JEE Main & Advanced

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   more...
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 more...

Two angles are said to be allied when their sum or difference is either zero or a multiple of \[{{90}^{o}}\].    
Allied angles \[\to \] \[\sin \theta \] \[cos\theta \] \[tan\theta \]
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\[\mathbf{\theta }\] 0 \[\mathbf{\pi /6}\] \[\mathbf{\pi /4}\] \[\mathbf{\pi /3}\] more...
 
  \[\theta \] \[7\frac{{{1}^{o}}}{2}\] \[{{15}^{o}}\] \[22\frac{{{1}^{o}}}{2}\] \[{{18}^{o}}\] more...
   
  \[\mathbf{sin}\,\mathbf{\theta }\] \[\mathbf{cos}\,\mathbf{\theta }\] \[\mathbf{tan}\,\mathbf{\theta }\] \[\mathbf{cot}\,\,\mathbf{\theta }\] 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,                           more...

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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