Question

How do you use the half angle formula to solve `Tan( θ/2) = Sin θ`?

Answer 1

`theta = 2kpi and theta = (2k+-1/2)pi, k = 0, 1, 2, 3, ...`.

In `[0, 2pi], theta = 0, pi/2, 3/2pi and 2pi`.

Explanation:

Use `sin theta = (2 tan (theta/2))/(1+tan^2(theta/2))`

Here,

`tan(theta/2)=sin theta = (2 tan (theta/2))/(1+tan^2(theta/2)`

So, `tan( theta/2)((1+tan^2(theta/2)-2)=0`

And so, `tan(theta/2)=0` gives `theta/2=kpi, k=0, 1, 2, 3, ..`

the other factor = 0 gives# tan^(theta/2)=1 to tan(theta/2)=+-1 to

theta/2=kpi+-pi/4#.

Combining both for values of `theta`,.

`theta = 2kpi and theta = (2k+-1/2)pi, k = 0, 1, 2, 3, ...`

Answer 2

`pi/4; (3pi)/4; (5pi)/4; (7pi)/4`

Explanation:

Use the trig identity
sin a = 2sin (a/2).cos (a/2)
Re-write the equation:
`(sin (t/2))/(cos (t/2)) = 2 sin(t/2).cos (t/2)`
Simplify both sides by sin (t/2(, we get:
`1/(cos (t/2)) = 2cos (t/2)`
`2cos^2 (t/2) = 1`
`cos^2 (t/2) = 1/2`
`cos (t/2) = +- 1/sqrt2 = +- sqrt2/2`
Trig table of special arcs, and trig unit circle give 4 solution arcs:
a. `cos x = sqrt2/2` --> `x = +- pi/4`
Two solution arcs:
`x = pi/4` and `x = (7pi)/4` (arc co-terminal to arc `(-pi/4)`)
b. `cos x = - sqrt2/2` --> `x = +- (3pi)/4`
Two solution arcs:
`x = (3pi)/4`, and `x = (5pi)/4` (arc co-terminal to arc `(-3pi)/4`)
Answers for `(0, 2pi)`
`pi/4; (3pi)/4, (5pi)/4, (7pi)/4`