Homework Help: Sine in algebra

1. Apr 10, 2008

Ry122

in bsin30-85.7sin(theta)=0
How can I make theta the subject?

2. Apr 10, 2008

Dick

Solve for sin(theta) and use arcsin.

3. Apr 10, 2008

Ry122

how is that done?

4. Apr 10, 2008

rocomath

It's the inverse.

Recall ...

$$y=x^2,x\geq0$$

$$f^{-1}(x)=\sqrt x$$

So similarly ...

$$\sin x=y$$

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

5. Apr 10, 2008

Ry122

so is this correct
sin(theta)=-bsin30/-85.7
1/sin(theta)=(-85.7)/(-binsin30)

sin^-1(-bsin30/-85.7)=(theta)

Last edited: Apr 10, 2008
6. Apr 10, 2008

Dick

That doesn't make it the subject, now does it? If sin(theta)=a, then theta=arcsin(a). That's the definition of arcsin.

7. Apr 10, 2008

rocomath

I forgot to add that arcsine is defined in Quadrants I & IV. It's domain is: $$-\frac{\pi}{2}\leq y\leq\frac{\pi}{2}$$.

8. Apr 10, 2008

rocomath

I think what you did was implied something like ... $$\frac{1}{x}=x^{-1}$$

It's not the same, the reciprocal of sine is cosecant. Refer to Dick's post.

9. Apr 10, 2008

Dick

Yes. That's it. arcsin(b*sin(30)/85.7)=theta.

10. Apr 11, 2008

Ry122

That's it. arcsin(b*sin(30)/85.7)=theta.
How do you work this out on a calculator?

11. Apr 11, 2008

Dick

To work it out on a calculator you need to know b, right? The arcsin function is usually labeled sin^(-1).