Basic Acute Angle

❀       The acute angle between the terminal side and the X - axis is called the basic acute angle.

❀       The basic acute angle is a positive acute angle.

❀       X-axis ႏွင့္ terminal side ၾကားတြင္ ရွိေသာ ေထာင့္က်ဥ္းကို basic acute angle ဟု ေခၚသည္။

❀       basic acute angle ကို အေပါင္းေထာင့္ အျဖစ္ အၿမဲသတ္မွတ္သည္။

❀       ေအာက္ပါဥပမာမ်ားကို ၾကည့္ပါ။

သတ္မွတ္ထားေသာ (ေပးထားေသာ) ေထာင့္ကို principal angle ဟုေခၚသည္။

Principal Angle ၏ trigonometric ratios မ်ား ရွာရန္

➤       principal angle ကို coordinate system (cartesian plane) တြင္ ေနရာခ်ပါ။

➤       သက္ဆိုင္ေသာ basic acute angle ကို ရွာပါ။

➤       basic acute angle ၏ sin ratio ႏွင့္ cos ratio ကို ရွာပါ။ (sin ႏွင့္ cos ကို သိလွ်င္ က်န္ေသာအခ်ိဳးမ်ားကို အလြယ္တကူ သိႏိုင္ပါသည္။)

➤       principal angle ႏွင့္ ၎ႏွင့္သက္ဆိုင္ေသာ basic acute angle တို႔၏ trigonometric ratio မ်ားသည္ ကိန္းဂဏန္းပမာဏ အားျဖင့္ (numerically) တူညီၾကပါသည္။

➤       သို႔ေသာ္ principal angle ၏ trigonometric ratio မ်ားအတြက္ သက္ဆိုင္ရာ quadrant ႏွင့္ ညိႇ၍ လကၡဏာ သတ္မွတ္ေပးရပါမည္။

Example : Find the six trigonometric ratios of 120°.

$ \displaystyle \ \ \ \sin 120{}^\circ =\sin 60{}^\circ \ \text{and}\ \cos 120{}^\circ =\cos 60{}^\circ \ \text{numerically}\text{.}$

$ \displaystyle \ \ \ \text{But }120{}^\circ \ \text{lies in the second quadrand}\text{.}$

$ \displaystyle \therefore \ \sin 120{}^\circ =\sin 60{}^\circ =\frac{{\sqrt{3}}}{2}$

$ \displaystyle \ \ \ \cos 120{}^\circ =-\cos 60{}^\circ =-\frac{1}{2}$

$ \displaystyle \therefore \ \tan 120{}^\circ =-\sqrt{3}$

$ \displaystyle \ \ \ \cot 120{}^\circ =-\frac{{\sqrt{3}}}{3}$

$ \displaystyle \ \ \ \sec 120{}^\circ =-2$

$ \displaystyle \ \ \ \operatorname{cosec}120{}^\circ =-\frac{{2\sqrt{3}}}{3}$

Trigonometric Ratios of ( – θ)

$ \displaystyle \ \ \ \text{In}\ \vartriangle PON,$

$ \displaystyle \ \ \ \sin \theta =y$

$ \displaystyle \ \ \ \cos \theta =x$

$ \displaystyle \ \ \ \tan \theta =\frac{y}{x}$

$ \displaystyle \ \ \ \cot \theta =\frac{x}{y}$

$ \displaystyle \ \ \ \sec \theta =\frac{1}{x}$

$ \displaystyle \ \ \ \operatorname{cosec}\theta =\frac{1}{y}$

$ \displaystyle \ \ \ \text{Since}\ \vartriangle {P}'O{N}'\cong \vartriangle PON,$

$ \displaystyle \ \ \ {y}'=y\ \text{and }{x}'=x\ \text{numerically}\text{.}$

$ \displaystyle \begin{array}{l}\ \ \ \text{But it is clear that, and }{y}'\ \text{have }y\ \text{opposite signs}\\\ \ \text{ }{x}'\ \text{and }x\ \text{have the same sign,}\end{array}$

$ \displaystyle \therefore {y}'=-y\ \text{and }{x}'=x.$

$ \displaystyle \therefore \sin (-\theta )={y}'=-y=-\sin \theta $

$ \displaystyle \ \ \ \cos (-\theta )={x}'=x=\cos \theta $

$ \displaystyle \ \ \ \tan (-\theta )=\frac{{{y}'}}{{{x}'}}=-\frac{y}{x}=-\tan \theta $

$ \displaystyle \ \ \ \cot (-\theta )=\frac{{{y}'}}{{{x}'}}=-\frac{x}{y}=-\cot \theta $

