Class 9: Trigonometrical Ratios of Standard Angles

$\displaystyle \textbf{Trigonometrical Ratios of Angles }30^\circ\textbf{ and }60^\circ$

$\displaystyle \text{Consider an equilateral triangle }ABC\text{ with each side }2a.$
$\displaystyle \text{Draw }AD\perp BC.$
$\displaystyle \text{Since }AD\text{ bisects }BC,\quad BD=DC=a.$
$\displaystyle \text{In }\triangle ABD,\quad AD^2=AB^2-BD^2$
$\displaystyle =(2a)^2-a^2=3a^2$
$\displaystyle \therefore AD=\sqrt{3}a$
$\displaystyle \text{Since each angle of an equilateral triangle is }60^\circ,$
$\displaystyle \angle B=60^\circ$
$\displaystyle \angle BAD=180^\circ-(60^\circ+90^\circ)=30^\circ$
$\displaystyle \textbf{For }60^\circ\textbf{:}$
$\displaystyle \sin60^\circ=\frac{AD}{AB}=\frac{\sqrt{3}a}{2a}=\frac{\sqrt{3}}{2}$
$\displaystyle \cos60^\circ=\frac{BD}{AB}=\frac{a}{2a}=\frac{1}{2}$
$\displaystyle \tan60^\circ=\frac{AD}{BD}=\frac{\sqrt{3}a}{a}=\sqrt{3}$
$\displaystyle \textbf{For }30^\circ\textbf{:}$
$\displaystyle \sin30^\circ=\frac{BD}{AB}=\frac{a}{2a}=\frac{1}{2}$
$\displaystyle \cos30^\circ=\frac{AD}{AB}=\frac{\sqrt{3}a}{2a}=\frac{\sqrt{3}}{2}$
$\displaystyle \tan30^\circ=\frac{BD}{AD}=\frac{a}{\sqrt{3}a}=\frac{1}{\sqrt{3}}$
$\displaystyle \textbf{Standard Values:}$

$\displaystyle \sin30^\circ=\frac{1}{2},\qquad \cos30^\circ=\frac{\sqrt{3}}{2},\qquad \tan30^\circ=\frac{1}{\sqrt{3}}$
$\displaystyle \sin60^\circ=\frac{\sqrt{3}}{2},\qquad \cos60^\circ=\frac{1}{2},\qquad \tan60^\circ=\sqrt{3}$

$\displaystyle \textbf{Trigonometrical Ratios of Angle }45^\circ$

$\displaystyle \text{Consider a right-angled isosceles triangle }ABC$
$\displaystyle \text{in which }\angle B=90^\circ\text{ and }AB=BC=a.$
$\displaystyle \therefore \angle A=\angle C=45^\circ$
$\displaystyle \text{By Pythagoras theorem,}$
$\displaystyle AC^2=AB^2+BC^2$
$\displaystyle =a^2+a^2=2a^2$
$\displaystyle \therefore AC=\sqrt{2}a$
$\displaystyle \sin45^\circ=\frac{BC}{AC}=\frac{a}{\sqrt{2}a}=\frac{1}{\sqrt{2}}$
$\displaystyle \cos45^\circ=\frac{AB}{AC}=\frac{a}{\sqrt{2}a}=\frac{1}{\sqrt{2}}$
$\displaystyle \tan45^\circ=\frac{BC}{AB}=\frac{a}{a}=1$
$\displaystyle \textbf{Trigonometrical Ratios of }0^\circ\textbf{ and }90^\circ$

$\displaystyle \sin0^\circ=0,\qquad \cos0^\circ=1$
$\displaystyle \sin90^\circ=1,\qquad \cos90^\circ=0$
$\displaystyle \textbf{Trigonometrical Ratios of Standard Angles}$

$\displaystyle \begin{array}{|c|c|c|c|c|c|} \hline \text{Angle}\rightarrow & 0^\circ & 30^\circ & 45^\circ & 60^\circ & 90^\circ \\ \hline \sin & 0 & \frac{1}{2} & \frac{1}{\sqrt{2}} & \frac{\sqrt{3}}{2} & 1 \\ \hline \cos & 1 & \frac{\sqrt{3}}{2} & \frac{1}{\sqrt{2}} & \frac{1}{2} & 0 \\ \hline \tan & 0 & \frac{1}{\sqrt{3}} & 1 & \sqrt{3} & \infty\;(\text{not defined}) \\ \hline \cot & \infty\;(\text{not defined}) & \sqrt{3} & 1 & \frac{1}{\sqrt{3}} & 0 \\ \hline \sec & 1 & \frac{2}{\sqrt{3}} & \sqrt{2} & 2 & \infty\;(\text{not defined}) \\ \hline \mathrm{cosec} & \infty\;(\text{not defined}) & 2 & \sqrt{2} & \frac{2}{\sqrt{3}} & 1 \\ \hline \end{array}$