Trigonometric Formulas

These are the common formulas used in Chapter 9 to 14 of Textbook of Algebra and Trigonometry Class XI, Punjab Textbook Board Lahore. This handout is very helpful to remember the formulas. All these formulas are given for real valued and defined trigonometric functions. A PDF file can be downloaded for high quality printing and a word file is also given if you wish to modify the contents or credit as you need.

  • ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
  • $1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta$
  • $1+{{\cot }^{2}}\theta ={{\csc }^{2}}\theta$
  • $\sin (-\theta )=-\sin \theta$
  • $\cos (-\theta )=\cos \theta$
  • $\tan (-\theta )=-\tan \theta$
  • $\sin \left( \alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta$
  • $\sin \left( \alpha -\beta \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta$
  • $\cos \left( \alpha +\beta \right)=\cos \alpha \cos \beta -\sin \alpha \sin \beta$
  • $\cos \left( \alpha -\beta \right)=\cos \alpha \cos \beta +\sin \alpha \sin \beta$
  • $\tan \left( \alpha +\beta \right)=\dfrac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$
  • $\tan \left( \alpha -\beta \right)=\dfrac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta }$
  • $\sin 2\theta =2\sin \theta \cos \theta$
  • $\cos 2\theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta$
  • $\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }$
  • ${{\sin }^{2}}\frac{\theta }{2}=\dfrac{1-\cos \theta }{2}$
  • ${{\cos }^{2}}\frac{\theta }{2}=\dfrac{1+\cos \theta }{2}$
  • ${{\tan }^{2}}\frac{\theta }{2}=\dfrac{1-\cos \theta }{1+\cos \theta }$
  • $\sin 3\theta =3\sin \theta -4{{\sin }^{3}}\theta$
  • $\cos 3\theta =4{{\cos }^{3}}\theta -3\cos \theta$
  • $\tan 3\theta =\dfrac{3\tan \theta -{{\tan }^{3}}\theta }{1-3{{\tan }^{2}}\theta}$
  • $\sin 2\theta =\dfrac{2\tan \theta }{1+{{\tan }^{2}}\theta }$
  • $\cos 2\theta \,=\dfrac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }$
  • $\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }$
  • $\sin \left( \alpha +\beta \right)\,+\,\sin \left( \alpha -\beta \right)\,=\,2\sin \alpha \cos \beta$
  • $\sin \left( \alpha +\beta \right)-\sin \left( \alpha -\beta \right)\,=\,2\cos \alpha \sin \beta$
  • $\cos \left( \alpha +\beta \right)\,+\,\cos \left( \alpha -\beta \right)=2\cos \alpha \cos \beta$
  • $\cos \left( \alpha +\beta \right)\,-\,\cos \left( \alpha -\beta \right)=-2\sin \alpha \sin \beta$
  • $\sin \theta +\sin \phi \,=2\sin \frac{\theta +\phi }{2}\,\,\cos \frac{\theta -\phi }{2}$
  • $\sin \theta -\sin \phi \,=2\cos \frac{\theta +\phi }{2}\,\,\sin \frac{\theta -\phi }{2}$
  • $\cos \theta +\cos \phi \,=2\cos \frac{\theta +\phi }{2}\,\,\cos \frac{\theta -\phi }{2}$
  • $\cos \theta -\cos \phi \,=-2\sin \frac{\theta +\phi }{2}\,\,\sin \frac{\theta -\phi }{2}$
  • ${{\sin }^{-1}}A+{{\sin }^{-1}}B={{\sin }^{-1}}\left( A\sqrt{1-{{B}^{2}}}+B\sqrt{1-{{A}^{2}}} \right)$
  • ${{\sin }^{-1}}A-{{\sin }^{-1}}B={{\sin }^{-1}}\left( A\sqrt{1-{{B}^{2}}}-B\sqrt{1-{{A}^{2}}} \right)$
  • ${{\cos }^{-1}}A+{{\cos }^{-1}}B={{\cos }^{-1}}\left( AB-\sqrt{\left( 1-{{A}^{2}} \right)\left( 1-{{B}^{2}} \right)} \right)$
  • ${{\cos }^{-1}}A-{{\cos }^{-1}}B={{\cos }^{-1}}\left( AB+\sqrt{\left( 1-{{A}^{2}} \right)\left( 1-{{B}^{2}} \right)} \right)$
  • ${{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\dfrac{A+B}{1-AB}$
  • ${{\tan }^{-1}}A-{{\tan }^{-1}}B={{\tan }^{-1}}\dfrac{A-B}{1+AB}$

Three Steps to solve $\sin \left( n\cdot \frac{\pi }{2}\pm \theta \right)$

Step I: First check that $n$ is even or odd.

Step II: If $n$ is even then the answer will be in $\sin$ and if the $n$ is odd then $\sin$ will be converted to $\cos$ and vice versa (i.e. cos will be converted to sin).

Step III: Now check in which quadrant $n\cdot \frac{\pi }{2}\pm \theta $ is lying if it is in Ist or IInd quadrant the answer will be positive as sin is positive in these quadrants and if it is in the IIIrd or IVth quadrant the answer will be negative.
e.g. $\sin {{667}^{{}^\circ }}$ $=\sin \left( 7(90)+37 \right)$
Since $n=7$ is odd so answer will be in $\cos$ and 667 is in IVth quadrant and sin is –ive in IVth quadrant therefore answer will be in negative. i.e $\sin {{667}^{\circ}}=-\cos {{37}^{\circ}}$.
Similar technique is used for other trigonometric ratios. i.e. $\tan \rightleftarrows \cot$ and $\sec \rightleftarrows \csc$.