The Student Room Group

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Reply 1
the 3rd line of your working is wrong, it should be:
tan^2 +1 = Sec^2 (as 1/cos = sec)
Reply 2
steve2005
solve cosec²?+1=3cot?

This is what i have got so far?

I think I have done it wrong. More in a moment



cosec²θ+1=3cotθ

and since cosec²θ = 1+cot²θ

so 1+cot²θ +1=3cotθ
now let cotθ=c
c²-3c+2=0
(c-2)(c-1)=0

Aitch
Reply 3
The values that are given in the book are : 20.9, 69.1, 201, 249?
Reply 4
turtle2
The values that are given in the book are : 20.9, 69.1, 201, 249?


From my post above, 26.57, 45...etc. seem to check OK...

Aitch
Reply 5
Aitch
cosec²θ+1=3cotθ

and since cosec²θ = 1+cot²θ

so 1+cot²θ +1=3cotθ
now let cotθ=c
c²-3c+2=0
(c-2)(c-1)=0

Aitch


c = 1
cotθ = 1
1/tanθ = 1
θ = π/4

c = 2
cotθ = 2
1/tanθ = 2
θ = 26.57°
Reply 6
Oh thanks. I actually have another Q: cotθ=1-cosec²θ, 0«θ«2π
Reply 7
turtle2
Oh thanks. I actually have another Q: cotθ=1-cosec²θ, 0«θ«2π

Use the identity you have that connects cotθ and cosecθ to form a quadratic in cotθ. Find cotθ and hence solve to find θ.
Reply 8
turtle2
Oh thanks. I actually have another Q: cot?=1-cosec²?, 0«?«2?


Here is my solution to the question that has already been answered.
( In case ppl are confused the original question was sent to me as a PM but I could not upload to image... so had to post)
Reply 9
steve2005
Here is my solution to the question that has already been answered.
( In case ppl are confused the original question was sent to me as a PM but I could not upload to image... so had to post)


But you've worked out: 1 + cot²θ = cosec²θ

turtle wanted to work out cotθ = 1-cosecθ²

I think ...
Reply 10
Ricki
But you've worked out: 1 + cot²? = cosec²?

turtle wanted to work out cot? = 1-cosec?²

I think ...


I don't know... I get so confused... turtle if I have answered the wrong question ... please say so.
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Ricki
But you've worked out: 1 + cot²? = cosec²?

turtle wanted to work out cot? = 1-cosec?²

I think ...


Ricki on my computer theta does not appear in my messages but it does appear in yours. The theta disappears to be replaced by a question mark in the quoted part.... even if a type nothing.

Do you know how to correct this. I use the option Enhanced Mode...and not Guided Mode. I have experimented in the past but ....

Testing ? ? µ ? ? ½ µ
Reply 11
well, I think you need to download javascript.
Reply 12
steve2005
Ricki on my computer theta does not appear in my messages but it does appear in yours. The theta disappears to be replaced by a question mark in the quoted part.... even if a type nothing.

Do you know how to correct this. I use the option Enhanced Mode...and not Guided Mode. I have experimented in the past but ....


I'm using the guided mode, the default. I've experienced this problem before, I'm not sure how it is corrected though ...
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BCHL85
well, I think you need to download javascript.


:ditto:

I've got the javascript. :smile:
Reply 13
Ricki
I'm using the guided mode, the default. I've experienced this problem before, I'm not sure how it is corrected though ...
--------------


:ditto:

I've got the javascript. :smile:


cos? I have just typed cos theta using Guided Mode. To get theta I clicked on theta in the menu . I am now doing pi ?
Reply 14
steve2005
cos? I have just typed cos theta using Guided Mode. To get theta I clicked on theta in the menu . I am now doing pi ?


It can be that your browser does not support the use of these symbols.

Or .. you need to download Java ... from ... http://java.sun.com/j2se/1.4.2/download.html
Reply 15
Ricki
It can be that your browser does not support the use of these symbols.

Or .. you need to download Java ... from ... http://java.sun.com/j2se/1.4.2/download.html


But, how come I can see your symbols but not the ones I produce. Question: In my last post there should be a symbol for theta .... can you see the symbol or do you see a question mark
Reply 16
steve2005
But, how come I can see your symbols but not the ones I produce. Question: In my last post there should be a symbol for theta .... can you see the symbol or do you see a question mark


A question mark.

It seems that your browser can read the symbols yet not produce them.
Reply 17
I have two more questions that I'm unsure about...

f) (secθ-cos&#952:wink:²=tanθ-sin²θ , 0«θ«π

g) tan²2θ=sec2θ-1 , 0«θ«180
Reply 18
tan²2θ=sec2θ-1

let 2θ be W

tan²W=secW-1

sin²W/cos²W = 1/cosW -1

sin²W = cosW - cos²W

1 - cos²W = cosW - cos²W

cosW = 1

W = 0°,360°, 720° etc

θ = 0°, 180°, 360° etc
Reply 19
turtle2
f) (secθ-cosθ )²=tanθ-sin²θ , 0«θ«π


f) (secθ-cosθ )²=tanθ-sin²θ
.. (1/cosθ - cosθ = sinθ/cosθ - sin²θ
... 1/cos²θ - 2 + cos²θ = sinθ/cosθ - sin²θ
... 1/cos²θ - 2 + cos²θ = sinθ/cosθ - (1-cos²θ )
... 1/cos²θ + cos²θ - 2 = cos²θ + sinθ/cosθ - 1
... 1/cos²θ + cos²θ - 2 - cos²θ - sinθ/cosθ + 1 = 0
... sec²θ - tanθ - 1 = 0

sub sec²θ = tan²θ + 1, following the identity:

... tan²θ + 1 - tanθ - 1 = 0
... tan²θ - tanθ = 0
... tanθ - 1 = 0
... tanθ = 1
... θ = arctan[1], θ = π/4 etc.