How do you derive the half angle formula for sin,cos and tan?

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Answer 1 · 2015-06-12T18:06:20

You can start from $sin^2(x)$ .

$sin^2x = (1-cos(2x))/2$

Similarly:
$sin^2(x/2) = (1-cosx)/2$

Thus:
$sin(x/2) = pmsqrt((1-cosx)/2)$
$+$ if in quadrant I or II
$-$ if in quadrant III or IV

Analogously:
$cos(x/2) = pmsqrt((1+cosx)/2)$
$+$ if in quadrant I or IV
$-$ if in quadrant II or III

Therefore:
$tan(x/2) = sin(x/2)/cos(x/2) = [(pmsqrt((1-cosx)/2)) / (pmsqrt((1+cosx)/2))]$

$= pmsqrt((1-cosx)/(1+cosx))$
$+$ if in quadrant I or III
$-$ if in quadrant II or IV

In quadrant I, the division gives $(+/+) = +$

In quadrant III, the division gives $(-/-) = +$

In quadrant II, the division gives $(+/-) = -$

In quadrant IV, the division gives $(-/+) = -$