Rush147 said:
I believe so as this gives "sin(a + b)" (or in my case sin(2x + x)) and this is equal (or can then be substituted with the identity) "sinacosb + cosasinb", BUT i really don't know if I'm heading the correct way as I've only just this week come across trigonometrical identities so it could be totally different. I know the last equation i did gave me a double angle (2x) which gave 2 angles within 360 degrees. Any help you can give would be most appreciated.
There are actually 3 ways to solve the above problem.
1. The first way, the most straightforward, and require the most calculation is to trace 3x down to x, by using
Triple-Angle Formulae: sin(3x) = 3 sin(x) - 4sin
3(x) (You can arrive to this formula by using the
Sum-Angle Identity twice). So, your equation becomes
a cubic equation in sin(x), which is pretty easy to solve. :)
2. The second way, the easiest way, is to use the properties of sin function:
\sin \alpha = \sin \beta
\Leftrightarrow \left[ \begin{array}{lcr} \alpha & = & \beta + k 360 ^ o \\ \alpha & = & 180 ^ o - \beta + k' 360 ^ o \end{array} \right., k, and k' are both integers.
One can isolate sin(
x) to the other side of the equation: sin(3
x) = sin(
x), and use method mentioned above. Then choose, k, and k' wisely so that your solution is on the interval [0, 360]
3. The 3rd way, the final one, is to use the Sum-To-Product Identities:
(You can arrive to this Itentity by using
neutrino's hint)
\sin \alpha - \sin \beta = 2 \cos \left( \frac{\alpha + \beta}{2} \right) \sin \left( \frac{\alpha - \beta}{2} \right)
Hopefully, you can go from here. :)
You can pick up one of the 3 ways mentioned above, or try all 3, and compare the result. :)