^\circ = 1 + \sqrt{2} \sin 65^\circ$

To Prove That (TPT) $\;\;\;$ $\cos 20^\circ + 2 \sin^2 55^\circ = 1 + \sqrt{2} \sin 65^\circ$

i.e. $\;$ TPT $\;\;\;$ $\cos 20^\circ + 2 \sin^2 55^\circ - \sqrt{2} \sin 65^\circ = 1$

$\begin{aligned} LHS & = \cos 20^\circ + 1 - \cos 110^\circ - \sqrt{2} \sin 65^\circ \;\;\; \left(\because \; 2 \sin^2 \theta = 1 - \cos 2 \theta\right) \\\\ & = \cos 20^\circ + 1 - \cos \left(90^\circ + 20^\circ\right) - \sqrt{2} \sin 65^\circ \\\\ & = \cos 20^\circ + 1 - \left(- \sin 20^\circ\right) - \sqrt{2} \sin 65^\circ \\\\ & = \left(\cos 20^\circ + \sin 20^\circ\right) + 1 - \sqrt{2} \sin 65^\circ \\\\ & = \sqrt{2} \left(\dfrac{1}{\sqrt{2}} \sin 20^\circ + \dfrac{1}{\sqrt{2}} \cos 20^\circ\right) - \sqrt{2} \sin 65^\circ + 1 \\\\ & = \sqrt{2} \left(\sin 20^\circ \cos 45^\circ + \cos 20^\circ \sin 45^\circ\right) - \sqrt{2} \sin 65^\circ + 1 \\\\ & = \sqrt{2} \sin \left(20^\circ + 45^\circ\right) - \sqrt{2} \sin 65^\circ + 1 \\\\ & = \sqrt{2} \sin 65^\circ - \sqrt{2} \sin 65^\circ + 1 \\\\ & = 1 = RHS \end{aligned}$

Hence verified.

Trigonometry - Simplification of Trigonometric Expressions

Verify the equality without using tables: $\;\;\;$ $\dfrac{1}{\cos 290^\circ} + \dfrac{1}{\sqrt{3} \sin 250^\circ} = \dfrac{4}{\sqrt{3}}$

$\cos 290^\circ = \cos \left(270^\circ + 20^\circ\right) = \sin 20^\circ$

$\sin 250^\circ = \sin \left(270^\circ - 20^\circ\right) = - \cos 20^\circ$

$\begin{aligned} LHS & = \dfrac{1}{\cos 290^\circ} + \dfrac{1}{\sqrt{3} \sin 250^\circ} \\\\ & = \dfrac{1}{\sin 20^\circ} - \dfrac{1}{\sqrt{3} \cos 20^\circ} \\\\ & = \dfrac{\sqrt{3} \cos 20^\circ - \sin 20^\circ}{\sqrt{3} \sin 20^\circ \cos 20^\circ} \\\\ & = \dfrac{2 \left(\dfrac{\sqrt{3}}{2} \cos 20^\circ - \dfrac{1}{2} \sin 20^\circ\right)}{\dfrac{\sqrt{3}}{2} \times 2 \sin 20^\circ \cos 20^\circ} \\\\ & = \dfrac{4 \left(\sin 60^\circ \cos 20^\circ - \cos 60^\circ \sin 20^\circ\right)}{\sqrt{3} \sin 40^\circ} \\\\ & = \dfrac{4 \sin \left(60^\circ - 20^\circ\right)}{\sqrt{3} \sin 40^\circ} \\\\ & = \dfrac{4 \sin 40^\circ}{\sqrt{3} \sin 40^\circ} \\\\ & = \dfrac{4}{\sqrt{3}} = RHS \end{aligned}$

Hence verified.

