Maths C3 - Trigonometry... Help??

Announcements Posted on
How helpful is our apprenticeship zone? Have your say with our short survey 02-12-2016
    • Thread Starter
    Online

    3
    ReputationRep:
    (Original post by RDKGames)
    Multiply top and bottom by cosine.
    Thank you!!! How did you know to multiply top and bottom by cos(x)?


    (Original post by notnek)
     \displaystyle \frac{1-tan x}{1+tan x} = \frac{1-\frac{\sin x}{\cos x}}{1+\frac{\sin x}{\cos x}}

    The next step is to multiply top and bottom of the fraction by \cos x.
    Thank you. I never would have known to multiply the top and bottom by cos(x)
    Online

    3
    ReputationRep:
    (Original post by Philip-flop)
    Thank you!!! How did you know to multiply top and bottom by cos(x)?



    Thank you. I never would have known to multiply the top and bottom by cos(x)
    Whenever you see a fraction on the top/bottom of another fraction then it's often useful to get rid of it as soon as possible by multiplying top and bottom by the denominator.

    E.g.

    \displaystyle \frac{y-x}{\frac{y}{x}-1}

    If you were trying to simplify this then you could get rid of \frac{y}{x} by multiplying top and bottom of the fraction by x. This gives

    \displaystyle \frac{y-x}{\frac{y}{x}-1} = \frac{yx-x^2}{y-x}
    • Thread Starter
    Online

    3
    ReputationRep:
    (Original post by notnek)
    Whenever you see a fraction on the top/bottom of another fraction then it's often useful to get rid of it as soon as possible by multiplying top and bottom by the denominator.

    E.g.

    \displaystyle \frac{y-x}{\frac{y}{x}-1}

    If you were trying to simplify this then you could get rid of \frac{y}{x} by multiplying top and bottom of the fraction by x. This gives

    \displaystyle \frac{y-x}{\frac{y}{x}-1} = \frac{yx-x^2}{y-x}
    Thanks! That's definitely helped fill a gap in my knowledge I really appreciate it!
    • Thread Starter
    Online

    3
    ReputationRep:
    notnek but for the next bit after I get...

     \frac{cos x - sinx}{cos x + sinx}

    how am I meant to know to do \frac{cos x - sinx}{cos x + sinx} \times \frac{cos x - sinx}{cos x - sinx}??
    Online

    3
    ReputationRep:
    (Original post by Philip-flop)
    notnek but for the next bit after I get...

     \frac{cos x - sinx}{cos x + sinx}

    how am I meant to know to do \frac{cos x - sinx}{cos x + sinx} \times \frac{cos x - sinx}{cos x - sinx}??
    This comes a bit from practice/experience.

    It's similar to why you multiply top and bottom of e.g. \displaystyle \frac{3}{2+\sqrt{3}} by 2-\sqrt{3} to rationalise it.

    The difference of two squares method can sometimes simplify the denominator or put it in another form.

    In this case I noticed that (\cos x + \sin x)(\cos x - \sin x) = \cos^2 - \sin^2x = \cos 2x

    Using difference of two squares like this in trig can be a bit of a last resort method if you can't get anywhere from standard manipulation.
    • Thread Starter
    Online

    3
    ReputationRep:
    Name:  C3 EXE7C Q6.png
Views: 29
Size:  3.5 KB
    (Exe7C Question 6)

    For part (b) how do I know which solutions from...
     \theta = \frac{\pi}{24}, \frac{5 \pi}{24}, \frac{13\pi}{24}, \frac{17\pi}{24}

    ...satisfy the equation...
     cos2 \theta - sin2 \theta = \frac{1}{\sqrt 2} ?? <<I'm guessing it has something to do with the square root on the RHS of this equation.

    Do I just have to test each "solution" of theta by subbing them into the equation above using my calculator or is there a quicker way?
    Offline

    3
    ReputationRep:
    (Original post by Philip-flop)
    Name:  C3 EXE7C Q6.png
Views: 29
Size:  3.5 KB
    (Exe7C Question 6)

    For part (b) how do I know which solutions from...
     \theta = \frac{\pi}{24}, \frac{5 \pi}{24}, \frac{13\pi}{24}, \frac{17\pi}{24}

    ...satisfy the equation...
     cos2 \theta - sin2 \theta = \frac{1}{\sqrt 2} ??

    Do I just have to test each "solution" of theta by subbing them into the equation above using my calculator or is there a quicker way?
    They all satisfy it, it even says "answers" (plural) in the question so you just list them.
    • Thread Starter
    Online

    3
    ReputationRep:
    (Original post by RDKGames)
    They all satisfy it, it even says "answers" (plural) in the question so you just list them.
    But the book says its only...
     \theta = \frac{\pi}{24}, \frac{17\pi}{24}

    Which is why I'm confused
    • Thread Starter
    Online

    3
    ReputationRep:
    Where the heck do I even start with part (b)i) ?? I'm not sure I even understand what the question is asking
    Name:  C3 EXE7C Q7.png
Views: 22
Size:  5.1 KB
    Offline

    3
    ReputationRep:
    (Original post by Philip-flop)
    Where the heck do I even start with part (b)i) ??
    Name:  C3 EXE7C Q7.png
Views: 22
Size:  5.1 KB
    For the first one if  \tan \frac{\theta }{2}=t we have  \displaystyle \frac{2t}{1+t^2}+\frac{2(1-t^2)}{1+t^2}=1 .
    SO just solve the quadratic equation as you normally would.
    Offline

