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

Ok so I find that the Edexcel Endorsed textbook doesn't really prepare you enough for these types of questions in the actual exam.

I understand questions that are presented in these videos and questions that are like...
Q) If 3sinθ+4cosθ≡5sin(θ+53)∘3sin \theta + 4cos \theta \equiv 5sin (\theta + 53)^{\circ} solve 3sinθ+4cosθ=33sin \theta + 4cos\theta = 3 and find the maximum and minimum values of y y and the first positive value of θ \theta for which this occur.

But I don't always understand the more complicated fractional types that are like...
Q) If 2cosθ+sinθ≡5cos(θ−26.6) 2cos \theta + sin \theta \equiv \sqrt{5} cos( \theta - 26.6) solve 22cosθ+sinθ−1=15 \frac{2}{2cos \theta + sin \theta -1} = 15 and find the maximum and minimum values and the first positive value of x x for which these occur.

Does anyone have some complex questions that they can guide me through? Or generally just help me?

P.s sorry for the dodgy questions that I made up to use as examples :colondollar:

Thanks in advance! :smile:
(edited 6 years ago)

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Original post by Philip-flop
Ok so I find that the Edexcel Endorsed textbook doesn't really prepare you enough for these types of questions in the actual exam.

I understand questions that are presented in these videos and questions that are like...
Q) If 3sinθ+4cosθ≡5sin(θ+53)∘3sin \theta + 4cos \theta \equiv 5sin (\theta + 53)^{\circ} solve 3sinθ+4cosθ=33sin \theta + 4cos\theta = 3 and find the maximum and minimum values of y y and the first positive value of θ \theta for which this occur.

But I don't always understand the more complicated fractional types that are like...
Q) If 2cosθ+sinθ≡5cos(θ−26.6) 2cos \theta + sin \theta \equiv \sqrt{5} cos( \theta - 26.6) solve 22cosθ+sinθ−1=15 \frac{2}{2cos \theta + sin \theta -1} = 15 and find the maximum and minimum values and the first positive value of x x for which these occur.

Does anyone have some complex questions that they can guide me through? Or generally just help me?

Thanks in advance! :smile:


What's your attempt of the complex question you've listed?
Original post by Kevin De Bruyne
What's your attempt of the complex question you've listed?


Normally I would tackle these fractional types similar to this...
Harmonic Identities Max and Min values example.jpg

But I know that the question I've stated above is more complex than that since I'm given two graphs (a cos graph, and a straight horizontal line). This means I'm going to have to rearrange the equation...
22cosθ+sinθ−1=15 \frac{2}{2cos \theta + sin \theta -1} = 15

to give...
cos(θ−26.6)=17155 cos( \theta - 26.6) = \frac{17}{15 \sqrt{5}}

From this, I can only see that the maximum value for the graph cos(θ−26.6) cos ( \theta - 26.6) will be cut off around the 0.5 mark on the y-axis but the minimum will still remain as -1. But I won't be able to tell you the max and min values for what I'm actually meant to be finding. :frown:
(edited 6 years ago)
Original post by Philip-flop
Normally I would tackle these fractional types similar to this...
Harmonic Identities Max and Min values example.jpg

But I know that the question I've stated above is more complex than that since I'm given two graphs (a cos graph, and a straight horizontal line). This means I'm going to have to rearrange the equation...
22cosθ+sinθ−1=15 \frac{2}{2cos \theta + sin \theta -1} = 15

to give...
cos(θ−26.6)=17155 cos( \theta - 26.6) = \frac{17}{15 \sqrt{5}}

From this, I can only see that the maximum value for the graph cos(θ−26.6) cos ( \theta - 26.6) will be cut off around the 0.5 mark on the y-axis but the minimum will still remain as -1. But I won't be able to tell you the max and min values for what I'm actually meant to be finding. :frown:


Rightly so, because the equation in the question is not one that varies!

Because it just says = 15, you can't say anything about minima and maxima. If that were = Y then you would have minimum and maximum values but it's just not a feature of the question as you've written it - can you see why?
(edited 6 years ago)
Original post by Kevin De Bruyne
Rightly so, because the equation in the question is not one that varies!

Because it just says = 15, you can't say anything about minima and maxima. If that were = Y then you would have minimum and maximum values but it's just not a feature of the question as you've written it - can you see why?


What do you mean by the question that I've used "is not one that varies" ?

But what if if was written as y=22cosθ+sinθ−1−15 y = \frac{2}{2cos \theta + sin \theta -1} - 15 ? I don't think I can see why :frown:

Or perhaps without the -1 in there so just.. y=22cosθ+sinθ−15 y = \frac{2}{2cos \theta + sin \theta} - 15 ???
(edited 6 years ago)
Original post by Philip-flop
What do you mean by the question that I've used "is not one that varies" ?