Trigonometry - Simplification of Trigonometric Expressions

Calculate without using tables: $\;\;\;$ $\dfrac{\sin 110^\circ \sin 250^\circ + \cos 540^\circ \cos 290^\circ \cos 430^\circ}{\cos^2 1260^\circ}$

$\sin 110^\circ = \sin \left(90^\circ + 20^\circ\right) = \cos 20^\circ$

$\sin 250^\circ = \sin \left(270^\circ - 20^\circ\right) = - \cos 20^\circ$

$\cos 540^\circ = \cos \left(2 \times 180^\circ\right) = -1$

$\cos 290^\circ = \cos \left(270^\circ + 20^\circ\right) = \sin 20^\circ$

$\cos 430^\circ = \cos \left(450^\circ - 20^\circ\right) = \sin 20^\circ$

$\cos 1260^\circ = \cos \left(8 \pi - 180^\circ\right) = \cos 180^\circ = -1$

The given expression is: $\;\;$ $\dfrac{\sin 110^\circ \sin 250^\circ + \cos 540^\circ \cos 290^\circ \cos 430^\circ}{\cos^2 1260^\circ}$

$= \dfrac{\cos 20^\circ \times \left(- \cos 20^\circ\right) + \left(-1\right) \times \sin 20^\circ \times \sin 20^\circ}{\left(-1\right)^2}$

$= \dfrac{- \cos^2 20^\circ - \sin^2 20^\circ}{1}$

$= - \left(\cos^2 20^\circ + \sin^2 20^\circ\right) = -1$

Trigonometry - Simplification of Trigonometric Expressions

Calculate without using tables: $\;\;\;$ $\tan 9^\circ - \tan 63^\circ + \tan 81^\circ - \tan 27^\circ$

The given expression is: $\;\;$ $\tan 9^\circ - \tan 63^\circ + \tan 81^\circ - \tan 27^\circ$

$\tan 9^\circ - \tan \left(90^\circ - 27^\circ\right) + \tan \left(90^\circ - 9^\circ\right) - \tan 27^\circ$

$= \dfrac{\sin 9^\circ}{\cos 9^\circ} - \dfrac{\cos 27^\circ}{\sin 27^\circ} + \dfrac{\cos 9^\circ}{\sin 9^\circ} - \dfrac{\sin 27^\circ}{\cos 27^\circ}$

$= \dfrac{\sin^2 9^\circ + \cos^2 9^\circ}{\sin 9^\circ \cos 9^\circ} - \dfrac{\cos^2 27^\circ + \sin^2 27^\circ}{\sin 27^\circ \cos 27^\circ}$

$= \dfrac{1}{\sin 9^\circ \cos 9^\circ} - \dfrac{1}{\sin 27^\circ \cos 27^\circ}$

$= \dfrac{2}{2 \sin 9^\circ \cos 9^\circ} - \dfrac{2}{2 \sin 27^\circ \cos 27^\circ}$

$= \dfrac{2}{\sin \left(2 \times 9^\circ\right)} - \dfrac{2}{\sin \left(2 \times 27^\circ\right)}$

$= 2 \left(\dfrac{1}{\sin 18^\circ} - \dfrac{1}{\sin 54^\circ}\right)$

$= 2 \left(\dfrac{\sin 54^\circ - \sin 18^\circ}{\sin 54^\circ \sin 18^\circ}\right)$

$= \dfrac{2 \times 2 \sin \left(\dfrac{54^\circ - 18^\circ}{2}\right) \cos \left(\dfrac{54^\circ + 18^\circ}{2}\right)}{\sin 54^\circ \sin 18^\circ}$

$= \dfrac{4 \sin 18^\circ \cos 36^\circ}{\sin 54^\circ \sin 18^\circ}$

$= \dfrac{4 \cos 36^\circ}{\sin 54^\circ}$

$= \dfrac{4 \cos \left(90^\circ - 54^\circ\right)}{\sin 54^\circ}$

$= \dfrac{4 \sin 54^\circ}{\sin 54^\circ} = 4$

Trigonometry - Simplification of Trigonometric Expressions

Calculate without using tables:
$\cos 10^\circ \cos 50^\circ \cos 70^\circ$

The given expression is: $\;\;$ $\cos 10^\circ \cos 50^\circ \cos 70^\circ$

$= \dfrac{1}{2} \times \left[2 \cos 10^\circ \cos 50^\circ\right] \cos 70^\circ$

$= \dfrac{1}{2} \left[\cos \left(10^\circ - 50^\circ\right) + \cos \left(10^\circ + 50^\circ\right)\right] \cos 70^\circ$

