Question

PLEASE ANSWER ASAP!!!!

Sin(theta/2)=1/2 I. Write an equation that expresses the value of (theta/2) in terms of an appropriate inverse trigonometric expression. II. On the interval [0, 2pi], what values of (theta/2) satisfy your equation in Part I? III. On the interval [0, 2pi], what values of theta does your answer to Part II produce? IV. Write an expression for all solutions to the equation.

Answer 1

We'll use the notation
`\textrm{Arcsin } x` for the principal value and
`\arcsin x` for all the values:

`\arcsin x = \textrm{Arcsin } x + 2\pi k \textrm{ or } (\pi - \textrm{Arcsin }x) + 2\pi k \quad \textrm{ integer } k`

Part I

`\sin(\theta/2) = \frac 1 2`

`\theta/2 = \arcsin \frac 1 2`

Part II. In our range we can write

`\sin(\theta/2) = \sin(\frac \pi 6)`

`\theta/2 = \frac \pi 6 \textrm{ or } \theta /2 = \pi - \frac \pi 6 = (5\pi)/(6)`

Part III.

`\theta = \frac \pi 3 \textrm{ or } \theta = (5\pi)/(3)`

Part IV.

`\theta/2 = \frac \pi 6 + 2\pi k \textrm{ or } \theta = (5 \pi)/(6) + 2\pi k \quad \textrm{integer } k`

`\theta = \frac \pi 3 + 4 \pi k \textrm{ or } \theta = (5 \pi)/(3) + 4\pi k \quad \textrm{integer } k`