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Posted

Could some kind soul explain the simplification of the following

 

sin(theta) + mucos(theta) =

cos(theta) - musin(theta)

 

tan(theta) + mu

1 - mutan(theta)

 

 

I understand that sin(theta)/cos(theta) = tan(theta), but cannot work out how the 1 - the rest comes from.

 

Thanks

Posted (edited)

What happens if you divide by cosQ ?

 

cosQ - usinQ goes to 1-utanQ no?

 

Also the sinQ + ucosQ will go to tanQ + u.

Edited by DrP
multiple post merged
Posted

Two little things: 1) you can't just "divide by [math]\cos\theta[/math]" You have ensure that whatever operations you do result in no change or that the same change is performed to both sides of the equation. For example, if you multiply a term by 1, that results in no change. And, multiplying by a clever choice of the form of 1 is what you are really doing here: [math]1=\frac{\frac{1}{\cos\theta}}{\frac{1}{\cos\theta}}[/math]. It is a small thing, but, technically you aren't "dividing by [math]\cos\theta[/math]."

 

and 2) you might want to write it in LaTeX next time. It is pretty easy to learn and makes reading the math many, many times easier.

 

[math] \frac{\sin\theta + \mu\cos\theta}{\cos\theta-\mu\sin\theta}[/math]

Posted

Thanks Bignose - I knew you couldn't just divide through across the whole thing by cosQ - but it gives the right result. I was thinking that dividing the top and bottom would be alright as the effect would cancel out. I just wasn't realy sure how to put it (thus the question marks in my post). Also - I think I need to work out how to use LaTeX as well. (I know there is a thread on it here somewhere - I just need to get the time to go through it properly).

Posted

It helps to hover over equations posted by others so you can see how they did it.

 

 

Oh yea - I'd not noticed that - thanks. It seems quite intuitive really.

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