$ \displaystyle \ \ \ \sec (-\theta )=\frac{1}{{{x}'}}=\frac{1}{x}=\sec \theta $

$ \displaystyle \ \ \ \operatorname{cosec}(-\theta )=\frac{1}{{{y}'}}=-\frac{1}{y}=-\operatorname{cosec}\theta $

Slider ကို ေ႐ႊ႕ၾကည့္ပါ။
 

Trigonometric Ratios of (270° + θ)

$ \displaystyle \ \ \ \text{In}\ \vartriangle PON,$

$ \displaystyle \ \ \ \sin \theta =y$

$ \displaystyle \ \ \ \cos \theta =x$

$ \displaystyle \ \ \ \tan \theta =\frac{y}{x}$

$ \displaystyle \ \ \ \cot \theta =\frac{x}{y}$

$ \displaystyle \ \ \ \sec \theta =\frac{1}{x}$

$ \displaystyle \ \ \ \operatorname{cosec}\theta =\frac{1}{y}$

$ \displaystyle \ \ \ \text{Since}\ \vartriangle {P}'O{N}'\cong \vartriangle OPN,$

$ \displaystyle \ \ \ {y}'=x\ \text{and }{x}'=y\ \text{numerically}\text{.}$

$ \displaystyle \ \ \ \text{But }{P}'({x}',{y}')\ \text{lies in the fourth quadrant}\text{.}$

$ \displaystyle \therefore {y}'=-x\ \text{and }{x}'=y.$

$ \displaystyle \ \ \ \sin (270{}^\circ +\theta )={y}'=-x=-\cos \theta $

$ \displaystyle \ \ \ \cos (270{}^\circ +\theta )={x}'=y=\sin \theta $

$ \displaystyle \ \ \ \tan (270{}^\circ +\theta )=\frac{{{y}'}}{{{x}'}}=-\frac{{x}}{{y}}=\cot \theta $

$ \displaystyle \ \ \ \cot (270{}^\circ +\theta )=\frac{{{x}'}}{{{y}'}}=-\frac{{y}}{{x}}=\tan \theta $

$ \displaystyle \ \ \ \sec (270{}^\circ +\theta )=\frac{1}{{{x}'}}=\frac{1}{y}=\operatorname{cosec}\theta $

$ \displaystyle \ \ \ \operatorname{cosec}(270{}^\circ +\theta )=\frac{1}{{{y}'}}=-\frac{1}{x}=-\sec\theta $

θ တန္ဖိုး ရိုက္ထည့္ပါ။

Trigonometric Ratios of (270° - θ)

$ \displaystyle \ \ \ \text{In}\ \vartriangle PON,$

$ \displaystyle \ \ \ \sin \theta =y$

$ \displaystyle \ \ \ \cos \theta =x$

$ \displaystyle \ \ \ \tan \theta =\frac{y}{x}$

$ \displaystyle \ \ \ \cot \theta =\frac{x}{y}$

$ \displaystyle \ \ \ \sec \theta =\frac{1}{x}$

$ \displaystyle \ \ \ \operatorname{cosec}\theta =\frac{1}{y}$

$ \displaystyle \ \ \ \text{Since}\ \vartriangle {P}'O{N}'\cong \vartriangle OPN,$

$ \displaystyle \ \ \ {y}'=x\ \text{and }{x}'=y\ \text{numerically}\text{.}$

$ \displaystyle \ \ \ \text{But }{P}'({x}',{y}')\ \text{lies in the third quadrant}\text{.}$

$ \displaystyle \therefore {y}'=-x\ \text{and }{x}'=-y.$

$ \displaystyle \ \ \ \sin (270{}^\circ -\theta )={y}'=-x=-\cos \theta $

$ \displaystyle \ \ \ \cos (270{}^\circ -\theta )={x}'=-y=-\sin \theta $

$ \displaystyle \ \ \ \tan (270{}^\circ -\theta )=\frac{{{y}'}}{{{x}'}}=\frac{{-x}}{{-y}}=\cot \theta $

$ \displaystyle \ \ \ \cot (270{}^\circ -\theta )=\frac{{{x}'}}{{{y}'}}=\frac{{-y}}{{-x}}=\tan \theta $

$ \displaystyle \ \ \ \sec (270{}^\circ -\theta )=\frac{1}{{{x}'}}=-\frac{1}{y}=-\operatorname{cosec}\theta $

$ \displaystyle \ \ \ \operatorname{cosec}(270{}^\circ -\theta )=\frac{1}{{{y}'}}=-\frac{1}{x}=-\sec\theta $

