Join TSR now and get all your revision questions answeredSign up now

Maths C3 - Trigonometry... Help?? Watch

    • Thread Starter
    Offline

    3
    ReputationRep:
    Ok. How do I even do part (c) and (d)??

    Name:  C3 Exe 7A Q12(a).png
Views: 16
Size:  3.3 KB

    For part (c)... Shall I start off by using the fact that...
     tan \theta = \frac{sin \theta}{cos \theta} ?
    • Thread Starter
    Offline

    3
    ReputationRep:
    ^ Anyone??
    • TSR Support Team
    • Study Helper
    Offline

    3
    ReputationRep:
    (Original post by Philip-flop)
    Ok. How do I even do part (c) and (d)??

    Name:  C3 Exe 7A Q12(a).png
Views: 16
Size:  3.3 KB

    For part (c)... Shall I start off by using the fact that...
     tan \theta = \frac{sin \theta}{cos \theta} ?
    Look back at question 5H that you did successfully a few days back.

    You need to use \tan 45 = 1 again to write the fraction in the form

    \displaystyle \frac{\tan A + \tan B}{1-\tan A \tan B}

    For d), use the fact that \displaystyle \sin 45 = \cos 45 = \frac{1}{\sqrt{2}}

    These are all quite tricky questions that you don't really see in exams. Don't worry too much if you find them hard.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by notnek)
    Look back at question 5H that you did successfully a few days back.

    You need to use \tan 45 = 1 again to write the fraction in the form

    \displaystyle \frac{\tan A + \tan B}{1-\tan A \tan B}

    For d), use the fact that \displaystyle \sin 45 = \cos 45 = \frac{1}{\sqrt{2}}

    These are all quite tricky questions that you don't really see in exams. Don't worry too much if you find them hard.
    Yeah I had a feeling that part (c) was similar to Question 5(h) that I did not long ago. Managed to figure that one out now. But am still stuck on part(d).

    Thanks a lot for your help
    • TSR Support Team
    • Study Helper
    Offline

    3
    ReputationRep:
    (Original post by Philip-flop)
    Yeah I had a feeling that part (c) was similar to Question 5(h) that I did not long ago. Managed to figure that one out now. But am still stuck on part(d).

    Thanks a lot for your help
    \displaystyle \frac{1}{\sqrt{2}}\left( \sin \theta + \cos \theta \right) = \frac{1}{\sqrt{2}} \sin \theta + \frac{1}{\sqrt{2}} \cos \theta

    \displaystyle sin 45 = cos 45 =  \frac{1}{\sqrt{2}}

    Try substituting these to get something of the form \sin A \cos B + \sin B \cos A
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by notnek)
    \displaystyle \frac{1}{\sqrt{2}}\left( \sin \theta + \cos \theta \right) = \frac{1}{\sqrt{2}} \sin \theta + \frac{1}{\sqrt{2}} \cos \theta

    \displaystyle sin 45 = cos 45 =  \frac{1}{\sqrt{2}}

    Try substituting these to get something of the form \sin A \cos B + \sin B \cos A
    Oh yeah of course! I actually started doing that then realised. So I've done...

     \frac{1}{\sqrt 2} (sin \theta + cos \theta)

     = \frac{1}{\sqrt 2} sin \theta + \frac{1}{\sqrt 2} cos \theta

     = sin \theta cos \frac{\pi}{4} + cos \theta sin \frac{\pi}{4}

     = sin( \theta + \frac{\pi}{4})

    Alternatively I could have done....
     \frac{1}{\sqrt 2} (sin \theta + cos \theta)

     = \frac{1}{\sqrt 2} sin \theta + \frac{1}{\sqrt 2} cos \theta

    = sin \frac{\pi}{4} sin \theta + cos \frac{\pi}{4} cos \theta

     = cos (\frac{\pi}{4} - \theta)

    Am I right??
    • TSR Support Team
    • Study Helper
    Offline

    3
    ReputationRep:
    (Original post by Philip-flop)
    Oh yeah of course! I actually started doing that then realised. So I've done...