But what if if was written as y=22cosθ+sinθ−1−15 y = \frac{2}{2cos \theta + sin \theta -1} - 15 ? I don't think I can see why :frown:

Or perhaps without the -1 in there so just.. y=22cosθ+sinθ−15 y = \frac{2}{2cos \theta + sin \theta} - 15 ???

That's a fair what if but not what you wrote so it's another issue.

My point is that you can't find a minimum of

22cosθ+sinθ−1=15 \frac{2}{2cos \theta + sin \theta -1} = 15

(What you wrote in your post)

Because you're not sure what to find, because it does not exist.

There is no meaningful minimum or maximum in the equation above as there is no variable output, only a variable input which is theta.
Original post by Kevin De Bruyne
That's a fair what if but not what you wrote so it's another issue.

My point is that you can't find a minimum of

22cosθ+sinθ−1=15 \frac{2}{2cos \theta + sin \theta -1} = 15

(What you wrote in your post)

Because you're not sure what to find, because it does not exist.

There is no meaningful minimum or maximum in the equation above as there is no variable output, only a variable input which is theta.

I guess I'm just having trouble distinguishing between the different types.

Starting to understand a bit now that you said "there is no variable output"

Can you recommend a few questions for me to get used to using harmonic identities in this way? I've done all the questions related to this topic from the Edexcel textbook but they seem much simpler :frown:
Original post by Philip-flop
I guess I'm just having trouble distinguishing between the different types.

Starting to understand a bit now that you said "there is no variable output"

Can you recommend a few questions for me to get used to using harmonic identities in this way? I've done all the questions related to this topic from the Edexcel textbook but they seem much simpler :frown:


Mate you're confusing a function that has a variable outcome with a function that has a fixed output.

By saying f(x)=constant you are basically allowing only certain values of x which satisfy the equation.

By saying f(x)=variable (pick y) then you are allowing for any x to go into the function to give you your y. So for certain f there is a point where they attain their max/min values of the output variable.

Anyway, what you've said above is kinda like saying "I've tried finding the min value of x2=1x^2=1." It just doesnt make sense.
(edited 6 years ago)
Original post by RDKGames
Mate you're confusing a function that has a variable outcome with a function that has a fixed output.

By saying f(x)=constant you are basically allowing only certain values of x which satisfy the equation.

By saying f(x)=variable (pick y) then you are allowing for any x to go into the function to give you your y. So for certain f there is a point where they attain their max/min values of the output variable.

Anyway, what you've said above is kinda like saying "I've tried finding the min value of x2=1x^2=1." It just doesnt make sense.

I guess it's just another flaw in my knowledge about the subject. This is something that seems to hold me back quite a bit especially where I seem to have skipped over some of the very basics of Maths (downside to being self-taught after having not studied Maths for so many years). Sorry if what I say next may sound a little daft but...

Is x2=1 x^2 = 1 different to y=x2−1 y = x^2 - 1 ?

Does x2=1 x^2 = 1 basically mean there are two graphs, a parabola and a straight horizontal line, which intersect each other at two points meaning that the variable x can only be x=−1,x=1 x = -1, x = 1 ?

But then y=x2−1 y = x^2 -1 means a parabola (that has been translated and crosses through the y-axis at -1) where x x can be any real number to give an output of y?
Original post by Philip-flop
I guess it's just another flaw in my knowledge about the subject. This is something that seems to hold me back quite a bit especially where I seem to have skipped over some of the very basics of Maths (downside to being self-taught after having not studied Maths for so many years). Sorry if what I say next may sound a little daft but...

Is x2=1 x^2 = 1 different to y=x2−1 y = x^2 - 1 ?

Does x2=1 x^2 = 1 basically mean there are two graphs, a parabola and a straight horizontal line, which intersect each other at two points meaning that the variable x can only be x=−1,x=1 x = -1, x = 1 ?

But then y=x2−1 y = x^2 -1 means a parabola (that has been translated and crosses through the y-axis at -1) where x x can be any real number to give an output of y?


Thats right
Original post by RDKGames
Thats right


Thank god for that. I'm just not sure why I'm struggling so much with these Min/Max values using Harmonic Identities :frown:

Do you know of any decent questions that can test my overall knowledge with this?
Original post by Philip-flop
Thank god for that. I'm just not sure why I'm struggling so much with these Min/Max values using Harmonic Identities :frown:

Do you know of any decent questions that can test my overall knowledge with this?