$ \displaystyle \theta$ တန္ဖိုး ရိုက္ထည့္ပါ။

Trigonometric Ratios of (90° – θ)

$ \displaystyle \ \ \ \text{In}\ \vartriangle PON,$

$ \displaystyle \ \ \ \sin \theta =y$

$ \displaystyle \ \ \ \cos \theta =x$

$ \displaystyle \ \ \ \tan \theta =\frac{y}{x}$

$ \displaystyle \ \ \ \cot \theta =\frac{x}{y}$

$ \displaystyle \ \ \ \sec \theta =\frac{1}{x}$

$ \displaystyle \ \ \ \operatorname{cosec}\theta =\frac{1}{y}$


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

$ \displaystyle \ \ \ \cos (90{}^\circ -\theta )=y=\sin \theta $

$ \displaystyle \ \ \ \tan (90{}^\circ -\theta )=\frac{x}{y}=\cot \theta $

$ \displaystyle \ \ \ \cot (90{}^\circ -\theta )=\frac{y}{x}=\tan \theta $

$ \displaystyle \ \ \ \sec (90{}^\circ -\theta )=\frac{1}{y}=\operatorname{cosec}\theta $

$ \displaystyle \ \ \ \operatorname{cosec}(90{}^\circ -\theta )=\frac{1}{x}=\sec\theta $

$ \displaystyle \theta$ တန္ဖိုး ရိုက္ထည့္ပါ။

Trigonometric Ratios of (360° – θ)

$ \displaystyle \ \ \ \text{In}\ \vartriangle PON,$

$ \displaystyle \ \ \ \sin \theta =y$

$ \displaystyle \ \ \ \cos \theta =x$

$ \displaystyle \ \ \ \tan \theta =\frac{y}{x}$

$ \displaystyle \ \ \ \cot \theta =\frac{x}{y}$

$ \displaystyle \ \ \ \sec \theta =\frac{1}{x}$

$ \displaystyle \ \ \ \operatorname{cosec}\theta =\frac{1}{y}$

$ \displaystyle \ \ \ \text{Since}\ \vartriangle {P}'O{N}'\cong \vartriangle PON,$

$ \displaystyle \ \ \ {y}'=y\ \text{and }{x}'=x\ \text{numerically}\text{.}$

$ \displaystyle \ \ \ \text{But }{P}'({x}',{y}')\ \text{lies in the fourth quadrant}\text{.}$

$ \displaystyle \therefore {y}'=-y\ \text{and }{x}'=x.$

$ \displaystyle \therefore \sin (360{}^\circ -\theta )={y}'=-y=-\sin \theta $

$ \displaystyle \ \ \ \cos (360{}^\circ -\theta )={x}'=x=\cos \theta $

$ \displaystyle \ \ \ \tan (360{}^\circ -\theta )=\frac{{{y}'}}{{{x}'}}=-\frac{{y}}{{x}}=-\tan \theta $

$ \displaystyle \ \ \ \cot (360{}^\circ -\theta )=\frac{{{y}'}}{{{x}'}}=-\frac{{x}}{{y}}=-\cot \theta $

$ \displaystyle \ \ \ \sec (360{}^\circ -\theta )=\frac{1}{{{x}'}}=\frac{1}{x}=\sec \theta $

$ \displaystyle \ \ \ \operatorname{cosec}(360{}^\circ -\theta )=\frac{1}{{{y}'}}=-\frac{1}{y}=-\operatorname{cosec}\theta $

Dynamic Presentation

Trigonometric Ratios of (180° + θ)

$ \displaystyle \ \ \ \text{In}\ \vartriangle PON,$

$ \displaystyle \ \ \ \sin \theta =y$

$ \displaystyle \ \ \ \cos \theta =x$

$ \displaystyle \ \ \ \tan \theta =\frac{y}{x}$

$ \displaystyle \ \ \ \cot \theta =\frac{x}{y}$

$ \displaystyle \ \ \ \sec \theta =\frac{1}{x}$

$ \displaystyle \ \ \ \operatorname{cosec}\theta =\frac{1}{y}$

$ \displaystyle \ \ \ \text{Since}\ \vartriangle {P}'O{N}'\cong \vartriangle PON,$

$ \displaystyle \ \ \ {y}'=y\ \text{and }{x}'=x\ \text{numerically}\text{.}$

$ \displaystyle \ \ \ \text{But }{P}'({x}',{y}')\ \text{lies in the third quadrant}\text{.}$

