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Maths C3 - Trigonometry... Help??

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Original post by Philip-flop
An identity with a '2' in it? I've never come across that yet :frown:


Well that was a really mess up with the latex lol. I meant an identity with a cot2\cot^2 in it.
Original post by RDKGames
Are you sure that is the question? The two are not identical.

OMG, my brain really is fried today!! I must have been looking at the answers to a complete different question :colondollar:

It's definitely one of those days where I should have just crawled back into bed as soon as I got home from work :frown: Seriously FML!!

Sorry for wasting everyone's time. I will refrain from posting for the rest of the night.
(edited 7 years ago)
Original post by Philip-flop
OMG, my brain really is fried today!! I must have been looking at the answers to a complete different question :colondollar:

It's definitely one of those days where I should have just crawled back into bed as soon as I got home from work :frown: Seriously FML!!

Sorry for wasting everyone's time. I will refrain from posting for the rest of the night.


Relax :tongue:
These Trig Identities are seriously testing my patience!

I've tried absolutely everything with this question to the point where I've started making up my own rules :frown:

C3 - EXE6D Q(7) Trig Identities.png

So I'm pretty certain I can start off by using the Trig Identity 1+tan2θ=sec2θ1+ tan^2 \theta = \sec ^2 \theta

to give me...
3tan2θ+4(1+tan2θ)=53tan^2 \theta +4(1+tan^2 \theta) = 5

Then I expand the brackets and re-arrange the equation to give me...
7tan2θ=1 7tan^2 \theta =1 <<<have I done something wrong already? :frown:
Original post by Philip-flop
These Trig Identities are seriously testing my patience!

I've tried absolutely everything with this question to the point where I've started making up my own rules :frown:

C3 - EXE6D Q(7) Trig Identities.png

So I'm pretty certain I can start off by using the Trig Identity 1+tan2θ=sec2θ1+ tan^2 \theta = \sec ^2 \theta

to give me...
3tan2θ+4(1+tan2θ)=53tan^2 \theta +4(1+tan^2 \theta) = 5

Then I expand the brackets and re-arrange the equation to give me...
7tan2θ=1 7tan^2 \theta =1 <<<have I done something wrong already? :frown:


You're correct. Why is it wrong??
Original post by RDKGames
You're correct. Why is it wrong??


Because I feel like I get stuck from there on :frown:

I know I can use the Identity tanθ=sinθcosθ tan \theta = \frac{sin \theta}{cos \theta} but then I have no idea how to solve just for sinθ sin \theta Things just get too messy :frown: :frown:
(edited 7 years ago)
Original post by Philip-flop
Because I feel like I get stuck from there on :frown:

I know I can use the Identity tanθ=sinθcosθ tan \theta = \frac{sin \theta}{cos \theta} but then I have no idea how to solve just for sinθ sin \theta


So you got the value of 7tan2(θ)=17\tan^2(\theta)=1 which means that 7sin2(θ)=cos2(θ)7\sin^2(\theta)=\cos^2(\theta) (mult. both sides by cosine squared after expressing tan in terms of sine and cosine) then express cosine squared in terms of sine squared then square root sine term and get the correct value for an obtuse angle.
(edited 7 years ago)
Original post by RDKGames
So you got the value of 7tan2(θ)=17\tan^2(\theta)=1 which means that 7sin2(θ)=cos2(θ)7\sin^2(\theta)=\cos^2(\theta) (mult. both sides by cosine squared after expressing tan in terms of sine and cosine) then express cosine squared in terms of sine squared then square root sine term and get the correct value for an obtuse angle.


Ok so from...
7sin2θ=cos2θ 7sin^2 \theta = cos^2 \theta

This gives me...
7sin2θ=1sin2θ 7sin^2 \theta = 1- sin^2 \theta

Then add sin^2(theta) to both sides? to give...
8sin2θ=1 8sin^2 \theta = 1

sin2θ=18 sin^2 \theta = \frac{1}{8}

sinθ=18 sin \theta = \sqrt \frac{1}{8}

To which I end up having to rationalise the denominator?
Original post by Philip-flop
Ok so from...
7sin2θ=cos2θ 7sin^2 \theta = cos^2 \theta

This gives me...
7sin2θ=1sin2θ 7sin^2 \theta = 1- sin^2 \theta

Then add sin^2(theta) to both sides? to give...
8sin2θ=1 8sin^2 \theta = 1

sin2θ=18 sin^2 \theta = \frac{1}{8}

sinθ=18 sin \theta = \sqrt \frac{1}{8}

To which I end up having to rationalise the denominator?


