PlanetMath: derivation of half-angle formulae for tangent

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derivation of half-angle formulae for tangent

(Derivation)

Start with the angle duplication formula

$\displaystyle \tan (x) = {2 \tan (x/2) \over 1 - \tan^2 (x/2)}.$

Cross-multiply and move terms around:

$\displaystyle \tan (x) \tan^2 (x/2) + 2 \tan (x/2) = \tan (x)$

Divide by $\tan (x)$ :

$\displaystyle \tan^2 (x/2) + {2 \tan (x/2) \over \tan x} = 1$

Add $1 / \tan^2 (x)$ to both sides:

$\displaystyle \tan^2 (x/2) + {2 \tan (x/2) \over \tan x} + {1 \over \tan^2 (x)} = 1 + {1 \over \tan^2 (x)}$

Complete the square:

$\displaystyle \left (\tan (x/2) + {1 \over \tan (x)} \right)^2 = 1 + {1 \over \tan^2 (x)}$

Take a square root and move a term to obtain the half-angle formula:

$\displaystyle \tan (x/2) = \sqrt{ 1 + {1 \over \tan^2 (x)} } - {1 \over \tan (x)}$

To derive the other forms of the formula, we start by substituting $\sin (x) / \cos (x)$ for $\tan (x)$ :

$\displaystyle \tan (x/2) = \sqrt{ 1 + {\cos^2 (x) \over \sin^2 (x)}} - {\cos (x) \over \sin (x)}$

Put the stuff inside the square root over a common denominator:

$\displaystyle \sqrt {\sin^2 (x) + \cos^2 (x) \over \sin^2 (x)} - {\cos (x) \over \sin (x)}$

Recall that $\sin^2 (x) + \cos^2 (x) = 1$ . Hence, we may get rid of the square root:

$\displaystyle {1 \over \sin x} - {\cos (x) \over \sin (x)}$

Putting the terms over a common denominator, we obtain our formula:

$\displaystyle \tan (x/2) = {1 - \cos (x) \over \sin (x)}$

To obtain the next formula, multiply both numerator and denominator by $1 + \cos (x)$ :

$\displaystyle {(1 - \cos (x)) (1 + \cos (x)) \over \sin (x) (1 + \cos (x))}$

Multiply out the numerator and simplify:

$\displaystyle {1 - \cos^2 (x) \over \sin (x) (1 + \cos (x))}$

Note that the numerator equals $\sin^2 (x)$ :

$\displaystyle {\sin^2 (x) \over \sin (x) (1 + \cos (x))}$

Cancel a common factor of $\sin (x)$ to obtain the formula

$\displaystyle \tan (x/2) = {\sin (x) \over 1 + \cos (x)} .$

To obtain the last formula, multiply the previous two formulae:

$\displaystyle \tan^2 (x/2) = {1 - \cos (x) \over \sin (x)}\cdot {\sin (x) \over 1 + \cos (x)}$

Cancel the common factor of $\sin (x)$ :

$\displaystyle \tan^2 (x/2) = {1 - \cos (x) \over 1 + \cos (x)}$

Take the square root of both sides to obtain the formula

$\displaystyle \tan{\frac{x}{2}}\; = \;\pm\sqrt{1 - \cos{x} \over 1 + \cos{x}};$

here the sign ( $\pm$ ) has to be chosen according to the quadrant where the angle $\displaystyle\frac{x}{2}$ is.

"derivation of half-angle formulae for tangent" is owned by rspuzio. [ full author list (3) ]

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See Also: tangent of halved angle

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Cross-references: quadrant, factor, numerator, denominator, square root, sides, divide, terms, angle

This is version 6 of derivation of half-angle formulae for tangent, born on 2007-04-27, modified 2007-04-28.
Object id is 9287, canonical name is DerivationOfHalfAngleFormulaeForTangent.
Accessed 1102 times total.

Classification:

26A09 (Real functions :: Functions of one variable :: Elementary functions)

Pending Errata and Addenda

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