Hi, Can anyone give me an idea how can I solve this?
Find sinά, if sin(45- ά)= - 2/3
I'd appreciate your help.Thanks
The arcsin function is also written $\displaystyle \sin^{-1}(x).$ It is the functional inverse of sin. Generally speaking, if you have $\displaystyle \sin(x)=y,$ then $\displaystyle x=\arcsin(y)=\sin^{-1}(y),$ modulo the symmetries of the sin function. You follow?
It's just like "undoing" any other function. To undo the exponential function, you might have this:
$\displaystyle e^{x}=y$
$\displaystyle \ln(e^{x})=\ln(y)$ (applied logarithm to both sides)
$\displaystyle x=\ln(y).$ (simplified)
Same thing here:
$\displaystyle \sin(x)=y$
$\displaystyle \sin^{-1}(\sin(x))=\sin^{-1}(y)$
$\displaystyle x=\sin^{-1}(y).$
So how could you use this in your situation?
The idea is sound, but you've made a mistake or two in carrying it out. Going from
$\displaystyle \displaystyle\frac{\sqrt{2}}{2}[\cos(\alpha)-\sin(\alpha)]=-\frac{2}{3}$ to
$\displaystyle \displaystyle\cos(\alpha)-\sin(\alpha)=\frac{2\sqrt{2}}{3},$
you lost the minus sign. You should have gotten
$\displaystyle \displaystyle\cos(\alpha)-\sin(\alpha)=-\frac{2\sqrt{2}}{3}.$
This is a mistake that, due to another sign error, you later correct! This is a bit sloppy.
So, I'd say I agree with your last step, but only as the result of two self-correcting sign errors.
I should also point out that you have done one (what we call) irreversible step, when you squared the equation. Squaring an equation can sometimes introduce spurious results. I think in your case it does. So, when you get your answers, double-check them against your original equation, and make sure they satisfy the original equation. You should always double-check the answers to problems anyway, to make sure they make sense. Make sense?