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If \[f:X\to Y\] is a function, defined on the set \[X,\] then the domain of the function \[f,\] written as Domf is the set of all independent variables \[x,\] for which the image \[f(x)\] is well defined element of \[Y,\] called the co-domain of \[f\].   Range of \[f:X\to Y\]is the set of all images \[{{72}^{o}}\] which belongs to \[Y,\] i.e., Range \[{{67.5}^{o}}\]\[\{f(x)\in Y:x\in X\}\,\subseteq Y\].   The domain and range of trigonometrical functions are tabulated as follows :
Trigonometrical Function   Domain   Range
\[\sin x\] \[R\] \[-1\le \sin x\le 1\]
\[\cos x\] \[R\] \[-1\le \cos x\le 1\]
\[\tan x\] \[R-\left\{ (2n+1)\frac{\pi }{2},\,n\in I \right\}\] \[R\]
\[\text{cosec}\,x\] \[R-\{n\,\pi ,\,n\in I\}\] \[R-\{x:-1<x<1\}\]
\[\sec x\] \[R-\left\{ (2n+1)\,\frac{\pi }{2},\,n\in I \right\}\]   \[R-\{x\,:\,-1<x<1\}\]
\[\cot x\] \[R-\{n\,\pi ,\,n\in I\}\] \[R\]

In the right angled triangle \[OMP,\] we have base \[=OM=x,\] perpendicular \[=PM=y\] and hypotenuse \[=OP=r\]. We define the following trigonometric ratio which are also known as trigonometric function.       \[\sin \theta =\frac{\text{Perpendicular}}{\text{Hypotenues}}=\frac{y}{r}\]   \[\frac{2n\pi \pm A}{2}\]   \[\tan \theta =\frac{\text{Perpendicular}}{\text{Base}}=\frac{y}{x}\]   \[\cot \theta =\frac{\text{Base}}{\text{Perpendicular}}=\frac{x}{y}\]     \[\sec \theta =\frac{\text{Hypotenues}}{\text{Base}}=\frac{r}{x}\]   \[\text{cosec}\theta =\frac{\text{Hypotenues}}{\text{Perpendicular}}=\frac{r}{y}\]   (1) Relation between trigonometric ratios (functions)   (i) \[\frac{\sqrt{4-\sqrt{2}-\sqrt{6}}}{2\sqrt{2}}\]              (ii) \[\tan \theta .\cot \theta =1\]   (iii) \[\cos \theta .\sec \theta =1\]            (iv) \[\tan \frac{A}{2}\]  (v) \[\cot \theta =\frac{\cos \theta }{\sin \theta }\]   (2) Fundamental trigonometric identities   (i) \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]                        (ii) \[1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \]   (iii) \[1+{{\cot }^{2}}\theta =\text{cose}{{\text{c}}^{2}}\theta \]   (3) Sign of trigonometrical ratios or functions : Their signs depends on the quadrant in which the terminal side of the angle lies.   In brief: A crude aid to memorise the signs of trigonometrical ratio in different quadrant.  "Add Sugar To Coffee".     Algorithm : First determine the sign of the trigonometric function.   If \[\theta \] is measured from\[{X}'OX\] i.e., {(p ± q, 2p – q)} then retain the original name of the function.   If \[\theta \] is measured from \[{Y}'OY\] i.e.,\[\left\{ \frac{\pi }{2}\pm \theta ,\,\frac{3\pi }{2}\pm \theta  \right\}\], then change sine to cosine, cosine to sine, tangent to cotangent, cot to tan, sec to cosec and cosec to sec.   (4) Variations in values of trigonometric functions in different quadrants : Let \[X'OX\] and \[YOY'\] be the coordinate axes. Draw a circle with centre at origin O and radius unity.   Let \[M(x,y)\] be a point on the circle such that \[\angle AOM=\theta \] then \[x=\cos \theta \] and \[y=\sin \theta \]; \[-1\le \cos \theta \le \]1 and \[-1\le \sin \theta \le 1\] for all values of \[\theta \].    
II-Quadrant (S) I-Quadrant (A)
\[\sin \theta \to \] decreases from 1 to 0 \[\sin \theta \to \] increases from 0 to 1
\[\cos \theta \to \] decreases from 0 to - 1 more...
Two angles are said to be allied when their sum or difference is either zero or a multiple of \[{{90}^{o}}\].    
Allied angles \[\to \] \[\sin \theta \] \[cos\theta \] \[tan\theta \]
Trigo. Ratio
\[\downarrow \,\,(-\theta )\] \[-\sin \theta \] \[cos\theta \] \[-tan\theta \]
\[(90-\theta )\] or \[\left( \frac{\pi }{2}-\theta  \right)\] \[cos\theta \] \[\sin \theta \] \[\cot \,\theta \]
\[(90-\theta )\] or \[\left( \frac{\pi }{2}-\theta  \right)\] \[\cos \theta \] \[-\,\sin \theta \] \[-\cot \,\theta \]
\[(180-\theta )\] or\[(\pi -\theta )\] \[\sin \theta \] \[-\,\cos \theta \] more...
 
