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maths trig a level

any help would be appreciated
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(edited 2 years ago)
Reply 1
Original post by abovethecl0uds
any help would be appreciated
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This is clearly related to the question you posted a little time earlier, and will rely on your answer to part (a). Did you find a way to re-write the expression in the denominator in a way that might help here?
Original post by Pangol
This is clearly related to the question you posted a little time earlier, and will rely on your answer to part (a). Did you find a way to re-write the expression in the denominator in a way that might help here?

yep! i'm not sure if this is right though
20211014_094457.jpg
Reply 3
That seems OK, although you have some strange choices of variable names which make things unclear. You should use 2x instead of θ all the way through, and α is a constant rather than a variable, so it should stay as an α, it is the θ that is playing the role of the 2x. You should also have tan α rather than tan x.

But leacing that aside, does it help to consider the same function with your version of the denominator rather than the one printed in the question?
Reply 4
Original post by abovethecl0uds
yep! i'm not sure if this is right though
20211014_094457.jpg

You need to treat 4cos2x and 3sin2x as two separate compound angle equations and work it out from there…
(edited 2 years ago)
Reply 5
Original post by 14mut64
You need to treat 4cos2x and 3sin2x as two separate compound angle equations and work it out from there…

Is that going to help to do this part? I think the harmonic form is the way to go.
Original post by Pangol
That seems OK, although you have some strange choices of variable names which make things unclear. You should use 2x instead of θ all the way through, and α is a constant rather than a variable, so it should stay as an α, it is the θ that is playing the role of the 2x. You should also have tan α rather than tan x.

But leacing that aside, does it help to consider the same function with your version of the denominator rather than the one printed in the question?

Oh I see, I've changed that now thanks for the advice.
If f(x) is 1/ (4cos2x + 3sin2x), which is the reciprocal of 4cos2x + 3sin2x or 5sin(2x+0.927), does f(x) is the inverse function of 4cos2x + 3sin2x ?
And if f(x) is the inverse function, does this mean the domain of the original sinx graph will become the range of the inverse function?
Reply 7
Original post by abovethecl0uds
Oh I see, I've changed that now thanks for the advice.
If f(x) is 1/ (4cos2x + 3sin2x), which is the reciprocal of 4cos2x + 3sin2x or 5sin(2x+0.927), does f(x) is the inverse function of 4cos2x + 3sin2x ?
And if f(x) is the inverse function, does this mean the domain of the original sinx graph will become the range of the inverse function?

You're over-thinking this problem - this is nothing to do with function domains or inverse functions :smile:

If I said what is the range of the function y = 5sin(3x) then the answer would be -5 <= y <= 5 because sin(3x) can take all values between -1 and +1 so 5sin(3x) can take all values between -5 and +5.

You need to apply similar logic here - what values can Rsin(ax + b) take, and so what values can its reciprocal take?
Original post by davros
You're over-thinking this problem - this is nothing to do with function domains or inverse functions :smile:

If I said what is the range of the function y = 5sin(3x) then the answer would be -5 <= y <= 5 because sin(3x) can take all values between -1 and +1 so 5sin(3x) can take all values between -5 and +5.

You need to apply similar logic here - what values can Rsin(ax + b) take, and so what values can its reciprocal take?

5sin(2x+0.927) has a range of -5 <= y <= 5
So 1 / (5sin(2x+0.927)) has a range of -1/5 <= y <= 1/5 ?
But that can't be right because if that were true, the function would be 1/5 (5sin(2x+0.927))
Reply 9
(edited 2 years ago)
Reply 10
Original post by abovethecl0uds
5sin(2x+0.927) has a range of -5 <= y <= 5
So 1 / (5sin(2x+0.927)) has a range of -1/5 <= y <= 1/5 ?
But that can't be right because if that were true, the function would be 1/5 (5sin(2x+0.927))

Think about what happens to 1/(5sin(2x + 0.927)) as the value of the sine part falls from 1 to 0. The fraction starts off as 1/5. What happens to the fraction as the denominator gets smaller and smaller? Once you've got that, consider what happens when the sine part varies between -1 and 0.
Original post by abovethecl0uds
5sin(2x+0.927) has a range of -5 <= y <= 5
So 1 / (5sin(2x+0.927)) has a range of -1/5 <= y <= 1/5 ?
But that can't be right because if that were true, the function would be 1/5 (5sin(2x+0.927))

Write w for 5sin(2x+0.927).

Then w has a range of -5 >= w < = 5 (*).
Your problem is to give the range of y = 1/w. This is not -1/5 <= 1/w <= 1/5. (E.g. suppose 1/w = 1/10, then w must be 10, which is impossible according to (*)).
(edited 2 years ago)
Original post by Pangol
Think about what happens to 1/(5sin(2x + 0.927)) as the value of the sine part falls from 1 to 0. The fraction starts off as 1/5. What happens to the fraction as the denominator gets smaller and smaller? Once you've got that, consider what happens when the sine part varies between -1 and 0.

Thanks for your help, I get it now!

Original post by DFranklin
Write w for 5sin(2x+0.927).

Then w has a range of -5 >= w < = 5 (*).
Your problem is to give the range of y = 1/w. This is not -1/5 <= 1/w <= 1/5. (E.g. suppose 1/w = 1/10, then w must be 10, which is impossible according to (*)).

So are you saying the answer is not -1/5 <= 1/w <= 1/5?
Original post by abovethecl0uds
Thanks for your help, I get it now!


So are you saying the answer is not -1/5 <= 1/w <= 1/5?

What do you think the answer is to the original question is?
Original post by abovethecl0uds
Thanks for your help, I get it now!


So are you saying the answer is not -1/5 <= 1/w <= 1/5?

The answer is *not* -1/5 <= 1/w <= 1/5.

Repeating myself, suppose 1/w = 1/10. Then w = 10. Is this possible?
Original post by mqb2766
What do you think the answer is to the original question is?

I thought the answer was -1/5 <= y <= 1/5 ? But I don't have the mark scheme so I can't check
Original post by DFranklin
The answer is *not* -1/5 <= 1/w <= 1/5.

Repeating myself, suppose 1/w = 1/10. Then w = 10. Is this possible?

No that's not possible, so I guess the answer is not -1/5 <= y <= 1/5
Original post by abovethecl0uds
I thought the answer was -1/5 <= y <= 1/5 ? But I don't have the mark scheme so I can't check

No. Did you look at the prevoius graph in #10?
Think what happens when sin() is small so 1/(5sin()) is ...
See if you can write your own explanation with the numerous hints above.
Reply 18
Original post by abovethecl0uds
No that's not possible, so I guess the answer is not -1/5 <= y <= 1/5

You need to do some thinking / testing based on the hints we have given. Certainly 5sin(something) has a range of -5 to +5 but don't forget that sin(something) could be very small - e,g, 1/100 or 1/1000 or ... - in which case, what happens to its reciprocal? Try to get some intuition for how the function is going to behave as its argument changes - or failing that, try to plot it in an online graphing tool and see what happens :smile:

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