What is the Half-Angle Identities?

1 Answer
Dec 18, 2014

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.