$= \dfrac{1}{2} \left[\cos \left(-40^\circ\right) + \cos 60^\circ\right] \cos 70^\circ$

$= \dfrac{1}{2} \cos 70^\circ \cos 40^\circ + \dfrac{1}{2} \cos 60^\circ \cos 70^\circ$

$= \dfrac{1}{4} \left[2 \cos 70^\circ \cos 40^\circ\right] + \dfrac{1}{2} \times \dfrac{1}{2} \times \cos 70^\circ$

$= \dfrac{1}{4} \left[\cos \left(70^\circ - 40^\circ\right) + \cos \left(70^\circ + 40^\circ\right)\right] + \dfrac{1}{4} \times \cos 70^\circ$

$= \dfrac{1}{4} \left[\cos 30^\circ + \cos 110^\circ\right] + \dfrac{1}{4} \cos 70^\circ$

$= \dfrac{1}{4} \times \dfrac{\sqrt{3}}{2} + \dfrac{1}{4} \times \cos 110^\circ + \dfrac{1}{4} \cos 70^\circ$

$= \dfrac{\sqrt{3}}{8} + \dfrac{1}{4} \left[\cos 110^\circ + \cos 70^\circ\right]$

$= \dfrac{\sqrt{3}}{8} + \dfrac{1}{4} \left[\cos \left(180^\circ - 70^\circ\right) + \cos 70^\circ\right]$

$= \dfrac{\sqrt{3}}{8} + \dfrac{1}{4} \left[\cos \left(- 70^\circ\right) + \cos 70^\circ\right]$

$= \dfrac{\sqrt{3}}{8} + \dfrac{1}{4} \left[- \cos 70^\circ + \cos 70 ^\circ\right]$

$= \dfrac{\sqrt{3}}{8}$

Trigonometry - Simplification of Trigonometric Expressions

Calculate without using tables:
$\tan 20^\circ \tan 40^\circ \tan 80^\circ$

The given expression is: $\;\;$ $\tan 20^\circ \tan 40^\circ \tan 80^\circ$

$= \dfrac{\sin 20^\circ \sin 40^\circ \sin 80^\circ}{\cos 20^\circ \cos 40^\circ \cos 80^\circ}$

$= \dfrac{\sin 20^\circ \times \dfrac{1}{2} \left[2 \sin 80^\circ \sin 40^\circ\right]}{\cos 20^\circ \times \dfrac{1}{2} \left[2 \cos 80^\circ \cos 40^\circ\right]}$

$= \dfrac{\sin 20^\circ \times \dfrac{1}{2} \left[\cos \left(80^\circ - 40^\circ\right) - \cos \left(80^\circ + 40^\circ\right)\right]}{\cos 20^\circ \times \dfrac{1}{2} \left[\cos \left(80^\circ - 40^\circ\right) + \cos \left(80^\circ + 40^\circ\right)\right]}$

$= \dfrac{\sin 20^\circ \left[\cos 40^\circ - \cos 120^\circ\right]}{\cos 20^\circ \left[\cos 40^\circ + \cos 120^\circ\right]}$

$= \dfrac{\sin 20^\circ \left[\cos 40^\circ - \left(\dfrac{-1}{2}\right)\right]}{\cos 20^\circ \left[\cos 40^\circ + \left(\dfrac{-1}{2}\right)\right]}$

$= \dfrac{2 \sin 20^\circ \cos 40^\circ + \sin 20^\circ}{2 \cos 20^\circ \cos 40^\circ - \cos 20^\circ}$