Correct working, and you don't have to rationalise it, but remember that square rooting gives ±\pm the answer and you need to state the reason for your choice.
Original post by RDKGames
Correct working, and you don't have to rationalise it, but remember that square rooting gives ±\pm the answer and you need to state the reason for your choice.

Oh yeah of course!! θ\theta only appears between 90 and 180 (in the 2nd quadrant) when sinsin is +ve

Therefore...
sinθ24 sin \theta \not = - \frac {\sqrt 2}{4}

so...
sinθ=+24 sin \theta = + \frac {\sqrt 2}{4}

Thanks again for your help!! I can't even begin to explain how much I appreciate it! Maybe I should donate money to you haha. Or pay you to tutor me :P
(edited 7 years ago)
Original post by Philip-flop
Oh yeah of course!! θ\theta only appears between 90 and 180 (in the 2nd quadrant) when sinsin is +ve

Therefore...
sinθ24 sin \theta \not = - \frac {\sqrt 2}{4}

so...
sinθ=+24 sin \theta = + \frac {\sqrt 2}{4}

Thanks again for your help!! I can't even begin to explain how much I appreciate it! Maybe I should donate money to you haha. Or pay you to tutor me :P


Haha no problem, well done :h:
Ok so I realise I absolutely suck at trigonometry as I'm stuck on yet another question :frown: ...

Q) Solve the equation for the following interval...
C3 Trignometry EXE2D Q8d.png

Am I right in thinking that the trig identity... 1+cot2θ=cosec2θ1 + \cot^2 \theta = \mathrm{cosec}^2 \theta... can be rearranged to give...
1cosec2θ=cot2θ1- \mathrm{cosec}^2 \theta = -\cot ^2 \theta

So for this question I would do...
cotθ=1cosec2θ \cot \theta = 1-\mathrm{cosec}^2 \theta

cotθ=cot2θ \cot \theta = - \cot ^2 \theta
Original post by Philip-flop
Ok so I realise I absolutely suck at trigonometry as I'm stuck on yet another question :frown: ...

Q) Solve the equation for the following interval...
C3 Trignometry EXE2D Q8d.png

Am I right in thinking that the trig identity... 1+cot2θ=cosec2θ1 + \cot^2 \theta = \mathrm{cosec}^2 \theta... can be rearranged to give...
1cosec2θ=cot2θ1- \mathrm{cosec}^2 \theta = -\cot ^2 \theta

So for this question I would do...
cotθ=1cosec2θ \cot \theta = 1-\mathrm{cosec}^2 \theta

cotθ=cot2θ \cot \theta = - \cot ^2 \theta


Okay so you cannot divide otherwise you'd be losing solutions.

From moving RHS onto LHS, you get cot2θ+cotθ=0\cot^2 \theta + \cot \theta = 0 at which point you can factorise LHS and solve two equations for θ\theta in the given range.
(edited 7 years ago)
Original post by RDKGames
Okay so you cannot divide otherwise you'd be losing solutions.

From moving RHS onto LHS, you get cot2θ+cotθ=0\cot^2 \theta + \cot \theta = 0 at which point you can factorise LHS and solve two equations for θ\theta in the given range.


You'll also notice why cotθ≢1tanθ\cot\theta \not \equiv \dfrac{1}{\tan\theta} from solving this equation.
Original post by RDKGames
Okay so you cannot divide otherwise you'd be losing solutions.

From moving RHS onto LHS, you get cot2θ+cotθ=0\cot^2 \theta + \cot \theta = 0 at which point you can factorise LHS and solve two equations for θ\theta in the given range.


Ok I'm not sure how I still ended up losing 2 of the solutions :frown:

This is what I got...
Photo 21-09-2016, 14 25 47.jpg

Original post by IrrationalRoot
You'll also notice why cotθ≢1tanθ\cot\theta \not \equiv \dfrac{1}{\tan\theta} from solving this equation.

