abovethecl0uds
Badges: 11
Rep:
?
#1
Report Thread starter 3 days ago
#1
any help would be appreciated
Name:  Screenshot_20211014-093101_Gallery.jpg
Views: 11
Size:  18.7 KB
Last edited by abovethecl0uds; 3 days ago
0
reply
Pangol
Badges: 15
Rep:
?
#2
Report 3 days ago
#2
(Original post by abovethecl0uds)
any help would be appreciated
Name:  Screenshot_20211014-093101_Gallery.jpg
Views: 11
Size:  18.7 KB
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?
0
reply
abovethecl0uds
Badges: 11
Rep:
?
#3
Report Thread starter 3 days ago
#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
Name:  20211014_094457.jpg
Views: 11
Size:  86.9 KB
0
reply
Pangol
Badges: 15
Rep:
?
#4
Report 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?
0
reply
14mut64
Badges: 5
Rep:
?
#5
Report 3 days ago
#5
(Original post by abovethecl0uds)
yep! i'm not sure if this is right though
Name:  20211014_094457.jpg
Views: 11
Size:  86.9 KB
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
0
reply
Pangol
Badges: 15
Rep:
?
#6
Report 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.
0
reply
abovethecl0uds
Badges: 11
Rep:
?
#7
Report Thread starter 2 days ago
#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?
0
reply
davros
  • Study Helper
Badges: 16
Rep:
?
#8
Report 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?
0
reply
abovethecl0uds
Badges: 11
Rep:
?
#9
Report Thread starter 2 days ago
#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))
0
reply
mqb2766
Badges: 19
Rep:
?
#10
Report 2 days ago
#10
https://www.desmos.com/calculator/3vrq2fot5j
1/sin = cosec
Last edited by mqb2766; 2 days ago
0
reply
Pangol
Badges: 15
Rep:
?
#11
Report 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.
0
reply
DFranklin
Badges: 18
Rep:
?
#12
Report 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
0
reply
abovethecl0uds
Badges: 11
Rep:
?
#13
Report Thread starter 13 minutes ago
#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?
0
reply
mqb2766
Badges: 19
Rep:
?
#14
Report 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?
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

How are you feeling now you've started university?

I'm loving it! (54)
14.29%
I'm enjoying it but I'm still settling in (95)
25.13%
I'm a bit unsure (66)
17.46%
I'm finding things difficult (130)
34.39%
Something else (let us know in the thread!) (33)
8.73%

Watched Threads

View All
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise