Derivation of the Double Angle Formulas

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The Double Angle Formulas can be derived from Sum of Two Angles listed below:
$ \sin (A + B) = \sin A \, \cos B + \cos A \, \sin B $   →   Equation (1)

$ \cos (A + B) = \cos A \, \cos B - \sin A \, \sin B $   →   Equation (2)

$ \tan (A + B) = \dfrac{\tan A + \tan B}{1 - \tan A \, \tan B} $   →   Equation (3)
 

Let θ = A = B; Equation (1) will become

$ \sin (\theta + \theta) = \sin \theta \, \cos \theta + \cos \theta \, \sin \theta $

$ \sin 2\theta = 2\sin \theta \, \cos \theta $

 

Let θ = A = B; Equation (2) will become
$ \cos (\theta + \theta) = \cos \theta \, \cos \theta - \sin \theta \, \sin \theta $

$ \cos 2\theta = \cos^2 \theta - \sin^2 \theta $   →   Equation (4)
 

The Pythagorean Identity sin2 θ + cos2 θ = 1 can be taken as sin2 θ = 1 - cos2 θ and Equation (4) will become...
$ \cos 2\theta = \cos^2 \theta - (1 - \cos^2 \theta) $

$ \cos 2\theta = 2\cos^2 \theta - 1 $
 

sin2 θ + cos2 θ = 1 can also be taken as cos2 θ = 1 - sin2 θ and Equation (4) will become...
$ \cos 2\theta = (1 - \sin^2) - \sin^2 \theta $

$ \cos 2\theta = 1 - 2\sin^2 \theta $
 

For easy reference, below is the summary for cos 2θ

$ \cos 2\theta = \cos^2 \theta - \sin^2 \theta $

$ \cos 2\theta = 2\cos^2 \theta - 1 $

$ \cos 2\theta = 1 - 2\sin^2 \theta $

 

Let θ = A = B; Equation (3) will become
$ \tan (\theta + \theta) = \dfrac{\tan \theta + \tan \theta}{1 - \tan \theta \, \tan \theta} $

$ \tan 2\theta = \dfrac{2\tan \theta}{1 - \tan^2 \theta} $

 



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