Yeah because it'll give you no solution right? How do i work around that?
(edited 7 years ago)
Original post by Philip-flop
Ok I'm not sure how I still ended up losing 2 of the solutions :frown:

This is what I got...


Correct solutions.

That is the pitfall you made as the question aims to throw people off, and IrrationalRoot pointed out :smile:

cotθ≢1tanθ\cot \theta \not\equiv \frac{1}{\tan \theta}

If you turn cotθcosθsinθ=0\cot \theta \equiv \frac{\cos \theta}{\sin \theta} = 0 then you can see if cosθ=0\cos \theta = 0 then your equation still holds and you get more solutions.
(edited 7 years ago)
Original post by IrrationalRoot
You'll also notice why cotθ≢1tanθ\cot\theta \not \equiv \dfrac{1}{\tan\theta} from solving this equation.


I find it absurd how so many place say that it is an IDENTITY, when this question is a perfect example of why it is not.
Original post by RDKGames
Correct solutions.

That is the pitfall you made as the question aims to throw people off, and IrrationalRoot pointed out :smile:

cotθ≢1tanθ\cot \theta \not\equiv \frac{1}{\tan \theta}

If you turn cotθcosθsinθ=0\cot \theta \equiv \frac{\cos \theta}{\sin \theta} = 0 then you can see if cosθ=0\cos \theta = 0 then your equation still holds and you get more solutions.


Oh right so I should have known that... cotθ1tanθ\cot \theta \equiv \frac{1}{\tan \theta} ... isn't actually an identity. I seriously thought it was! Why are the books so misleading at times? :frown:

Ok so now that I know. I should have done this...

cotθ=0 \cot \theta = 0

1tanθ=0 \frac{1}{tan \theta} = 0

cosθsinθ=0 \frac{cos \theta}{sin \theta} = 0

Then I times both sides by sinθsin \theta to give...

cosθ=0 cos \theta = 0

Then just work out the other solutions from there?
Original post by Philip-flop
Oh right so I should have known that... cotθ1tanθ\cot \theta \equiv \frac{1}{\tan \theta} ... isn't actually an identity. I seriously thought it was! Why are the books so misleading at times? :frown:

Ok so now that I know. I should have done this...

cotθ=0 \cot \theta = 0

1tanθ=0 \frac{1}{tan \theta} = 0

cosθsinθ=0 \frac{cos \theta}{sin \theta} = 0

Then I times both sides by sinθsin \theta to give...

cosθ=0 cos \theta = 0

Then just work out the other solutions from there?


Essentially yeah, but your second line wouldn't be correct because if I were to multiply both sides by tanθ\tan \theta then I would be left with 1=0tanθ=01=01=0\cdot \tan \theta = 0 \Rightarrow 1=0 which doesn't make sense.
Again, you made the same mistake by assuming cotθ1tanθ\cot \theta \equiv \frac{1}{\tan \theta}. If you ignore line 2, then that working would be fine.

Also it would see more straight forward if from your fractorised form you went:

cosθsinθ(1+cotθ)=0cosθ(1+cotθ)=0\frac{\cos \theta}{\sin \theta}(1+\cot \theta)=0 \Rightarrow \cos \theta (1+\cot \theta)=0

From multiplying both sides by sine.
(edited 7 years ago)
Original post by Philip-flop
Oh right so I should have known that... cotθ1tanθ\cot \theta \equiv \frac{1}{\tan \theta} ... isn't actually an identity. I seriously thought it was! Why are the books so misleading at times? :frown:

Ok so now that I know. I should have done this...

cotθ=0 \cot \theta = 0

1tanθ=0 \frac{1}{tan \theta} = 0

cosθsinθ=0 \frac{cos \theta}{sin \theta} = 0

Then I times both sides by sinθsin \theta to give...

cosθ=0 cos \theta = 0

Then just work out the other solutions from there?


Question for my own interest (not meant to be patronising and I don't know the answer to it) but how easily do you feel that you give up trying to solve a question, in particular for when you post here?

This is to kind of understand A. How helpful this is and B. Whether you can change your approach to questions slightly. I
(edited 7 years ago)

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