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Sine in algebra

  • Thread starter Ry122
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565
2
in bsin30-85.7sin(theta)=0
How can I make theta the subject?
 

Answers and Replies

Dick
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Solve for sin(theta) and use arcsin.
 
565
2
how is that done?
 
1,750
1
how is that done?
It's the inverse.

Recall ...

[tex]y=x^2,x\geq0[/tex]

[tex]f^{-1}(x)=\sqrt x[/tex]

So similarly ...

[tex]\sin x=y[/tex]

[tex]\sin^{-1}y=x[/tex]
 
565
2
so is this correct
sin(theta)=-bsin30/-85.7
1/sin(theta)=(-85.7)/(-binsin30)


sin^-1(-bsin30/-85.7)=(theta)
 
Last edited:
Dick
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That doesn't make it the subject, now does it? If sin(theta)=a, then theta=arcsin(a). That's the definition of arcsin.
 
1,750
1
I forgot to add that arcsine is defined in Quadrants I & IV. It's domain is: [tex]-\frac{\pi}{2}\leq y\leq\frac{\pi}{2}[/tex].
 
1,750
1
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 ... [tex]\frac{1}{x}=x^{-1}[/tex]

It's not the same, the reciprocal of sine is cosecant. Refer to Dick's post.
 
Dick
Science Advisor
Homework Helper
26,258
618
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.
 
565
2
That's it. arcsin(b*sin(30)/85.7)=theta.
How do you work this out on a calculator?
 
Dick
Science Advisor
Homework Helper
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618
To work it out on a calculator you need to know b, right? The arcsin function is usually labeled sin^(-1).
 

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