1. ## The double angle

Hey guys, was just revising some trig work and came across a problem.

If you were told that sinx is pie/2 < x < pie, is sin3x still the same domain, or would it become 3pie/2 < 3x < 3pie. If this is the case then does that mean sin3x can have 2 values, one positive one negative?

2. Given that the domain for sin x is $\displaystyle \frac {\pi}{2}<x<\pi \Rightarrow$ the domain for sin 3x will be $\displaystyle \frac {3\pi}{2}<3x<3\pi$

Lets say that when you solve for sin x, you get x = $\displaystyle \alpha, \beta$ for the domain $\displaystyle \frac {\pi}{2}<x<\pi$

Now to find for 3x, this means 3 revolutions, that is,

1st revolution, x = $\displaystyle \alpha, \beta$
2nd revolution, 2x = $\displaystyle \alpha, \beta, \alpha+2\pi, \beta+2\pi$ , (where 2\pi radians is one revolution or 360 degrees.)
3rd revolution, 3x = $\displaystyle \alpha, \beta, \alpha+2\pi, \beta+2\pi, \alpha+2\pi+2\pi, \beta+2\pi+2\pi$

So this 3rd revolution, that is for 3x, would be for the domain $\displaystyle \frac {3\pi}{2}<3x<3\pi$

To find once again for x all you need to do is divide all the six answers that was found in the 3rd revolution by 3, ensuring that it now lies in the domain $\displaystyle \frac {\pi}{2}<x<\pi$.