     \frac{1}{\sqrt 2} (sin \theta + cos \theta)

     = \frac{1}{\sqrt 2} sin \theta + \frac{1}{\sqrt 2} cos \theta

     = sin \theta cos \frac{\pi}{4} + cos \theta sin \frac{\pi}{4}

     = sin( \theta + \frac{\pi}{4})

    Alternatively I could have done....
     \frac{1}{\sqrt 2} (sin \theta + cos \theta)

     = \frac{1}{\sqrt 2} sin \theta + \frac{1}{\sqrt 2} cos \theta

    = sin \frac{\pi}{4} sin \theta + cos \frac{\pi}{4} cos \theta

     = cos (\frac{\pi}{4} - \theta)

    Am I right??
    Yes that's all correct.

    And another way you can see that those two forms are equivalent is by using

    \displaystyle \sin A = \cos \left(\frac{\pi}{2}-A \right) :

    \displaystyle \sin \left( \theta + \frac{\pi}{4} \right) = \cos \left[ \frac{\pi}{2} - \left(\theta + \frac{\pi}{4}\right) \right] = \cos \left(\frac{\pi}{4} - \theta \right)
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by notnek)
    Yes that's all correct.

    And another way you can see that those two forms are equivalent is by using

    \displaystyle \sin A = \cos \left(\frac{\pi}{2}-A \right) :

    \displaystyle \sin \left( \theta + \frac{\pi}{4} \right) = \cos \left[ \frac{\pi}{2} - \left(\theta + \frac{\pi}{4}\right) \right] = \cos \left(\frac{\pi}{4} - \theta \right)
    Oh yeah! Because sin is a translation of cos by 90 degrees (to the right) parallel to the x-axis, correct?

    Thanks again! Things are definitely making sense now
    • TSR Support Team
    • Study Helper
    Offline

    3
    ReputationRep:
    (Original post by Philip-flop)
    Oh yeah! Because sin is a translation of cos by 90 degrees (to the right) parallel to the x-axis, correct?

    Thanks again! Things are definitely making sense now
    That's basically correct although a shift of \cos x by \frac{\pi}{2} to the right gives

    \cos \left(x-\frac{\pi}{2} \right)

    But as you discovered in an older post, cos is an even function so you have

    \displaystyle \cos \left(x-\frac{\pi}{2}\right) = \cos \left[-\left(x-\frac{\pi}{2}\right) \right] = \cos \left(\frac{\pi}{2}-x\right)
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by notnek)
    That's basically correct although a shift of \cos x by \frac{\pi}{2} to the right gives

    \cos \left(x-\frac{\pi}{2} \right)

    But as you discovered in an older post, cos is an even function so you have

    \displaystyle \cos \left(x-\frac{\pi}{2}\right) = \cos \left[-\left(x-\frac{\pi}{2}\right) \right] = \cos \left(\frac{\pi}{2}-x\right)
    Yes, cos is an "even function" due to it's symmetry on either side of the y-axis. Whereas sin is an odd function.
    • Thread Starter
    Offline

    3
    ReputationRep:
    Ok so I'm stuck again already

    Name:  C3 EXE7A Q13.png
Views: 20
Size:  6.6 KB

    Have I even been doing this one right??
    Attachment 584984584986

    Then I don't really know where to go from there.
    Attached Images
     
    • TSR Support Team
    • Study Helper
    Offline

    3
    ReputationRep:
    (Original post by Philip-flop)
    Ok so I'm stuck again already

    Name:  C3 EXE7A Q13.png
Views: 20
Size:  6.6 KB

    Have I even been doing this one right??
    Attachment 584984584986

    Then I don't really know where to go from there.
    You have 3\cos \theta on one side of the equation and \sqrt{3}\cos \theta on the other.

    You can move them to the same side and combine them.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by notnek)
    You have 3\cos \theta on one side of the equation and \sqrt{3}\cos \theta on the other.

    You can move them to the same side and combine them.
    Yeah I thought about doing that, so...

