Turn on thread page Beta
    • Thread Starter
    Offline

    20
    ReputationRep:
    I've been asked to prove a trig identity. It involves cos 4x

    Since sin is on the left hand side, I know I must use:

    Cos 2x = 1 - 2sin²x

    But how can i change it to cos 4x? I never learnt this is C3

    Do you just square everything so it becomes:

    cos 4x = 1-4sin^4 2x
    Offline

    19
    Squaring would give you cos^2 2x = (1-2sin^2x)^2

    Try cos (A+B)
    Offline

    16
    ReputationRep:
    Say we have this:
    \cos{2A} = 1 - 2\sin^2{A}

    Where A could be x or 2x...

    Can you do it now?
    Offline

    0
    ReputationRep:
    If you replace x by 2x you obtain cos4x=1-2(sin2x)^2. Squaring everything doesn't give the expression you've written down.
    • Thread Starter
    Offline

    20
    ReputationRep:
    (Original post by TheTallOne)
    Say we have this:
    \cos{2A} = 1 - 2\sin^2{A}

    Where A could be x or 2x...

    Can you do it now?

    (Original post by EierVonSatan)
    Squaring would give you cos^2 2x = (1-2sin^2x)^2

    Try cos (A+B)
    Ok, using both your methods:

    cos(A+B)=cos A cos B - sin A sin B

    Therefore Cos (2x + 2x)= Cos 2x cos 2x - sin2x sin2x

    Therefore:  Cos (4x) = Cos^2 (2x) - sin^2(2x)

    Thus: Cos (4x) = 1 - 2sin^2(2x) ???
    Offline

    16
    ReputationRep:
    (Original post by djpailo)
    Therefore:  Cos (4x) = Cos^2 (2x) - sin^2(2x)

    Thus: Cos (4x) = 1 - 2sin^2(2x) ???
    And remember:
    \cos^2{x} + \sin^2{x}=1
    • Thread Starter
    Offline

    20
    ReputationRep:
    I don't know where to go from there
    Offline

    19
    What's the identity you're trying to prove?
    • Thread Starter
    Offline

    20
    ReputationRep:
     (1 + sin (2x) )^2 = 1/2((3 + 4sin (2x) - cos (4x))
    Offline

    19
    So using your new formula for cos 4x and rewriting sin 2x = S you should be able to do it
    • Thread Starter
    Offline

    20
    ReputationRep:
    (Original post by EierVonSatan)
    So using your new formula for cos 4x and rewriting sin 2x = S you should be able to do it
    Cool. I thought

    Cos (4x) = 1 - 2sin^2(2x)

    was wrong lol. Okay, I'll try it out now.
    • Thread Starter
    Offline

    20
    ReputationRep:
    I got it to:

     1 + 2sin (2x) + 2 sin^2 (2x) but got stuck.

    I could use sin^2...=1 but I don't think it'd get be anywhere :s

    Edit: NVM! I found out where I went wrong. When I got rid of the 1/2 outside the bracket, I carelessly let the 2 remain on 2 sin^2 (2x) . Thanks for all your help guys
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: March 18, 2009

University open days

  • University of Lincoln
    Mini Open Day at the Brayford Campus Undergraduate
    Wed, 19 Dec '18
  • University of East Anglia
    UEA Mini Open Day Undergraduate
    Fri, 4 Jan '19
  • Bournemouth University
    Undergraduate Mini Open Day Undergraduate
    Wed, 9 Jan '19
Poll
Were you ever put in isolation at school?
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

Equations

Best calculators for A level Maths

Tips on which model to get

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.