$ \displaystyle \therefore {y}'=-y\ \text{and }{x}'=-x.$

$ \displaystyle \therefore \sin (180{}^\circ +\theta )={y}'=-y=-\sin \theta $

$ \displaystyle \ \ \ \cos (180{}^\circ +\theta )={x}'=-x=-\cos \theta $

$ \displaystyle \ \ \ \tan (180{}^\circ +\theta )=\frac{{{y}'}}{{{x}'}}=\frac{{-y}}{{-x}}=\tan \theta $

$ \displaystyle \ \ \ \cot (180{}^\circ +\theta )=\frac{{{y}'}}{{{x}'}}=\frac{{-x}}{{-y}}=\cot \theta $

$ \displaystyle \ \ \ \sec (180{}^\circ +\theta )=\frac{1}{{{x}'}}=-\frac{1}{x}=-\sec \theta $

$ \displaystyle \ \ \ \operatorname{cosec}(180{}^\circ +\theta )=\frac{1}{{{y}'}}=-\frac{1}{y}=-\operatorname{cosec}\theta $

Dynamic Presentation

Trigonometric Ratios of (90° + θ)

$ \displaystyle \ \ \ \text{In}\ \vartriangle PON,$

$ \displaystyle \ \ \ \sin \theta =y$

$ \displaystyle \ \ \ \cos \theta =x$

$ \displaystyle \ \ \ \tan \theta =\frac{y}{x}$

$ \displaystyle \ \ \ \cot \theta =\frac{x}{y}$

$ \displaystyle \ \ \ \sec \theta =\frac{1}{x}$

$ \displaystyle \ \ \ \operatorname{cosec}\theta =\frac{1}{y}$

$ \displaystyle \ \ \ \text{Since}\ \vartriangle {P}'O{N}'\cong \vartriangle OPN,$

$ \displaystyle \ \ \ {y}'=x\ \text{and }{x}'=y\ \text{numerically}\text{.}$

$ \displaystyle \ \ \ \text{But }{P}'({x}',{y}')\ \text{lies in the second quadrant}\text{.}$

$ \displaystyle \therefore {y}'=x\ \text{and }{x}'=-y.$

$ \displaystyle \ \ \ \sin (90{}^\circ +\theta )={y}'=x=\cos \theta $

$ \displaystyle \ \ \ \cos (90{}^\circ +\theta )={x}'=-y=-\sin \theta $

$ \displaystyle \ \ \ \tan (90{}^\circ +\theta )=\frac{{{y}'}}{{{x}'}}=-\frac{x}{y}=\cot \theta $

$ \displaystyle \ \ \ \cot (90{}^\circ +\theta )=\frac{{{x}'}}{{{y}'}}=-\frac{y}{x}=\tan \theta $

$ \displaystyle \ \ \ \sec (90{}^\circ +\theta )=\frac{1}{{{x}'}}=-\frac{1}{y}=-\operatorname{cosec}\theta $

$ \displaystyle \ \ \ \operatorname{cosec}(90{}^\circ +\theta )=\frac{1}{{{y}'}}=\frac{1}{x}=sec\theta $

 $ \displaystyle \theta$ တန္ဖိုး႐ိုက္ထည့္ၾကည့္ပါ။

Trigonometric Ratios of (180° – θ)

$ \displaystyle \begin{array}{l}\ \ \ \text{In}\ \vartriangle PON,\\\\\ \ \ \sin \theta =y\\\\\ \ \ \cos \theta =x\end{array}$

$ \displaystyle \ \ \ \tan \theta =\frac{y}{x}$

$ \displaystyle \ \ \ \cot \theta =\frac{x}{y}$

$ \displaystyle \ \ \ \sec \theta =\frac{1}{x}$

$ \displaystyle \ \ \ \operatorname{cosec}\theta =\frac{1}{y}$

$ \displaystyle \begin{array}{l}\ \ \ \text{Since}\ \vartriangle {P}'O{N}'\cong \vartriangle PON,\\\\\ \ \ {y}'=y\ \text{and }{x}'=x\ \text{numerically}\text{.}\\\\\ \ \ \text{But }{P}'({x}',{y}')\ \text{lies in the second quadrant}\text{.}\\\\\therefore {y}'=y\ \text{and }{x}'=-x.\\\\\therefore \sin (180{}^\circ -\theta )={y}'=y=\sin \theta \\\\\ \ \ \cos (180{}^\circ -\theta )={x}'=-x=-\cos \theta \end{array}$

$ \displaystyle \ \ \ \tan (180{}^\circ -\theta )=\frac{{{y}'}}{{{x}'}}=-\frac{y}{x}=-\tan \theta $

$ \displaystyle \ \ \ \cot (180{}^\circ -\theta )=\frac{{{y}'}}{{{x}'}}=-\frac{x}{y}=-\cot \theta $