     (3- \sqrt 3) cos\theta = sin \theta

    But then I had no idea where to go from there so I just assumed I was wrong
    • Community Assistant
    • Welcome Squad
    Offline

    3
    ReputationRep:
    (Original post by Philip-flop)
    Yeah I thought about doing that, so...

     (3- \sqrt 3) cos\theta = sin \theta

    But then I had no idea where to go from there so I just assumed I was wrong
    Now divide by cosine to get tan and solve it that way.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by RDKGames)
    Now divide by cosine to get tan and solve it that way.
    Omg, I'm actually going to kick myself!! Managed to get there in the end.

     tan \theta = 3- \sqrt 3

    and then solved it from there to get...  \theta = 51.7,  231.7







    Thank you RDKGames and notnek I'm forever grateful for your help!!
    • Thread Starter
    Offline

    3
    ReputationRep:
    Name:  C3 EXE7A Q13.png
Views: 19
Size:  6.6 KB

    For part (c) should I start by doing?...

     cos (\theta + 25) + sin (\theta +65) = 1

     (cos \theta cos 25 - sin \theta sin 25) + (sin \theta cos 65 +cos \theta sin65) = tan 45

     (cos \theta cos 25 - sin \theta sin 25) + (sin \theta cos 65 +cos \theta sin65) = \frac{sin45}{cos45}

    and then should I times by cos(45)?? Or am I being completely retarded here? actually have no idea what I'm doing.
    Why do I suck at trigonometry?
    • TSR Support Team
    • Study Helper
    Offline

    3
    ReputationRep:
    (Original post by Philip-flop)
    Name:  C3 EXE7A Q13.png
Views: 19
Size:  6.6 KB

    For part (c) should I start by doing?...

     cos (\theta + 25) + sin (\theta +65) = 1

     (cos \theta cos 25 - sin \theta sin 25) + (sin \theta cos 65 +cos \theta sin65) = tan 45

     (cos \theta cos 25 - sin \theta sin 25) + (sin \theta cos 65 +cos \theta sin65) = \frac{sin45}{cos45}

    and then should I times by cos(45)?? Or am I being completely retarded here? actually have no idea what I'm doing.
    Why do I suck at trigonometry?
    Since 25+65 = 90 we have

    \cos 65 = \sin 25 and \cos 25 = \sin 65

    Try using this here and look out for angle sum of 90 in future questions - you need it in different types of trig questions.

    Also, don't immediately change 1 into \tan 45 unless you think it will be useful. It was useful for a couple of questions that you've done but not here.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by notnek)
    Since 25+65 = 90 we have

    \cos 65 = \sin 25 and \cos 25 = \sin 65

    Try using this here and look out for angle sum of 90 in future questions - you need it in different types of trig questions.

    Also, don't immediately change 1 into \tan 45 unless you think it will be useful. It was useful for a couple of questions that you've done but not here.
    Ohhhh I see.

    I never knew that if...  cos(a) and sin(b) where a and b sum up to 90 degrees that...  cos(a) = sin(b)
    Again, the book fails to explain any of that!

    Ok, so for this question I would do?...

     cos (\theta + 25) + sin (\theta +65) = 1

     (cos \theta cos 25 - sin \theta sin 25) + (sin \theta cos 65 +cos \theta sin65) = 1

     (cos \theta sin 65 - sin \theta cos 65) + (sin \theta cos 65 +cos \theta sin65) = 1

     2(cos \theta sin65) = 1

    But then where from there? :/
    • TSR Support Team
    • Study Helper
    Offline

    3
    ReputationRep:
    (Original post by Philip-flop)
    Ohhhh I see.

    I never knew that if...  cos(a) and sin(b) where a and b sum up to 90 degrees that...  cos(a) = sin(b)
    Again, the book fails to explain any of that!

    Ok, so for this question I would do?...

     cos (\theta + 25) + sin (\theta +65) = 1

     (cos \theta cos 25 - sin \theta sin 25) + (sin \theta cos 65 +cos \theta sin65) = 1

     (cos \theta sin 65 - sin \theta cos 65) + (sin \theta cos 65 +cos \theta sin65) = 1

     2(cos \theta sin65) = 1

    But then where from there? :/
    From there \sin 65 is just a number you can work out on your calculator. So you can move it to the other side along with the 2.

    The 'adding to 90' rule is just an application of the identity \sin x = \cos \left(90-x\right) e.g. where x is 25.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by notnek)
    From there \sin 65 is just a number you can work out on your calculator. So you can move it to the other side along with the 2.

    The 'adding to 90' rule is just an application of the identity \sin x = \cos \left(90-x\right) e.g. where x is 25.
    Thank you! Not sure why I panicked when I saw...
     2(cos \theta sin65) = 1

 
 
 
Poll
What is your favourite film genre?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

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

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.