$= \dfrac{\sin \left(40^\circ + 20^\circ\right) + \sin \left(20^\circ - 40^\circ\right) + \sin 20^\circ}{\cos \left(40^\circ - 20^\circ\right) + \cos \left(40^\circ + 20^\circ\right) - \cos 20^\circ}$

$= \dfrac{\sin 60^\circ + \sin \left(-20^\circ\right) + \sin 20^\circ}{\sin 20^\circ + \cos 60^\circ - \cos 20^\circ}$

$= \dfrac{\sin 60^\circ - \sin 20^\circ + \sin 20^\circ}{\cos 60^\circ}$

$= \dfrac{\sin 60^\circ}{\cos 60^\circ}$

$= \tan 60^\circ = \sqrt{3}$

Trigonometry - Simplification of Trigonometric Expressions

Calculate without using tables:
$128 \sin^2 20^\circ \sin^2 40^\circ \sin^2 60^\circ \sin^2 80^\circ$

The given expression is: $\;\;$ $128 \sin^2 20^\circ \sin^2 40^\circ \sin^2 60^\circ \sin^2 80^\circ$ $\;\;\; \cdots \; (1)$

Consider the expression: $\;\;$ $\sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ$

$= \sin 20^\circ \times \left(\sin 80^\circ \sin 40^\circ\right) \times \dfrac{\sqrt{3}}{2}$

$= \dfrac{\sqrt{3}}{2} \times \sin 20^\circ \times \dfrac{1}{2} \left[\cos \left(80^\circ - 40^\circ\right) - \cos \left(80^\circ + 40^\circ\right)\right]$

$= \dfrac{\sqrt{3}}{4} \times \sin 20^\circ \left[\cos 40^\circ - \cos 120^\circ\right]$

$= \dfrac{\sqrt{3}}{4} \times \sin 20^\circ \left[\cos 40^\circ - \left(\dfrac{-1}{2}\right)\right]$

$= \dfrac{\sqrt{3}}{8} \left[2 \sin 20^\circ \cos 40^\circ + \sin 20^\circ\right]$

$= \dfrac{\sqrt{3}}{8} \left[\sin \left(20^\circ + 40^\circ\right) + \sin \left(20^\circ - 40^\circ\right) + \sin 20^\circ\right]$

$= \dfrac{\sqrt{3}}{8} \left[\sin 60^\circ + \sin \left(-20^\circ\right) + \sin 20^\circ\right]$

$= \dfrac{\sqrt{3}}{8} \left[\dfrac{\sqrt{3}}{2} - \sin 20^\circ + \sin 20^\circ\right]$

$\therefore$ $\sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ = \dfrac{3}{16}$ $\;\;\; \cdots \; (2)$

In view of expression $(2)$, expression $(1)$ becomes

$128 \times \left(\dfrac{3}{16}\right)^2 = \dfrac{9}{2}$

Trigonometry - Simplification of Trigonometric Expressions

Calculate without using tables:
$\dfrac{96 \sin 80^\circ \sin 65^\circ \sin 35^\circ}{\sin 20^\circ + \sin 50^\circ + \sin 110^\circ}$

The given expression is: $\;\;$ $\dfrac{96 \sin 80^\circ \sin 65^\circ \sin 35^\circ}{\sin 20^\circ + \sin 50^\circ + \sin 110^\circ}$ $\;\;\; \cdots \; (1)$

Denominator of the given expression is

$\sin 20^\circ + \sin 50^\circ + \sin 110^\circ$

$= \sin \left(2 \times 10^\circ\right) + \left(\sin 50^\circ + \sin 110^\circ\right)$

$= 2 \sin 10^\circ \cos 10^\circ + 2 \sin \left(\dfrac{110^\circ + 50^\circ}{2}\right) \cos \left(\dfrac{110^\circ - 50^\circ}{2}\right)$

