# maths trig a level

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any help would be appreciated

Last edited by abovethecl0uds; 3 days ago

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#2

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(Original post by

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?

**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?

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#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?

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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#5

Last edited by 14mut64; 3 days ago

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#6

(Original post by

You need to treat 4cos2x and 3sin2x as two separate compound angle equations and work it out from there…

**14mut64**)You need to treat 4cos2x and 3sin2x as two separate compound angle equations and work it out from there…

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(Original post by

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?

**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?

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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#8

(Original post by

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?

**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?

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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(Original post by

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?

**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?

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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#10

https://www.desmos.com/calculator/3vrq2fot5j

1/sin = cosec

1/sin = cosec

Last edited by mqb2766; 2 days ago

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#11

(Original post by

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))

**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))

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#12

**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))

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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(Original post by

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.

**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.

(Original post by

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

**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 (*)).
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#14

(Original post by

Thanks for your help, I get it now!

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

**abovethecl0uds**)Thanks for your help, I get it now!

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

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