C3 trig -- max and min value help!

So, I need help on not a particular question but on understanding max and min values for trig and what value of theta these max and min values occur out, after the rcos(x-a) or rsin(x-a) questions...

I know that:
- max value at which sin(x-a) = 1
- min value at which sin (x-a) = 0
I don't understand why...

Also, what would:
- max value at which cos(x-a) = ?
- min value at which cos(x-a) = ?

Thank you.

No the minimum of sin(anything) is -1. I have a feeling I know why you're confused but you'll need to post the whole question.

Ok, I'm very confused now...

So this is the question(part c) by the way) which gave the answers as sin(x-a) = 0 for the minimum... I don't understand why though...

Also, I was doing another question and it gave cos(x-a) = -1 as the minimum so I'm definitely cofuggled.

Ok, I'm very confused now...

So this is the question(part c) by the way) which gave the answers as sin(x-a) = 0 for the minimum... I don't understand why though...

Also, I was doing another question and it gave cos(x-a) = -1 as the minimum so I'm definitely cofuggled.

I thought it might be this question that confused you.

Notice that the trig expression is squared in this function H. So replacing it with $R\sin ...$ you get something of the form

$\displaystyle H(\theta) = 4 +5\left(R\sin(3\theta - \alpha)\right)^2$

The minimum value of $\sin(3\theta - \alpha)$ is $-1$ so the minimum of the stuff in the brackets is $-R$. This is what you're used to.

But if you square $-R$ you get $R^2$ and it's no longer small anymore. So if you want to find the minimum of $H(\theta)$ then you want the squared expression to be as small as possible. This happens when the stuff in the brackets is equal to 0.

Does that make sense?

The value of $[sin(3\theta - \alpha)]^2$ is always positive, so what is always greater than or equal to 0, as it is squared, so what is the smallest value it can be?

I see that, since the bracket is squared, it could never be a negative. But I don't understand the terms used when you want the expression to be as small as possible?

Part ci) asks you to find the minimum value of $H(\theta)$. In other words it's asking you to find the smallest that $H(\theta)$ could possibly be.

I don't really understand what you mean by the smallest possible value? I see how you have worked it out, but that statement I don't get.

Would that mean, to find the maximum, we have to use the greatest possible value, i.e = 1?

So, in a normal context where nothing is squared...
Rsin(x-a) = 1 -- maximum
Rsin(x-a) = -1 -- minimum??