Maths C3 - Harmonic Identities - Finding Max and Min Values... HELP???

Seems right.

So do you now know where $\sin(3\theta-1.1071...)=0$ comes from? I'm asking because this hasn't been explained to you since you said that you don't know where it comes from.

Just want to thank you guys (@Kevin De Bruyne, @notnek and @RDKGames) for helping me out and being patient with me with a topic that I always seemed to struggle with! Hopefully it makes more sense to me from now on.

Just gonna say it in case he doesn't...

Your expression is $H(\theta)=4+5(\sqrt{20}\sin(3 \theta +\alpha))^2=4+100\sin^2(3 \theta+ \alpha)$

To minimise $H$, you need to minimise $\sin^2(3\theta+\alpha)$. Clearly the minimum value of this is 0, and it is achieved when $\sin(3\theta+\alpha)=0$ (because if it were equal anything other than 0, the square of it would not be 0)

Yeah so the expression is...
$H(\theta)=4+5[\sqrt{20}\sin(3 \theta +\alpha)]^2$

And to find it's minimum value we let the $sin(3 \theta - 1.10714...) = 0$ because we want it to give the smallest possible value for this minimum, thus giving...
$H(\theta) = 4+5 [\sqrt{20} (0) ]^2 = 4$

Right?

You're not wrong, but I don't like the way your argument goes from $\sin$ to $\sin^2$, rather than the other way round.

The minimum value of $\sin$ is clearly -1 when minimised but that would mean $\sin^2$ is 1.

But if you start your argument by minimising $\sin^2$, so it is 0, then $\sin$ is 0.

Oh I see, so it's better to expand the bracket first so...
$H(\theta)=4+5[\sqrt{20}\sin(3 \theta +1.10714...)]^2$

Becomes...
$H(\theta)=4+100sin^2(3 \theta +1.10714...)$

$H(\theta)=4+100sin(3 \theta +1.10714...)^2$

So for the minimum value $sin(3 \theta + 1.10714...) = 0$

Therefore min value of $H(\theta) = 4 + 100(0)^2 = 4$

Is that the way I should do it?

$H(\theta)=4+100sin(3 \theta +1.10714...)^2$

So for the minimum value $sin(3 \theta + 1.10714...) = 0$

All in all, this bit just lacks argument as to WHY $\sin(3 \theta + 1.10714...)$ is picked as 0 and not its minimum which would be -1.

I guess, there's nothing wrong with it... but again, you go from $\sin$ to $\sin^2$ when it makes more sense to do it the other way around since $H(\theta)$ is dependent on $\sin^2$, so you want THIS minimised which means $\sin^2 =0$ and this IMPLIES $\sin =0$

Maybe I'm just being pedantic here and I doubt you would lose ANY marks for this, but to minimise $H$ you should notice that first saying "min of $\sin$" then going onto "min of $\sin^2$" doesn't flow very nicely because the min of $\sin$ is -1 (not 0 as you claim) and this value does NOT minimise $H$. Whereas if you went from saying first "min of $\sin^2$" (which is 0 and minimised H first of all) IMPLIES "min of $\sin$" (which is also 0 as a result) then it makes sense. It's more logical this way.

All in all, this bit just lacks argument as to WHY $\sin(3 \theta + 1.10714...)$ is picked as 0 and not its minimum which would be -1.

I don't think I completely follow I don't even know what the minimum of $sin^2$ is

I don't think I completely follow I don't even know what the minimum of $sin^2$ is

All I can see is that it has to be $sin(3 \theta - 1.10714...) = 0$ since if it is -1 we would get $H(\theta) = 4+100(-1)^2 = 104$ which is a bigger number than if we have it as 0, therefore $H(\theta) = 4 + 100(0)^2 = 4$ which is the smallest possible value we can make for the minimum.

Philip-flop Just adding to what RDKGames said here:

A large percentage of students would think that $\sin(3 \theta + 1.10714...)$ should be -1 to have a minimum because this is what they're used to in most exam questions. It's important that you understand why in this case it should be 0.

You REALLY need to know that the square of anything that can be 0, IS 0...

To me this looks like you understand.

$\sin^2 x = (\sin x)^2$

The minimum of $\sin x$ is -1 but the minimum of $(\sin x)^2$ is 0 for the reasons you say above.

In this case you have $sin(3 \theta - 1.10714...)^2$ and the minmum of this will be the same as the minimum of $(\sin x)^2$.