What is the Half-Angle Identities?

The half-angle identities are defined as follows:

#\mathbf(sin(x/2) = pmsqrt((1-cosx)/2))#

#(+)# for quadrants I and II
#(-)# for quadrants III and IV

#\mathbf(cos(x/2) = pmsqrt((1+cosx)/2))#

#(+)# for quadrants I and IV
#(-)# for quadrants II and III

#\mathbf(tan(x/2) = pmsqrt((1-cosx)/(1+cosx)))#

#(+)# for quadrants I and III
#(-)# for quadrants II and IV

We can derive them from the following identities:

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

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

#color(blue)(sin(x/2) = pmsqrt((1-cos(x))/2))#

Knowing how #sinx# is positive for #0-180^@# and negative for #180-360^@#, we know that it is positive for quadrants I and II and negative for III and IV.

#cos^2x = (1+cos(2x))/2#

#cos^2(x/2) = (1+cos(x))/2#

#color(blue)(cos(x/2) = pmsqrt((1+cos(x))/2))#

Knowing how #cosx# is positive for #0-90^@# and #270-360^@#, and negative for #90-270^@#, we know that it is positive for quadrants I and IV and negative for II and III.

#tan(x/2) = sin(x/2)/(cos(x/2)) = (pmsqrt((1-cos(x))/2))/(pmsqrt((1+cos(x))/2))#

#color(blue)(tan(x/2) = pmsqrt((1-cos(x))/(1+cos(x))))#

We can see that if we take the conditions for positive and negative values from #sinx# and #cosx# and divide them, we get that this is positive for quadrants I and III and negative for II and IV.