Hopefully you can see why you won't be asked a question like that which doesn't have an answer - I think you just created an example and then confused yourself.

You don't really need to know much else so just practice questions where they ask for min/max and finding trig solutions and that's it..
Original post by Kevin De Bruyne
Hopefully you can see why you won't be asked a question like that which doesn't have an answer - I think you just created an example and then confused yourself.

You don't really need to know much else so just practice questions where they ask for min/max and finding trig solutions and that's it..


Yeah I think I did confuse myself a little bit haha. Thanks for helping me though, I really appreciate it!
So these are the types of questions I seem to struggle with (part c). This has been taken from the Edexcel C3 June 2014 paper (Question 9).

C3 June 2014 Q9 - Harmonic Identities Min or Max values.png

I managed to work out part c)i) of the question but have been struggling with c)ii).

I don't understand where the 0,π,2π,3π 0, \pi, 2 \pi, 3 \pi has come from? Can someone please explain this? :frown:
Reply 14
Avatar for Notnek
Notnek
21
Original post by Philip-flop
So these are the types of questions I seem to struggle with (part c). This has been taken from the Edexcel C3 June 2014 paper (Question 9).

C3 June 2014 Q9 - Harmonic Identities Min or Max values.png

I managed to work out part c)i) of the question but have been struggling with c)ii).

I don't understand where the 0,π,2π,3π 0, \pi, 2 \pi, 3 \pi has come from? Can someone please explain this? :frown:

You have the equation sin⁡(3θ−1.1071)=0\sin (3\theta - 1.1071) = 0 and the first step to solving this is to do inverse sine and CAST (or use other methods) to get:

3θ−1.1071=π,2π,...3\theta - 1.1071 = \pi, 2\pi, ...

Just like you would do in a C2 trig question. Are you happy where the equation sin⁡(3θ−1.1071)=0\sin (3\theta - 1.1071) = 0 came from?
Original post by notnek
You have the equation sin⁡(3θ−1.1071)=0\sin (3\theta - 1.1071) = 0 and the first step to solving this is to do inverse sine and CAST (or use other methods) to get:

3θ−1.1071=π,2π,...3\theta - 1.1071 = \pi, 2\pi, ...

Just like you would do in a C2 trig question. Are you happy where the equation sin⁡(3θ−1.1071)=0\sin (3\theta - 1.1071) = 0 came from?


Ok I kind of glossed over the sin(3θ−1.1071...)=0 sin(3 \theta - 1.1071...) = 0 since I didn't really know where it's come from, from our original equation which was H(θ)=4+5[20sin(3θ−1.1071...)]2 H( \theta) = 4 + 5[ \sqrt{20} sin(3 \theta - 1.1071...)]^2

From this I worked out the minimum value of H(θ)=4+5[20(0)]2=4 H (\theta) = 4 + 5 [\sqrt{20}(0)]^2 = 4

But I'm a bit confused with where to go from there :frown: Why do we set sin(3θ−1.1071...)=0 sin(3 \theta - 1.1071...) = 0 ?
Original post by Philip-flop
Ok I kind of glossed over the sin(3θ−1.1071...)=0 sin(3 \theta - 1.1071...) = 0 since I didn't really know where it's come from, from our original equation which was H(θ)=4+5[20sin(3θ−1.1071...)]2 H( \theta) = 4 + 5[ \sqrt{20} sin(3 \theta - 1.1071...)]^2

From this I worked out the minimum value of H(θ)=4+5[20(0)]2=4 H (\theta) = 4 + 5 [\sqrt{20}(0)]^2 = 4

But I'm a bit confused with where to go from there :frown: Why do we set sin(3θ−1.1071...)=0 sin(3 \theta - 1.1071...) = 0 ?


You know the minimum value occurs when sin⁡(3θ−1.1071...)=0\sin(3\theta-1.1071...)=0

So the question asks for the largest value of θ\theta in the range 0≤θ<π0 \leq \theta < \pi when this. So you solve the equation above to get the values of θ\theta in that range and pick the largest one.
Original post by RDKGames
You know the minimum value occurs when sin⁡(3θ−1.1071...)=0\sin(3\theta-1.1071...)=0

So the question asks for the largest value of θ\theta in the range 0≤θ<π0 \leq \theta < \pi when this. So you solve the equation above to get the values of θ\theta in that range and pick the largest one.


So something along the lines of?...
Photo 19-05-2017, 13 20 35.jpg
Original post by Philip-flop
So something along the lines of?...


Seems right.
Reply 19
Avatar for Notnek
Notnek
21
Original post by Philip-flop
So something along the lines of?...

So do you now know where sin⁡(3θ−1.1071...)=0\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.

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