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A-Level Maths Proof question

Hey.

Could anyone help me solve the below please?

Using algebra, prove that for all real values x
(x-2)^2 > 8x -33 + 2 sin x cos x

Would you use substitute 2 sin x cos x for sin 2x, and then since you know that the max. and min. values of sin 2x (-1 and +1), rearrange to find x?

Edit: Probably should mention that the question is only 4 marks
(edited 4 years ago)

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Original post by Kaz_A-Level_2019
Hey.

Could anyone help me solve the below please?

Using algebra, prove that for all real values x
(x-2)^2 > 8x -33 + 2 sin x cos x

Would you use substitute 2 sin x cos x for sin 2x, and then since you know that the max. and min. values of sin 2x (-1 and +1), rearrange to find x?

Actually, that doesn't work because there are no real solutions XD

Can anyone offer any guidance?

Thanks in advance :smile:
Reply 2
Original post by Kaz_A-Level_2019
Hey.

Could anyone help me solve the below please?

Using algebra, prove that for all real values x
(x-2)^2 > 8x -33 + 2 sin x cos x

Would you use substitute 2 sin x cos x for sin 2x, and then since you know that the max. and min. values of sin 2x (-1 and +1), rearrange to find x?


Yes. Complete the square on the left and show it's > 1 (the trig terms)
Original post by mqb2766
Yes. Complete the square on the left and show it's > 1 (the trig terms)

Ahh yes!

Thank you so much!!
Original post by mqb2766
Yes. Complete the square on the left and show it's > 1 (the trig terms)

Does this look correct?

Or do I actually have to sub in the six and show that ... > ... before saying that last line?
Reply 5
Original post by Kaz_A-Level_2019
Does this look correct?

Or do I actually have to sub in the six and show that ... > ... before saying that last line?

Just about. Id probably prove that the quadratic was >= 1, with equality when x=6.
Then bring the trig terms in and show they were <= 1, and strictly < 1 when x=6. Then bring the two parts together.
Original post by mqb2766
Just about. Id probably prove that the quadratic was >= 1, with equality when x=6.
Then bring the trig terms in and show they were <= 1, and strictly < 1 when x=6. Then bring the two parts to you do together.

Why wouldn't you do that with -1?
Reply 7
Original post by Kaz_A-Level_2019
Why wouldn't you do that with -1?

-1 < 1
Unless I'm missing something?
Original post by mqb2766
-1 < 1
Unless I'm missing something?

Oh, so you have to prove that the quadratic is bigger than the higher possible value of sin 2x (i.e. 1) rather than the smaller value (i.e. -1)?
Original post by Kaz_A-Level_2019
Why wouldn't you do that with -1?

Oh wait, nooo

Did you mean that prove that the quadratic is >1 as well?
Original post by Kaz_A-Level_2019
Oh, so you have to prove that the quadratic is bigger than the higher possible value of sin 2x (i.e. 1) rather than the smaller value (i.e. -1)?

Yes ... Make sure you understand what you've written otherwise you'll have problems.

You're showing:
* the minimum value of the quadratic is >= max value of trig terms
With a bit of reasoning about the equality.
Original post by mqb2766
Yes ... Make sure you understand what you've written otherwise you'll have problems.

You're showing:
* the minimum value of the quadratic is >= max value of trig terms
With a bit of reasoning about the equality.

So, would you do the bit in orange?
Original post by Kaz_A-Level_2019
So, would you do the bit in orange?

Think you've made a mistake on the second orange line down ----> -36+37=1, not -1

If you keep following the working through after that, you should get a subtly different answer.

That being that (x-6)^2>0 which is clearly true other than for x=6, so you needn't sub sqrt2+6 back into your original equation
(edited 4 years ago)
Original post by anonymous1026
Think you've made a mistake on the second orange line down ----> -36+37=1, not -1

If you keep following the working through after that, you should get a subtly different answer.

That being that (x-6)^2>0 which is clearly true other than for x=6, so you needn't sub sqrt2+6 back into your original equation

So, this...?
Original post by Kaz_A-Level_2019
So, this...?

Yeah that's correct, but I believe you're doing unnecessary extra working at the end - when you sub in 6 on both sides at the end to show that the original statement is true, it's not necessary since you already showed it with all your prior working. Your line which has (x-6)^2>0) is sufficient to show that the original statement is true, without having to sub 6 back in.

But you can do so anyways to check if you'd like :smile:
Original post by anonymous1026
Yeah that's correct, but I believe you're doing unnecessary extra working at the end - when you sub in 6 on both sides at the end to show that the original statement is true, it's not necessary since you already showed it with all your prior working. Your line which has (x-6)^2>0) is sufficient to show that the original statement is true, without having to sub 6 back in.

But you can do so anyways to check if you'd like :smile:

Thank you so much :smile:
Original post by mqb2766
Yes ... Make sure you understand what you've written otherwise you'll have problems.
You're showing:
* the minimum value of the quadratic is >= max value of trig terms
With a bit of reasoning about the equality.

aren't they both equal to one tho?
Reply 17
Original post by aamina.hs
aren't they both equal to one tho?

Not sure what you mean by both ... ?
Original post by mqb2766
Not sure what you mean by both ... ?

Sorry haha mb.
So i expanded the left hand side, rearranged and then completed the square to get (x-6)2 +1, and then changed 2sinxcosx to sin2x.
(x-6)2 +1 > sin(2x)
The lowest value you can get for LHS is 1, and the highest value u can get for RHS is also 1. So surely that means the LHS is greater than or equal to the RHS, not just greater than? Unless I've made a mistake?
Reply 19
Original post by aamina.hs
Sorry haha mb.
So i expanded the left hand side, rearranged and then completed the square to get (x-6)2 +1, and then changed 2sinxcosx to sin2x.
(x-6)2 +1 > sin(2x)
The lowest value you can get for LHS is 1, and the highest value u can get for RHS is also 1. So surely that means the LHS is greater than or equal to the RHS, not just greater than? Unless I've made a mistake?

Thats spot on. When are the two sides (particularly the left) equal to 1?
Note it may be better to leave the > or ... unspecified until you make the final argument.
(edited 8 months ago)

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