Maximum value of 2sin^x - sinx + 1/3?

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Carman3
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What is the maximum value of 2sin^x - sinx + 1/3
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CheeseIsVeg
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(Original post by Carman3)
What is the maximum value of 2sin^x - sinx + 1/3
Draw it?
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Carman3
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(Original post by CheeseIsVeg)
Draw it?
I dont know how to draw it all together
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IrrationalRoot
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(Original post by Carman3)
What is the maximum value of 2sin^x - sinx + 1/3
Notice that the given expression is just a function of \sin x so the question is equivalent to 'What is the maximum value of 2y^2-y+\frac{1}{3}?' where y is restricted to a certain interval (which you know).
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CheeseIsVeg
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(Original post by Carman3)
I dont know how to draw it all together
Ok, first, in the question did they say to use any particular methods etc?
Is this A-level?
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Carman3
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(Original post by IrrationalRoot)
Notice that the given expression is just a function of \sin x so the question is equivalent to 'What is the maximum value of 2y^2-y+\frac{1}{3}?' where y is restricted to a certain interval (which you know).
You cant factorise the y equation so how do you do it... Plus its a positive quadratic so it wouldnt have a maximum would it?
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NotNotBatman
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First consider the variable factor sin(x)   -1\leq sin(x) \leq 1 usually you would consider sin(x) = 1, sin(x) =0 or sin(x) =-1 to see which one gives the maximum value.
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IrrationalRoot
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(Original post by Carman3)
You cant factorise the y equation so how do you do it... Plus its a positive quadratic so it wouldnt have a maximum would it?
Factorising has nothing to do with locating maxima/minima. You would complete the square to find the maximum/minimum of a quadratic (this is basic C1 knowledge, but not sure what stage you're at in school).

And yes it's a positive quadratic but I did mention that y i.e. \sin x is restricted to a certain interval that you know, so it has a maximum.
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Carman3
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(Original post by IrrationalRoot)
Factorising has nothing to do with locating maxima/minima. You would complete the square to find the maximum/minimum of a quadratic (this is basic C1 knowledge, but not sure what stage you're at in school).

And yes it's a positive quadratic but I did mention that y i.e. \sin x is restricted to a certain interval that you know, so it has a maximum.
Can you show me how to do it i cant get it
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IrrationalRoot
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(Original post by Carman3)
Can you show me how to do it i cant get it
You're trying to maximise the function 2y^2-y+\frac{1}{3} with the restriction that -1 \leq y \leq 1.

So first, complete the square. Then think about which value of y between -1 and 1 (inclusive) is going to maximise the function.
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Carman3
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(Original post by IrrationalRoot)
You're trying to maximise the function 2y^2-y+\frac{1}{3} with the restriction that -1 \leq y \leq 1.

So first, complete the square. Then think about which value of y between -1 and 1 (inclusive) is going to maximise the function.
ok i get it thanks. how do you know if they are talking about the max/min value which would be the vertex or the maximum/min overall value like the above
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IrrationalRoot
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(Original post by Carman3)
ok i get it thanks. how do you know if they are talking about the max/min value which would be the vertex or the maximum/min overall value like the above
It's not about which one they're talking about, it's about what the function is. 2\sin^2 x - \sin x + \frac{1}{3} is not the same function as 2x^2-x+\frac{1}{3} (which is a parabola with a vertex and such). It's just that the maximum of 2\sin^2 x - \sin x + \frac{1}{3} is going to be the same as that of the function 2x^2-x+\frac{1}{3} restricted to -1 \leq x \leq 1, since sine can only take those values.
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some-student
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(Original post by Carman3)
What is the maximum value of 2sin^x - sinx + 1/3
To get the maximum value:
we want the maximum value of 2\mathrm{sin}^2 \, x
we want the minimum value of \mathrm{sin} \, x as we are subtracting it from our maximum of 2\mathrm{sin}^2 \, x
we cannot change \frac{1}{3} as it is a constant

Solution
Spoiler:
Show
0 \leq \mathrm{sin}^2 \, x \leq 1 and hence 0 \leq 2\mathrm{sin}^2 \, x \leq 2, meaning that the maximum value of 2\mathrm{sin}^2 \, x is 2.

