Turn on thread page Beta

can someone please help me with this C3 trig question watch

    • Thread Starter
    Offline

    2
    ReputationRep:
    Hence or otherwise show that the equation:

    cosx (cos2x/cosx + sinx) = 1/2, can be written as:

    sin2x = cos2x


    afterwards, how do I re-arrange to solve 'sin2x = cos2x', 0< x < 2pi
    Offline

    8
    ReputationRep:
    clearly sin 2x =0 or cos 2x=0 are not solutions. so we can divide by one of them.
    Offline

    12
    ReputationRep:
    (Original post by W.H.T)
    Hence or otherwise show that the equation:

    cosx (cos2x/cosx + sinx) = 1/2, can be written as:

    sin2x = cos2x


    afterwards, how do I re-arrange to solve 'sin2x = cos2x', 0< x < 2pi
    Divide both sides by cos2x to get tan2x =1

    Offline

    0
    ReputationRep:
    (Original post by W.H.T)
    Hence or otherwise show that the equation:

    cosx (cos2x/cosx + sinx) = 1/2, can be written as:

    sin2x = cos2x

    and afterwards, how do I re-arrange to solve 'sin2x = cos2x', 0< x < 2pi
    multiply out the brackets...
    cos2x + sinx cosx = 1/2

    then use this identity...
    sin(2x) = 2 sinx cosx

    and then to solve divide both sides by cos(2x), so sin(2x)/cos(2x) = tan(2x) = 1.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by kat_s)
    multiply out the brackets...
    cos2x + sinx cosx = 1/2

    then use this identity...
    sin(2x) = 2 sinx cosx

    and then to solve divide both sides by cos(2x), so sin(2x)/cos(2x) = tan(2x) = 1.

    for the first step of multiplying out the brackets, would I get:

    cos^2 2x / cosx+sinx = 1/2
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Superman_Jr)
    Divide both sides by cos2x to get tan2x =1

    sin2x = cos2x

    how can I divide by cos2x, because on the 'cos2x' side there is nothing multiplying it. So surely to get 'cos2x' to the other side, I would have to subtract by 'cos2x' ?
    Offline

    2
    ReputationRep:
    (Original post by W.H.T)
    sin2x = cos2x

    how can I divide by cos2x, because on the 'cos2x' side there is nothing multiplying it. So surely to get 'cos2x' to the other side, I would have to subtract by 'cos2x' ?
    when you divide by cos2x, you'd be left with 1. And you can divide by cos2x as long as you do the same to both sides, the equation remains unchanged.
    Offline

    0
    ReputationRep:
    (Original post by W.H.T)
    for the first step of multiplying out the brackets, would I get:

    cos^2 2x / cosx+sinx = 1/2
    cosx (cos2x/cosx + sinx) = 1/2

    (cosx cos2x)/cosx + cosx sinx = 1/2

    the cos x at the top and bottom of the first fraction cancel.

    Unless you mean cosx (cos2x/(cosx +sinx)) for the original equation??
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by kat_s)
    cosx (cos2x/cosx + sinx) = 1/2

    (cosx cos2x)/cosx + cosx sinx = 1/2

    the cos x at the top and bottom of the first fraction cancel.

    Unless you mean cosx (cos2x/(cosx +sinx)) for the original equation??

    yeah my original equation is: cosx (cos2x/(cosx +sinx)) = 1/2

    I don;t understand how you got the part in bold here:
    (cosx cos2x)/cosx + cosx sinx = 1/2
    Offline

    2
    ReputationRep:
    how do you solve 2sin(2x)+4cos^2(x)=3
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: May 3, 2015

University open days

  1. University of Bradford
    University-wide Postgraduate
    Wed, 25 Jul '18
  2. University of Buckingham
    Psychology Taster Tutorial Undergraduate
    Wed, 25 Jul '18
  3. Bournemouth University
    Clearing Campus Visit Undergraduate
    Wed, 1 Aug '18
Poll
How are you feeling in the run-up to Results Day 2018?
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

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