$= 2 \sin 10^\circ \cos 10^\circ + 2 \sin 80^\circ \cos 30^\circ$

$= 2 \sin 10^\circ \cos 10^\circ + 2 \sin \left(90^\circ - 10^\circ\right) \cos \left(90^\circ - 60^\circ\right)$

$= 2 \sin 10^\circ \cos 10^\circ + 2 \cos 10^\circ \sin 60^\circ$

$= 2 \cos 10^\circ \left(\sin 60^\circ + \sin 10^\circ\right)$

$= 2 \cos \left(90^\circ - 80^\circ\right) \times 2 \sin \left(\dfrac{60^\circ + 10^\circ}{2}\right) \cos \left(\dfrac{60^\circ - 10^\circ}{2}\right)$

$= 4 \sin 80^\circ \sin 35^\circ \cos 25^\circ$

$= 4 \sin 80^\circ \sin 35^\circ \cos \left(90^\circ - 65^\circ\right)$

$= 4 \sin 80^\circ \sin 35^\circ \sin 65^\circ$ $\;\;\; \cdots \; (2)$

In view of $(2)$, the given expression $(1)$ becomes

$\dfrac{96 \sin 80^\circ \sin 65^\circ \sin 35^\circ}{4 \sin 80^\circ \sin 35^\circ \sin 65^\circ}$

$= \dfrac{96}{4} = 24$

Trigonometry - Simplification of Trigonometric Expressions

Calculate without using tables:
$4 \left(\cos 24^\circ + \cos 48^\circ - \cos 84^\circ - \cos 12^\circ\right)$

$4 \left(\cos 24^\circ + \cos 48^\circ - \cos 84^\circ - \cos 12^\circ\right)$

$= 4 \left[\left(\cos 24^\circ - \cos 84^\circ\right) + \left(\cos 48^\circ - \cos 12^\circ\right)\right]$

$= 4 \left[-2 \sin \left(\dfrac{24^\circ - 84^\circ}{2}\right) \sin \left(\dfrac{24^\circ + 84^\circ}{2}\right) - 2 \sin \left(\dfrac{48^\circ - 12^\circ}{2}\right) \sin \left(\dfrac{48^\circ + 12^\circ}{2}\right)\right]$

$= 4 \times \left(-2\right) \left[\sin \left(-30^\circ\right) \sin 54^\circ + \sin 18^\circ \sin 30^\circ\right]$

$= -8 \left[\dfrac{-1}{2} \times \sin 54^\circ + \dfrac{1}{2} \times \sin 18^\circ\right]$

$= -8 \times \left(\dfrac{-1}{2}\right) \left[\sin 54^\circ - \sin 18^\circ\right]$

$= 4 \left[\sin 54^\circ - \sin 18^\circ\right]$

$= 4 \times 2 \times \left(\dfrac{54^\circ - 18^\circ}{2}\right) \cos \left(\dfrac{54^\circ + 18^\circ}{2}\right)$

$= 4 \times 2 \times \sin 18^\circ \cos 36^\circ$ $\;\;\; \cdots \; (1)$

Now, $\;$ $\sin 36^\circ = \sin \left(2 \times 18^\circ\right) = 2 \sin 18^\circ \cos 18^\circ$

$\implies$ $\sin 18^\circ = \dfrac{\sin 36^\circ}{2 \cos 18^\circ}$ $\;\;\; \cdots \; (2)$

In view of $(2)$, expression $(1)$ becomes

$4 \times 2 \cos 36^\circ \times \dfrac{\sin 36^\circ}{2 \cos 18^\circ}$

$= \dfrac{2 \sin \left(2 \times 36^\circ\right)}{\cos 18^\circ}$

$= \dfrac{2 \sin 72^\circ}{\cos \left(90^\circ - 72^\circ\right)}$

$= \dfrac{2 \sin 72^\circ}{\sin 72^\circ}$

$= 2$