    3
    ReputationRep:
    (Original post by Philip-flop)
    Where the heck do I even start with part (b)i) ?? I'm not sure I even understand what the question is asking
    Name:  C3 EXE7C Q7.png
Views: 22
Size:  5.1 KB
    get  tan\theta from the previous part, using \frac{sin\theta}{cos\theta}, in terms of  tan\frac{\theta}{2}
    • Thread Starter
    Online

    3
    ReputationRep:
    (Original post by B_9710)
    For the first one if  \tan \frac{\theta }{2} we have  \displaystyle \frac{2t}{1+t^2}+\frac{2(1-t^2)}{1+t^2}=1 .
    SO just solve the quadratic equation as you normally would.
    But how do you get...  \tan \frac{\theta }{2} >>>  \displaystyle \frac{2t}{1+t^2}+\frac{2(1-t^2)}{1+t^2}=1

    from... (b)i)  sin \theta + 2cos \theta = 1 ?? Sorry if I'm being silly
    • Thread Starter
    Online

    3
    ReputationRep:
    (Original post by NotNotBatman)
    get  tan\theta from the previous part, using \frac{sin\theta}{cos\theta}, in terms of  tan\frac{\theta}{2}
    Oh right! It's starting to make sense now. But where have those 't's come from?
    Offline

    3
    ReputationRep:
    (Original post by Philip-flop)
    Oh right! It's starting to make sense now. But where have those 't's come from?
    The above user has shortened it, so  t = tan\frac{\theta}{2} and has done  sin\theta +2cos\theta = 1 in terms of t.
    • Thread Starter
    Online

    3
    ReputationRep:
    (Original post by NotNotBatman)
    The above user has shortened it, so  t = tan\frac{\theta}{2} and has done  sin\theta +2cos\theta = 1 in terms of t.
    Thank you so much
    • Thread Starter
    Online

    3
    ReputationRep:
    For part (a) can someone please explain how this goes from....
     2cos^2 \frac{\theta}{2}

    to this...
     = \frac{1+cos \theta}{2} ??

    Name:  C3 EXE7C Q10.png
Views: 19
Size:  3.6 KB
    Offline

    3
    ReputationRep:
    (Original post by Philip-flop)
    But the book says its only...
     \theta = \frac{\pi}{24}, \frac{17\pi}{24}

    Which is why I'm confused
    Ah yes, looked at the question and it makes sense. The two extra solutions come from the fact that when you square \frac{1}{\sqrt{2}} you create an answer which can be achieved from squaring -\frac{1}{\sqrt{2}} so two of your solutions satisfy \sin(2\theta)-\cos(2\theta)=-\frac{1}{\sqrt{2}} which is not what we want.

    The easiest way around is to plug the values back through the equation and collect the values you need. You can also work out for which angles \cos2 \theta - \sin2 \theta is positive and consider only the answers which fall into those domains.
    Offline

    3
    ReputationRep:
    (Original post by Philip-flop)
    For part (a) can someone please explain how this goes from....
     2cos^2 \frac{\theta}{2}

    to this...
     = \frac{1+cos \theta}{2} ??

    Name:  C3 EXE7C Q10.png
Views: 19
Size:  3.6 KB
    It doesn't.

    \cos(\theta)=\cos^2(\frac{\theta  }{2})-\sin^2(\frac{\theta}{2})=2\cos^2  (\frac{\theta}{2})-1 \implies \cos(\theta)+1=2\cos^2(\frac{ \theta }{2} )
    • Thread Starter
    Online

    3
    ReputationRep:
    (Original post by RDKGames)
    Ah yes, looked at the question and it makes sense. The two extra solutions come from the fact that when you square \frac{1}{\sqrt{2}} you create an answer which can be achieved from squaring -\frac{1}{\sqrt{2}} so two of your solutions satisfy \sin(2\theta)-\cos(2\theta)=-\frac{1}{\sqrt{2}} which is not what we want.

    The easiest way around is to plug the values back through the equation and collect the values you need. You can also work out for which angles \cos2 \theta - \sin2 \theta is positive and consider only the answers which fall into those domains.
    Thank you. That makes perfect sense to me now!

    (Original post by RDKGames)
    It doesn't.

    \cos(\theta)=\cos^2(\frac{\theta  }{2})-\sin^2(\frac{\theta}{2})=2\cos^2  (\frac{\theta}{2})-1 \implies \cos(\theta)+1=2\cos^2(\frac{ \theta }{2} )
    Thaaaank you!! I never thought I could re-arrange the double angle formulae in that way!! Seriously appreciate your help
    • Thread Starter
    Online

    3
    ReputationRep:
    Can someone help explain from where it says "Occurs when cos( \theta - 22.6) = 1 ......" <<Where has the number 1 come from?..
    I've worked out the rest of the question and can see that the maximum is 13 but I can't seem to understand the last bit. Why is the smallest value of  \theta 22.6 ??

    Name:  C3 - EXA 19.png
Views: 9
Size:  18.3 KB
 
 
 
Write a reply… Reply
Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. Oops, you need to agree to our Ts&Cs to register
  2. Slide to join now Processing…

Updated: November 6, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

Today on TSR
Poll
Would you rather have...?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read here first

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups
Study resources

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.