Question

Find all angles 0 (theta) such that zero degree is less than or equal to theta and less than or equal to 360 degrees for which the statement sin (-104 degrees) = sin 0 theta is true

Answer 1

Explanation

So basically we must find an angle between 0° and 360° that meets the following:

$\sin(-104^{\circ})=\sin\theta$

First of all it's important to recall a property of the sine:

$\sin x=\sin(x+360^{\circ})$

Then if we take x=-104° we get:

$\sin(-104^{\circ})=\sin(-104^{\circ}+360^{\circ})=\sin256^{\circ}$

So for now we know that 256° is one value of theta.

For all the angles between 0° and 360° there are always two angles that share the same sine. For any angle in the first quadrant (0° to 90°) there's another angle in the second quadrant (90° to 180°) that has the same sine. The same happens with the third (180° to 270°) and fourth quadrant (270° to 360°). 256° is in the third quadrant which means that there's another possible theta in the fourth quadrant. The difference between this angle and 270° must have the same absolute that the difference between 256° and 270°. Then for the new theta we have:

$\begin{gathered} \lvert{270^{\circ}}-256^{\circ}\rvert=\lvert{\theta-270^{\circ}}\rvert \\ 14^{\circ}=\lvert{\theta-270^{\circ}}\rvert \end{gathered}$

Since theta is part of the fourth quadrant is greater than 270° so the term inside the absolute value is positive and we can get rid of the module:

$14^{\operatorname{\circ}}=\theta-270^{\circ}$

Then we add 270° to both sides:

$\begin{gathered} 14^{\operatorname{\circ}}+270^{\operatorname{\circ}}=\theta-270^{\operatorname{\circ}}+270^{\operatorname{\circ}} \\ \theta=284^{\circ} \end{gathered}$Answer

Then the answers are 256° and 284°.