Rewriting cos function as sine function

[Rijad Hadzic]

Homework Statement

So in an AC circuit with inductance, I found

$I = (-ε_{max} cos(ωt)) / ωL$

I want to rewrite this as a sine function.

Homework Equations

The Attempt at a Solution

I can ignore $ε_{max} / ωL$ and work with -cos(ωt)

I set ωt = θ

I know cos(θ) = sin(θ+pi/2)

so -cosθ = -sin(θ+pi/2)

-sinθ = sin(-θ)

so shouldnt

-sin(θ+pi/2) = sin (-θ-pi/2) ?

but my book rewrote cos(ωt) as sin(ωt - pi/2)

How can you write -θ-pi/2 as θ-pi/2?

My hunch is telling me the step

-cosθ = -sin(θ+pi/2)
-cosθ = sin(-θ-pi/2)

is where I made my mistake but I don't understand why.

 

[Rijad Hadzic]

Is it perhaps I used the wrong identity to start with?

instead of using
cos(θ) = sin(θ+pi/2)

If I use sin(pi/2 - θ) = cosθ

then sin(- (-pi/2 + θ) ) = cosθ

then -sin(-pi/2 + θ) = cosθ

then sin(-pi/2 + θ) = -cosθ

does this method seem correct?

It seems logical to me now, but I'm still having doubts. Why would one identity work and the other wouldn't? Makes no sense to me.

 

[Rijad Hadzic]

Basically I think my question boils down to:

why are sin(x - pi/2), sin(-x -pi/2) and -sin(x+pi/2) all equal to each other?

if -sin(x) = sin(-x) wouldn't that imply -sin(x+pi/2) and sin(-x -pi/2) are equal to each other, but not sin(x - pi/2)?

 

[cnh1995]

if -sin(x) = sin(-x) wouldn't that imply -sin(x+pi/2) and sin(-x -pi/2) are equal to each other, but not sin(x - pi/2)?

Yes. You can verify it using sin(A±B)=sinAcosB±cosAsinB.

(And, this question belongs to the math homework section).