As we are subtracting \mathrm{sin} \, x we want its smallest value. -1\leq \mathrm{sin} \, x  \leq 1 and hence we are using the value of \mathrm{sin} \, x to be -1.

We cannot change \frac{1}{3} and hence our maximum value is 2 -(-1) + \frac{1}{3} = \frac{10}{3}
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the bear
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i would use differentiation :dontknow:
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IrrationalRoot
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(Original post by some-student)
To get the maximum value:
we want the maximum value of 2\mathrm{sin}^2 \, x
we want the minimum value of \mathrm{sin} \, x as we are subtracting it from our maximum of 2\mathrm{sin}^2 \, x
we cannot change \frac{1}{3} as it is a constant

Solution
Spoiler:
Show
0 \leq \mathrm{sin}^2 \, x \leq 1 and hence 0 \leq 2\mathrm{sin}^2 \, x \leq 2, meaning that the maximum value of 2\mathrm{sin}^2 \, x is 2.

As we are subtracting \mathrm{sin} \, x we want its smallest value. -1\leq \mathrm{sin} \, x  \leq 1 and hence we are using the value of \mathrm{sin} \, x to be -1.

We cannot change \frac{1}{3} and hence our maximum value is 2 -(-1) + \frac{1}{3} = \frac{10}{3}
Yep good approach for this particular problem but OP should remember that this method does not work in general for similar functions.
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Carman3
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(Original post by some-student)
To get the maximum value:
we want the maximum value of 2\mathrm{sin}^2 \, x
we want the minimum value of \mathrm{sin} \, x as we are subtracting it from our maximum of 2\mathrm{sin}^2 \, x
we cannot change \frac{1}{3} as it is a constant

Solution
Spoiler:
Show
0 \leq \mathrm{sin}^2 \, x \leq 1 and hence 0 \leq 2\mathrm{sin}^2 \, x \leq 2, meaning that the maximum value of 2\mathrm{sin}^2 \, x is 2.

As we are subtracting \mathrm{sin} \, x we want its smallest value. -1\leq \mathrm{sin} \, x  \leq 1 and hence we are using the value of \mathrm{sin} \, x to be -1.

We cannot change \frac{1}{3} and hence our maximum value is 2 -(-1) + \frac{1}{3} = \frac{10}{3}
Thanks...
Dont answer the question from the paper as i havent tried it yet but would you also do the same for this: and similar questions as well

https://s3-eu-west-1.amazonaws.com/w...56dfc5cd_1.pdf

Question 1E
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some-student
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(Original post by IrrationalRoot)
Yep good approach for this particular problem but OP should remember that this method does not work in general for similar functions.
Yeah - I should comment that this problem only happens as when \mathrm{sin}x = -1, it lines up with \mathrm{sin}^2x = 1, and this isn't always the case.
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RogerOxon
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(Original post by IrrationalRoot)
You're trying to maximise the function 2y^2-y+\frac{1}{3} with the restriction that -1 \leq y \leq 1.

So first, complete the square. Then think about which value of y between -1 and 1 (inclusive) is going to maximise the function.
I don't see a need to complete the square here, although it is a technique that you need to know and is useful for sketching the function.

As it's a parabola with a positive y^2 coefficient, you know that the maximum value over any range is going to be at the end of the range. For a symmetric range, the negative y coefficient tells you which end, or simply trying the two values.
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IrrationalRoot
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(Original post by RogerOxon)
I don't see a need to complete the square here, although it is a technique that you need to know and is useful for sketching the function.

As it's a parabola with a positive y^2 coefficient, you know that the maximum value over any range is going to be at the end of the range. For a symmetric range, the negative y coefficient tells you which end, or simply trying the two values.
Yeah that's true, but I was trying to show them explicitly how this works.
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RogerOxon
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(Original post by IrrationalRoot)
Yeah that's true, but I was trying to show them explicitly how this works.
Fair enough.
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