Question

What is the Half-Angle Identities?

Answer 1

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.