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Trigonometric Formulas

The textbook for algebra and trigonometry for class XI published by the Punjab Text Book Board in Lahore uses these formulas often from Chapter 9 to Chapter 14. To help you recall the formulas, use this page. These formulas are all provided for trigonometric functions that have real values and are defined.

  • \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 (\alpha+\beta)=\sin \alpha \cos \beta+\cos \alpha \sin \beta
  • \sin (\alpha-\beta)=\sin \alpha \cos \beta-\cos \alpha \sin \beta
  • \cos (\alpha+\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \beta
  • \cos (\alpha-\beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta
  • \tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}
  • \tan (\alpha-\beta)=\frac{\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=\frac{2 \tan \theta}{1-\tan ^2 \theta}
  • \sin ^2 \frac{\theta}{2}=\frac{1-\cos \theta}{2}
  • \cos ^2 \frac{\theta}{2}=\frac{1+\cos \theta}{2}
  • \tan ^2 \frac{\theta}{2}=\frac{1-\cos \theta}{1+\cos \theta}
  • \sin 3 \theta=3 \sin \theta-4 \sin ^3 \theta \quad\cos 3 \theta=4 \cos ^3 \theta-3 \cos \theta
  • \tan 3 \theta=\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}
  • \sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^2 \theta}
  • \cos 2 \theta=\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}
  • \sin (\alpha+\beta)+\sin (\alpha-\beta)=2 \sin \alpha \cos \beta
  • \sin (\alpha+\beta)-\sin (\alpha-\beta)=2 \cos \alpha \sin \beta
  • \cos (\alpha+\beta)+\cos (\alpha-\beta)=2 \cos \alpha \cos \beta
  • \cos (\alpha+\beta)-\cos (\alpha-\beta)=-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(A B-\sqrt{\left(1-A^2\right)\left(1-B^2\right)}\right)
  • \cos ^{-1} A-\cos ^{-1} B=\cos ^{-1}\left(A B+\sqrt{\left(1-A^2\right)\left(1-B^2\right)}\right)
  • \tan ^{-1} A+\tan ^{-1} B=\tan ^{-1} \frac{A+B}{1-A B}
  • \tan ^{-1} A-\tan ^{-1} B=\tan ^{-1} \frac{A-B}{1+A B}

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

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 virsa (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 in the IIIrd or IVth quadrant the answer will be negative.
e.g. \sin 667^{\circ}=\sin (7(90)+37)
Since \mathrm{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.

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