MHB Can Trig Identities be Derived from Easier Formulas?

[hatelove]

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.)

 

[SuperSonic4]

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.)

Using the addition formula for tan would be the easiest: $\tan(A+B) = \frac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)}$

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).

 

[CaptainBlack]

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.)

\[\tan(2a)=\frac{\sin(2a)}{\cos(2a)}=\frac{2\sin(a) \cos(a)}{\cos^2(a)-\sin^2(a)}\]

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

CB

 

[chisigma]

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.)

In...

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