Graph of Trigonometric Functions – Domain & Range

Here, you will learn graph of trigonometric functions and domain & range of trigonometric functions.

Graph of Trigonometric Functions :

Values of T-Ratio of some standard angles

Angles T-Ratio	0	$\pi\over 6$	$\pi\over 4$	$\pi\over 3$	$\pi\over 2$	$\pi$
$sin\theta$	0	$1\over 2$	$1\over \sqrt{2}$	$\sqrt{3}\over 2$	1	0
$cos\theta$	1	$\sqrt{3}\over 2$	$1\over \sqrt{2}$	$1\over 2$	0	-1
$tan\theta$	0	$1\over \sqrt{3}$	1	$\sqrt{3}$	N.D	0
$cot\theta$	N.D	$\sqrt{3}$	1	$1\over \sqrt{3}$	0	N.D
$sec\theta$	1	$2\over \sqrt{3}$	$\sqrt{2}$	2	N.D	-1
$cosec\theta$	N.D	2	$\sqrt{2}$	$2\over \sqrt{3}$	1	N.D

N.D = Not defined

Domain, Ranges and Periodicity of Trigonometric function

T-Ratio	Domain	Range	Period
sin x	R	[-1, 1]	$2\pi$
cos x	R	[-1, 1]	$2\pi$
tan x	R – {(2n+1)$\pi/2$; n $\in$ I}	R	$\pi$
cot x	R – {n$\pi$ : n $\in$ I}	R	$\pi$
sec x	R – {(2n+1)$\pi/2$; n $\in$ I}	(-$\infty$, -1] $\cup$ [1, $\infty$]	$2\pi$
cosec x	R – {n$\pi$ : n $\in$ I}	(-$\infty$, -1] $\cup$ [1, $\infty$]	$2\pi$

Trigonometric ratios of some standard angles :

(i)  sin$18^{\circ}$ = sin$\pi\over 10$ = $\sqrt{5}-1\over 4$ = cos$72^{\circ}$ = cos$2\pi\over 5$

(ii)  cos$36^{\circ}$ = cos$\pi\over 5$ = $\sqrt{5}+1\over 4$ = sin$54^{\circ}$ = sin$3\pi\over 10$

(iii)  sin$72^{\circ}$ = sin$2\pi\over 5$ = $\sqrt{10 + 2\sqrt{5}}\over 4$ = cos$18^{\circ}$ = cos$\pi\over 10$

(iv)  sin$36^{\circ}$ = sin$\pi\over 5$ = $\sqrt{10 – 2\sqrt{5}}\over 4$ = cos$54^{\circ}$ = cos$3\pi\over 10$

(v)  sin$15^{\circ}$ = sin$\pi\over 12$ = $\sqrt{3}-1\over {2\sqrt{2}}$ = cos$75^{\circ}$ = cos$5\pi\over 12$

(vi)  cos$15^{\circ}$ = sin$\pi\over 12$ = $\sqrt{3}+1\over {2\sqrt{2}}$ = sin$75^{\circ}$ = sin$5\pi\over 12$

(vii)  tan$15^{\circ}$ = tan$\pi\over 12$ = $2 – \sqrt{3}$ = $\sqrt{3}-1\over {\sqrt{3}+1}$ = cot$75^{\circ}$ = cot$5\pi\over 12$

(viii)  tan$75^{\circ}$ = tan$5\pi\over 12$ = $2 + \sqrt{3}$ = $\sqrt{3}+1\over {\sqrt{3}-1}$ = cot$15^{\circ}$ = cot$\pi\over 12$

(ix)  tan($22.5^{\circ}$) = tan$\pi\over 8$ = $\sqrt{2}-1$ = cot($67.5^{\circ}$) = cot$3\pi\over 8$

(x)  tan($67.5^{\circ}$) = tan$3\pi\over 8$ = $\sqrt{2}+1$ = cot($22.5^{\circ}$) = cot$\pi\over 8$

Example : Evaluate sin$78^{\circ}$ – sin$66^{\circ}$ – sin$42^{\circ}$ + sin$6^{\circ}$

Solution : (sin$78^{\circ}$ – sin$66^{\circ}$) – (sin$42^{\circ}$ – sin$6^{\circ}$-)

= 2cos($60^{\circ}$)sin($18^{\circ}$) – 2cos($36^{\circ}$)sin($30^{\circ}$)

= sin$18^{\circ}$ – cos$36^{\circ}$

= ($\sqrt{5}-1\over 4$) – ($\sqrt{5}+1\over 4$) = $-1\over 2$