# Math Help - Find sinά

1. ## Find sinά

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

2. Have you have any ideas so far?

3. I think it's about applying addition/reduction formulas, but nothing concrete

4. How about applying the arcsin function to both sides? What would that give you?

5. I'm sorry I dont know what arcsin function is:$:$

6. The arcsin function is also written $\sin^{-1}(x).$ It is the functional inverse of sin. Generally speaking, if you have $\sin(x)=y,$ then $x=\arcsin(y)=\sin^{-1}(y),$ modulo the symmetries of the sin function. You follow?

7. Im sorry I havent learned inverse functions..there isnt any other way right..?
Can u please tell me how arcsin function is applied to both sides?

I really appreciate ur help. Thanks

8. It's just like "undoing" any other function. To undo the exponential function, you might have this:

$e^{x}=y$

$\ln(e^{x})=\ln(y)$ (applied logarithm to both sides)

$x=\ln(y).$ (simplified)

Same thing here:

$\sin(x)=y$

$\sin^{-1}(\sin(x))=\sin^{-1}(y)$

$x=\sin^{-1}(y).$

So how could you use this in your situation?

9. 45- ά= arcsin-2/3 am I close..?:\$

10. Right. Now, with your notation, you should make sure you have ά in degrees. What do you get when you solve for it?

11. Sorry for disturbing again, but can you solve that for me please?

Thanks for your patience.I wish my maths techer was like you.

12. If you're doing trig, you should know how to solve an equation like a-x=b, for x. That's what you have here. What do you get?

13. [IMG][/IMG]

I'd like to know if this procedure is allowed in Maths.

Thanks a lot

14. The idea is sound, but you've made a mistake or two in carrying it out. Going from

$\displaystyle\frac{\sqrt{2}}{2}[\cos(\alpha)-\sin(\alpha)]=-\frac{2}{3}$ to

$\displaystyle\cos(\alpha)-\sin(\alpha)=\frac{2\sqrt{2}}{3},$

you lost the minus sign. You should have gotten

$\displaystyle\cos(\alpha)-\sin(\alpha)=-\frac{2\sqrt{2}}{3}.$

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?