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- Mar 1, 2012

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Using the addition formula for tan would be the easiest: $\tan(A+B) = \frac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)}$I know you can derive the double angle formulas for sin(2a) and cos(2a) from Euler's identity, but is there any way to derive the tan(2a) in a similar manner from an easier formula? What about the addition/subtraction formulas (i.e. sin(a+b), etc.)

Alternatively you can use the fact that $\tan(ax) = \frac{\sin(ax)}{\cos(ax)}$ (where a is a constant) together with your values for sin(2a) and cos(2a).

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- Jan 26, 2012

- 890

\[\tan(2a)=\frac{\sin(2a)}{\cos(2a)}=\frac{2\sin(a) \cos(a)}{\cos^2(a)-\sin^2(a)}\]I know you can derive the double angle formulas for sin(2a) and cos(2a) from Euler's identity, but is there any way to derive the tan(2a) in a similar manner from an easier formula? What about the addition/subtraction formulas (i.e. sin(a+b), etc.)

Now divide top and bottom by \(\cos^2(a)\)

CB

- Feb 13, 2012

- 1,704

In...I know you can derive the double angle formulas for sin(2a) and cos(2a) from Euler's identity, but is there any way to derive the tan(2a) in a similar manner from an easier formula? What about the addition/subtraction formulas (i.e. sin(a+b), etc.)

http://mathworld.wolfram.com/TrigonometricAdditionFormulas.html

... a purely geometric way to obtain the sine of the sum of two angles is given...

Kind regards

chi sigma