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double angle identity (Theorem)

The double angle identities are

\begin{eqnarray} \sin(2x) & = & 2\sin{x}\cos{x} \\ \cos(2x) & = & \cos^2{x}-\sin^2{x} = 2\cos^2{x}-1 = 1-2\sin^2{x} \\ \tan(2x) & = & \frac{2\tan{x}}{1-\tan^2{x}} \end{eqnarray} These are all derived from their respective trigonometric addition formulas. For example,

\begin{eqnarray*} \sin(2x) & = & \sin(x+x) \\ & = & \sin{x}\cos{x}+\cos{x}\sin{x} \\ & = & 2\sin{x}\cos{x} \end{eqnarray*} The formula for cosine follows similarly, and the formula tangent is derived by taking the ratio of sine to cosine, as always.

The double angle identities can also be derived from the de Moivre identity.




"double angle identity" is owned by Wkbj79. [ full author list (2) | owner history (1) ]
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See Also: de Moivre identity, angle sum identity, addition and subtraction formulas for sine and cosine

Other names:  double-angle identity, double angle formula, double-angle formula, double angle formulae, double-angle formulae

Attachments:
proof of double angle identity (Proof) by drini
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Cross-references: de Moivre identity, sine, ratio, tangent, cosine, trigonometric addition formulas
There are 10 references to this entry.

This is version 12 of double angle identity, born on 2002-01-30, modified 2007-06-25.
Object id is 1623, canonical name is DoubleAngleIdentity.
Accessed 34587 times total.

Classification:
AMS MSC26A09 (Real functions :: Functions of one variable :: Elementary functions)
 33B10 (Special functions :: Elementary classical functions :: Exponential and trigonometric functions)

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double angle identity by perucho on 2004-05-17 11:28:37
Also through Euler formula (equivalent to Moivre formula mentioned by Mr. Akrowne):
e^{2i\alpha}=cos{2\alpha}+isin{2\alpha} (1)
(e^{i\alpha})^2=(cos{\alpha}+isin{\alpha})^2=
=(cos{\alpha}^2-sin{\alpha}^2)+i(2sin{\alpha}cos{\alpha}) (2)
and equating.
 

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