$ \displaystyle \ \ \ \sec (180{}^\circ -\theta )=\frac{1}{{{x}'}}=-\frac{1}{x}=-\sec \theta $

$ \displaystyle \ \ \ \operatorname{cosec}(180{}^\circ -\theta )=\frac{1}{{{y}'}}=\frac{1}{y}=\operatorname{cosec}\theta $

Dynamic Presentation

Exercise (11.2) No.1 Solution

(a) $ \displaystyle \ \ \ \frac{{\tan \theta -1+\sec \theta }}{{\tan \theta +1-\sec \theta }}$

$ \displaystyle \text{=}\frac{{\tan \theta +\sec \theta -1}}{{\tan \theta -\sec \theta +1}}$

$ \displaystyle \text{=}\frac{{\left( {\tan \theta +\sec \theta } \right)-\left( {{{{\sec }}^{2}}\theta -{{{\tan }}^{2}}\theta } \right)}}{{\tan \theta -\sec \theta +1}}\ $
$ \displaystyle \ \ \ \left[ {\because {{{\tan }}^{2}}\theta +1={{{\sec }}^{2}}\theta } \right]$

$ \displaystyle \text{=}\frac{{\left( {\tan \theta +\sec \theta } \right)-\left( {\sec \theta +\tan \theta } \right)\left( {\sec \theta -\tan \theta } \right)}}{{\tan \theta +1-\sec \theta }}$

$ \displaystyle \text{=}\frac{{\left( {\tan \theta +\sec \theta } \right)\left( {1-\sec \theta +\tan \theta } \right)}}{{1-\sec \theta +\tan \theta }}$

$ \displaystyle =\tan \theta +\sec \theta $

$ \displaystyle =\frac{{\sin \theta }}{{\cos \theta }}+\frac{1}{{\cos \theta }}$

$ \displaystyle =\frac{{1+\sin \theta }}{{\cos \theta }}$


(b) $ \displaystyle \ \ \ \frac{{1-\cos x}}{{1+\cos x}}$

$ \displaystyle =\frac{{1-\cos x}}{{1+\cos x}}\times \frac{{1-\cos x}}{{1-\cos x}}$

$ \displaystyle =\frac{{{{{\left( {1-\cos x} \right)}}^{2}}}}{{1-{{{\cos }}^{2}}x}}$

$ \displaystyle =\frac{{1-2\cos x+{{{\cos }}^{2}}x}}{{{{{\sin }}^{2}}x}}$

$ \displaystyle =\frac{1}{{{{{\sin }}^{2}}x}}-\frac{2}{{\sin x}}\cdot \frac{{\cos x}}{{\sin x}}+\frac{{{{{\cos }}^{2}}x}}{{{{{\sin }}^{2}}x}}$

$ \displaystyle ={{\operatorname{cosec}}^{2}}x-2\operatorname{cosec}x\cot x+{{\cot }^{2}}x$

$ \displaystyle ={{\operatorname{cosec}}^{2}}x-2\operatorname{cosec}x\cot x+{{\operatorname{cosec}}^{2}}x-1\ $
$ \displaystyle \ \ \ \left[ {\because 1+{{{\cot }}^{2}}x={{{\operatorname{cosec}}}^{2}}x} \right]$

$ \displaystyle =2{{\operatorname{cosec}}^{2}}x-2\operatorname{cosec}x\cot x-1$


(c)$ \displaystyle \ \ \ (1+\tan x-\sec x)(1+\cot x+\operatorname{cosec}x)$

$ \displaystyle =\ \left( {1+\frac{{\sin x}}{{\cos x}}-\frac{1}{{\cos x}}} \right)\left( {1+\frac{{\cos x}}{{\sin x}}+\frac{1}{{\sin x}}} \right)$

$ \displaystyle =\frac{{\cos x+\sin x-1}}{{\cos x}}\ \times \frac{{\sin x+\cos x+1}}{{\sin x}}$

$ \displaystyle =\frac{{{{{(\cos x+\sin x)}}^{2}}-1}}{{\sin x\cos x}}$

$ \displaystyle =\frac{{{{{\cos }}^{2}}x+2\sin x\cos x+{{{\sin }}^{2}}x-1}}{{\sin x\cos x}}$

$ \displaystyle =\frac{{{{{\cos }}^{2}}x+{{{\sin }}^{2}}x+2\sin x\cos x-1}}{{\sin x\cos x}}$

$ \displaystyle =\frac{{1+2\sin x\cos x-1}}{{\sin x\cos x}}$

$ \displaystyle =2$