[macintologist]

Ok I have an equation

0.5 = Sin(4t)


Ok how do I isolate the 4t so that it is separated from Sin?

[deej5871]

Take the inverse sine of both sides (in case you don't know, inverse sine is just the inverse operation of sine, on your calculator it looks like sin to the -1 power). This leaves you with (pi/6)=4t. Then just divide both sides by 4 and there you go. Answer is pi/24.

Edit: That's incomplete. When you take inverse sine you have to take into account the other (infinite number of) angles where sine equals 1/2. It's not just going to be pi/24. I forget how exactly to the do pi*n thing. Anybody else wanna help out?

Edit 2: I think it might be [(12n+5)pi]/24 and [(12n+1)pi]/24. Maybe. Why am I even trying? I don't know...

[macintologist]

What's the difference between 1/sin(x) and sin^-1(x)

[turtle777]

My calculator hampstor says eleventy billion(tm) ?

-t

[Fyre4ce]

Use an inverse trig function, namely arcsin:

arcsin(0.5) = arcsin(sin(4t))

arcsin(0.5) = 4t

t = arcsin(0.5)/4 which is approx 0.1309

Check:

sin(0.1309*4) = 0.5

[Fyre4ce]

Originally Posted by macintologist

What's the difference between 1/sin(x) and sin^-1(x)

1/sin(x) is exactly as it looks - it's 1 divided by the value sin(x). So, if sin(x) = 1/2, then 1/(1/2) = 2.

sin^-1(x) is the inverse of the sine function. Read it as "the angle whose sine is...". It's also called "arcsine." So, what would sin^-1(1) be equal to? You might know that sin(pi/2) = 1, so you can see that intuitively sin^-1(1) = pi/2. But, sin(pi/2 + N*2*pi) = 1 for any integer values of N, so theoretically sin^-1(1) should be equal to pi/2 + N*2*pi. But, the sin^-1 function is limited in range from -pi/2 to +pi/2 to avoid the problem of multiple "outputs." For this reason, it's a good idea to check your answers in the original equations because there could be errors generated.

[wataru]

Originally Posted by macintologist

What's the difference between 1/sin(x) and sin^-1(x)

That is a kind of confusing convention. sin^2(x) is what you'd think it would be, but for any sin^n(x) where n is negative, most people just put it in the denominator. As has already been mentioned, sin^-1(x) actually refers to arcsin(x).

[Ghoser777]

I'd actually use substitution for that problem because it's so simple:

let x = 4t
then 0.5 = sin(x)
therefore x = π/6 + 2πk, 5π/6 + 2πk for even integer's k
Solving for t by using the original substitution, we have
t = π/24 + kπ/2, 5π/24 + kπ/2 for even non-zero integer's k

If t can only be in the range 0 to 2π, then t=π/24, 25π/24, 5π/24, and 27π/24.