\[\mathbf{\theta }\] 0 \[\mathbf{\pi /6}\] \[\mathbf{\pi /4}\] \[\mathbf{\pi /3}\] \[\mathbf{\pi /2}\] \[\mathbf{\pi }\] \[\mathbf{3\pi /2}\] \[\mathbf{2\pi }\]
\[\sin \theta \] 0 1/2 \[1/\sqrt{2}\] \[\sqrt{3}/2\] 1 0 \[-1\] 0
\[\cos \,\theta \] more...
 
  \[\theta \] \[7\frac{{{1}^{o}}}{2}\] \[{{15}^{o}}\] \[22\frac{{{1}^{o}}}{2}\] \[{{18}^{o}}\] \[{{36}^{o}}\]
\[\sin \,\theta \] \[\frac{\sqrt{4-\sqrt{2}-\sqrt{6}}}{2\sqrt{2}}\] \[\frac{\sqrt{3}-1}{2\sqrt{2}}\] \[\frac{1}{2}\sqrt{2-\sqrt{2}}\] \[\frac{\sqrt{5}-1}{4}\] \[\frac{1}{4}\sqrt{10-2\sqrt{5}}\]
\[\cos \theta \] \[\frac{\sqrt{4+\sqrt{2}+\sqrt{6}}}{2\sqrt{2}}\] \[\frac{\sqrt{3}+1}{2\sqrt{2}}\] \[\frac{1}{2}\sqrt{2+\sqrt{2}}\] \[\frac{1}{4}\sqrt{10+2\sqrt{5}}\] \[\frac{\sqrt{5}+1}{4}\]
\[\tan \theta \] more...
   
  \[\mathbf{sin}\,\mathbf{\theta }\] \[\mathbf{cos}\,\mathbf{\theta }\] \[\mathbf{tan}\,\mathbf{\theta }\] \[\mathbf{cot}\,\,\mathbf{\theta }\] \[\mathbf{sec}\,\,\mathbf{\theta }\] \[\mathbf{cosec}\,\,\mathbf{\theta }\]
\[\mathbf{sin}\,\mathbf{\theta }\] \[\sin \,\theta \] \[\sqrt{1-{{\cos }^{2}}\theta }\] \[\frac{\tan \theta }{\sqrt{1+{{\tan }^{2}}\theta }}\] \[\frac{1}{\sqrt{1+{{\cot }^{2}}\theta }}\] \[\frac{\sqrt{{{\sec }^{2}}\theta -1}}{\sec \theta }\] \[\frac{1}{\text{cosec}\theta }\]
\[\mathbf{cos}\,\mathbf{\theta }\] \[\sqrt{1-{{\sin }^{2}}\theta }\] \[\cos \theta \] \[\frac{1}{\sqrt{1+{{\tan }^{2}}\theta }}\] more...
(1) \[2\sin A\cos B=\sin (A+B)+\sin (A-B)\]   (2) \[2\cos A\sin B=\sin (A+B)-\sin (A-B)\]   (3) \[2\cos A\cos B=\cos (A+B)+\cos (A-B)\]   (4) \[2\sin A\sin B=\cos (A-B)-\cos (A+B)\]   Let \[A+B=C\] and \[A-B=D\]   Then, \[A=\frac{C+D}{2}\] and \[B=\frac{C-D}{2}\]   Therefore, we find out the formulae to transform the sum or difference into product.   (1) \[\sin C+\sin D=2\sin \frac{C+D}{2}\cos \frac{C-D}{2}\]   (2) \[\sin C-\sin D=2\cos \frac{C+D}{2}\sin \frac{C-D}{2}\]   (3) \[\cos C+\cos D=2\cos \frac{C+D}{2}\cos \frac{C-D}{2}\]   (4) \[\cos C-\cos D=2\sin \frac{C+D}{2}\sin \frac{D-C}{2}=-2\sin \frac{C+D}{2}\sin \frac{C-D}{2}\].

