Question

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

Answer 1

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 `(-/+) = -`