# Sine in algebra

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

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Dick
Homework Helper
Solve for sin(theta) and use arcsin.

how is that done?

how is that done?
It's the inverse.

Recall ...

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

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

So similarly ...

$$\sin x=y$$

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

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

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

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

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

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

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

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

sin^-1(-bsin30/-85.7)=(theta)
Yes. That's it. arcsin(b*sin(30)/85.7)=theta.

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

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