# maths trig a level

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#1
any help would be appreciated
Last edited by abovethecl0uds; 3 days ago
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3 days ago
#2
(Original post by abovethecl0uds)
any help would be appreciated
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?
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#3
(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
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3 days ago
#4
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?
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3 days ago
#5
(Original post by abovethecl0uds)
yep! i'm not sure if this is right though
You need to treat 4cos2x and 3sin2x as two separate compound angle equations and work it out from there…
Last edited by 14mut64; 3 days ago
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3 days ago
#6
(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.
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#7
(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?
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2 days ago
#8
(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 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?
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#9
(Original post by davros)
You're over-thinking this problem - this is nothing to do with function domains or inverse functions 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))
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2 days ago
#10
https://www.desmos.com/calculator/3vrq2fot5j
1/sin = cosec
Last edited by mqb2766; 2 days ago
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2 days ago
#11
(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.
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2 days ago
#12
(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 (*)).
Last edited by DFranklin; 2 days ago
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#13
(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?
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1 minute ago
#14
(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?
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