(1) \[\sin 2A=2\sin A\cos A\]\[=\frac{2\tan A}{1+{{\tan }^{2}}A}\]   (2) \[\frac{\sqrt{5}-1}{4}\]\[\frac{1}{4}\sqrt{10-2\sqrt{5}}\]   \[={{\cos }^{2}}A-{{\sin }^{2}}A\]\[2-\sqrt{3}\]; where \[A\ne (2n+1)\frac{\pi }{4}\].   (3) \[\tan 2A=\frac{2\tan A}{1-{{\tan }^{2}}A}\]   (4) \[\sin 3A=3\sin A-4{{\sin }^{3}}A\]\[=4\sin ({{60}^{o}}-A).\sin A.\sin ({{60}^{o}}+A)\]   (5) \[\cos 3A=4{{\cos }^{3}}A-3\cos A\]\[=4\cos ({{60}^{o}}-A).\cos A.\cos ({{60}^{o}}+A)\]   (6) \[\tan 3A=\frac{3\tan A-{{\tan }^{3}}A}{1-3{{\tan }^{2}}A}=\tan ({{60}^{o}}-A).\tan A.\tan ({{60}^{o}}+A)\], where \[A\ne n\pi +\pi /6\]   (7) \[\sin 4\theta =4\sin \theta .{{\cos }^{3}}\theta -4\cos \theta {{\sin }^{3}}\theta \]   (8) \[\cos 4\theta =8{{\cos }^{4}}\theta -8{{\cos }^{2}}\theta +1\]   (9) \[\tan 4\theta =\frac{4\tan \theta -4{{\tan }^{3}}\theta }{1-6{{\tan }^{2}}\theta +{{\tan }^{4}}\theta }\]   (10) \[\sin 5A=16{{\sin }^{5}}A-20{{\sin }^{3}}A+5\sin A\]   (11) \[\cos 5A=16{{\cos }^{5}}A-20{{\cos }^{3}}A+5\cos A\]

(1) \[\left| \,\sin \frac{A}{2}+\cos \frac{A}{2}\, \right|=\sqrt{1+\sin A}\]   or \[\sin \frac{A}{2}+\cos \frac{A}{2}=\pm \sqrt{1+\sin A}\]   i.e., \[\left\{ \begin{matrix} +,\,\text{If }2n\pi -\pi /4\le A/2\le 2n\pi +\frac{3\pi }{4}  \\ -,\,\text{otherwise}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,  \\ \end{matrix} \right.\]   (2) \[\left| \,\sin \frac{A}{2}-\cos \frac{A}{2}\, \right|=\sqrt{1-\sin A}\]   or  \[(\sin \frac{A}{2}-\cos \frac{A}{2})=\pm \sqrt{1-\sin A}\]   i.e., \[\left\{ \begin{matrix} +,\,\text{If }2n\pi +\pi /4\le A/2\le 2n\pi +\frac{5\pi }{4}  \\ -,\,\text{otherwise}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,  \\ \end{matrix} \right.\]   (3) (i) \[\tan \frac{A}{2}=\frac{\pm \sqrt{{{\tan }^{2}}A+1}-1}{\tan A}=\pm \,\sqrt{\frac{1-\cos A}{1+\cos A}}=\frac{1-\cos A}{\sin A}\],   where \[A\ne (2n+1)\pi \]   (ii) \[\cot \frac{A}{2}=\pm \,\sqrt{\frac{1+\cos A}{1-\cos A}}=\frac{1+\cos A}{\sin A}\], where \[A\ne 2n\pi \]   The ambiguities of signs are removed by locating the quadrants in which \[\frac{A}{2}\] lies or you can follow the following figure,                           (4) \[{{\tan }^{2}}\frac{A}{2}=\frac{1-\cos A}{1+\cos A}\] ; where \[A\ne (2n+1)\pi \]   (5) \[{{\cot }^{2}}\frac{A}{2}=\frac{1+\cos A}{1-\cos A}\]; where \[A\ne 2n\pi \]  

Let  \[a=r\cos \alpha \]    .....(i)   and   \[b=r\sin \alpha \]         .....(ii)   Squaring and adding (i) and (ii), then \[{{a}^{2}}+{{b}^{2}}={{r}^{2}}\] or, \[r=\sqrt{{{a}^{2}}+{{b}^{2}}}\]   \[\therefore \]          \[a\sin \theta +b\cos \theta =r(\sin \theta \cos \alpha +\cos \theta \sin \alpha )=r\sin (\theta +\alpha )\]   But \[-1\le \sin \theta <1\] So, \[-1\le \sin (\theta +\alpha )\le 1\];   Then \[-r\le r\sin (\theta +\alpha )\le r\]   Hence, \[\sqrt{2}-1\]   Then the greatest and least values of \[a\sin \theta +b\cos \theta \] are respectively \[\sqrt{{{a}^{2}}+{{b}^{2}}}\] and \[-\sqrt{{{a}^{2}}+{{b}^{2}}}\].   Therefore, \[{{\sin }^{2}}x+c\text{ose}{{\text{c}}^{\text{2}}}x\ge 2,\] for every real \[x\].   \[{{\cos }^{2}}x+{{\sec }^{2}}x\ge 2,\] for every real \[x\].   \[{{\tan }^{2}}x+{{\cot }^{2}}x\ge 2\], for every real \[x\].


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