Current Affairs JEE Main & Advanced

\[\mathbf{\theta }\] 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 \[x\].

We have certain trigonometric identities.   Like, \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]  and  \[1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \] etc.   Such identities are identities in the sense that they hold for all value of the angles which satisfy the given condition among them and they are called conditional identities.   (1) If \[A+B+C={{180}^{o}}\], then   (i) \[\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C\]   (ii) \[\sin 2A+\sin 2B-\sin 2C=4\cos A\cos B\sin C\]   (iii) \[\sin (B+C-A)+\sin (C+A-B)+\sin (A+B-C)\]\[=4\sin A\sin B\sin C\]   (iv) \[\cos 2A+\cos 2B+\cos 2C=-1-4\cos A\cos B\cos C\]   (v) \[\cos 2A+\cos 2B-\cos 2C=1-4\sin A\sin B\cos C\]   (2) If \[A+B+C={{180}^{o}}\], then   (i) \[\sin A+\sin B+\sin C=4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}\]   (ii) \[\sin A+\sin B-\sin C=4\sin \frac{A}{2}\sin \frac{B}{2}\cos \frac{C}{2}\]   (iii) \[\cos A+\cos B+\cos C=1+4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\]   (iv) \[\cos A+\cos B-\cos C=-1+4\cos \frac{A}{2}\cos \frac{B}{2}\sin \frac{C}{2}\]   (v) \[\frac{\cos A}{\sin B\sin C}+\frac{\cos B}{\sin C\sin A}+\frac{\cos C}{\sin A\sin B}=2\]   (3) If \[A+B+C=\pi \], then   (i) \[{{\sin }^{2}}A+{{\sin }^{2}}B-{{\sin }^{2}}C=2\sin A\sin B\cos C\]   (ii) \[{{\cos }^{2}}A+{{\cos }^{2}}B+{{\cos }^{2}}C=1-2\cos A\cos B\cos C\]   (iii) \[{{\sin }^{2}}A+{{\sin }^{2}}B+{{\sin }^{2}}C=1-2\sin A\sin B\cos C\]   (4) If \[A+B+C=\pi ,\] then   (i) \[{{\sin }^{2}}\frac{A}{2}+{{\sin }^{2}}\frac{B}{2}+{{\sin }^{2}}\frac{C}{2}=1-2\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\]   (ii) \[{{\cos }^{2}}\frac{A}{2}+{{\cos }^{2}}\frac{B}{2}+{{\cos }^{2}}\frac{C}{2}=2+2\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\]   (iii) \[{{\sin }^{2}}\frac{A}{2}+{{\sin }^{2}}\frac{B}{2}-{{\sin }^{2}}\frac{C}{2}=1-2\cos \frac{A}{2}\cos \frac{B}{2}\sin \frac{C}{2}\]   (iv) \[{{\cos }^{2}}\frac{A}{2}+{{\cos }^{2}}\frac{B}{2}-{{\cos }^{2}}\frac{C}{2}=2\cos \frac{A}{2}\cos \frac{B}{2}\sin \frac{C}{2}\]   (5) If \[x+y+z=\frac{\pi }{2}\], then   (i) \[{{\sin }^{2}}x+{{\sin }^{2}}y+{{\sin }^{2}}z=1-2\sin x\sin y\sin z\]   (ii) \[{{\cos }^{2}}x+{{\cos }^{2}}y+{{\cos }^{2}}z=2+2\sin x\sin y\sin z\]   (iii) \[\sin 2x+\sin 2y+\sin 2z=4\cos x\cos y\cos z\]   (6) If \[A+B+C=\pi \], then   (i) \[\tan A+\tan B+\tan C=\tan A\tan B\tan C\]   (ii) \[\cot B\cot C+\cot C\cot A+\cot A\cot B=1\]   (iii) \[\tan \frac{B}{2}\tan \frac{C}{2}+\tan \frac{C}{2}\tan \frac{A}{2}+\tan \frac{A}{2}\tan \frac{B}{2}=1\]   (iv) \[\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}=\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}\]

A sequence of numbers \[<{{t}_{n}}>\] is said to be in arithmetic progression (A.P.) when the difference \[{{t}_{n}}-{{t}_{n-1}}\] is a constant for all n Î N. This constant is called the common difference of the A.P. and is usually denoted by the letter d.     If \['a'\] is the first term and \['d'\] the common difference, then an A.P. can be represented as \[a,\,a+d,a+2d,\,a+3d,........\]     Example : 2, 7, 12, 17, 22, …… is an A.P. whose first term is 2 and common difference 5.     Algorithm to determine whether a sequence is an A.P. or not.     Step I: Obtain \[{{a}_{n}}\] (the \[{{n}^{th}}\] term of the sequence).     Step II: Replace \[n\] by \[n-1\] in \[{{a}_{n}}\] to get \[{{a}_{n-1}}\].     Step III: Calculate \[{{a}_{n}}-{{a}_{n-1}}\].     If \[{{a}_{n}}-{{a}_{n-1}}\] is independent from \[n,\] the given sequence is an A.P. otherwise it is not an A.P.     \[\therefore \] \[{{t}_{n}}=An+B\] represents the \[{{n}^{th}}\] term of an A.P. with common difference A.

(1) Let \['a'\] be the first term and \['d'\] be the common difference of an A.P. Then its \[{{n}^{th}}\] term is \[a+(n-1)d\]i.e., \[{{T}_{n}}=a+(n-1)d\].   (2) \[{{p}^{th}}\] term of an A.P. from the end : Let \['a'\] be the first term and \['d'\] be the common difference of an A.P. having \[n\] terms. Then  \[{{p}^{th}}\] term from the end is \[{{(n-p+1)}^{th}}\] term from the beginning   i.e., \[{{p}^{th}}\text{ term from the end }=\text{ }{{T}_{(n-p+1)}}=a+(n-p)d\].  
  • If last term of an A.P. is l then \[{{p}^{th}}\]term from end\[=l-(p-1)d\]

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