derivation of half-angle formulae for tangent
Start with the angle duplication formula
tan(x)=2tan(x/2)1-tan2(x/2). |
Cross-multiply and move terms around:
tan(x)tan2(x/2)+2tan(x/2)=tan(x) |
Divide by tan(x):
tan2(x/2)+2tan(x/2)tanx=1 |
Add 1/tan2(x) to both sides:
tan2(x/2)+2tan(x/2)tanx+1tan2(x)=1+1tan2(x) |
Complete the square (http://planetmath.org/CompletingTheSquare):
(tan(x/2)+1tan(x))2=1+1tan2(x) |
Take a square root and move a term to obtain the half-angle formula:
tan(x/2)=√1+1tan2(x)-1tan(x) |
To derive the other forms of the formula, we start by substituting sin(x)/cos(x) for tan(x):
tan(x/2)=√1+cos2(x)sin2(x)-cos(x)sin(x) |
Put the stuff inside the square root over a common denominator:
√sin2(x)+cos2(x)sin2(x)-cos(x)sin(x) |
Recall that sin2(x)+cos2(x)=1. Hence, we may get rid of the square root:
1sinx-cos(x)sin(x) |
Putting the terms over a common denominator, we obtain our formula:
tan(x/2)=1-cos(x)sin(x) |
To obtain the next formula, multiply both numerator and denominator by 1+cos(x):
(1-cos(x))(1+cos(x))sin(x)(1+cos(x)) |
Multiply out the numerator and simplify:
1-cos2(x)sin(x)(1+cos(x)) |
Note that the numerator equals sin2(x):
sin2(x)sin(x)(1+cos(x)) |
Cancel a common factor of sin(x) to obtain the formula
tan(x/2)=sin(x)1+cos(x). |
To obtain the last formula, multiply the previous two formulae:
tan2(x/2)=1-cos(x)sin(x)⋅sin(x)1+cos(x) |
Cancel the common factor of sin(x):
tan2(x/2)=1-cos(x)1+cos(x) |
Take the square root of both sides to obtain the formula
tanx2=±√1-cosx1+cosx; |
here the sign (±) has to be chosen according to the
quadrant where the angle x2 is.
Title | derivation of half-angle formulae for tangent |
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Canonical name | DerivationOfHalfangleFormulaeForTangent |
Date of creation | 2013-03-22 17:00:19 |
Last modified on | 2013-03-22 17:00:19 |
Owner | rspuzio (6075) |
Last modified by | rspuzio (6075) |
Numerical id | 9 |
Author | rspuzio (6075) |
Entry type | Derivation |
Classification | msc 26A09 |
Related topic | TangentOfHalvedAngle |