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HL Maths-trig

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  • 28-11-2009 7:30pm
    #1
    gaeilge-abú
    Closed Accounts Posts: 79 ✭✭


    any1 able to help me out with this problem?

    show that 2sin(135degree's + A)sin(45degree's + A)=cos2A

    thanks in advance!:D


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    #2
    blubloblu
    Registered Users Posts: 1,226 ✭✭✭blubloblu


    Look at page 15 of the new formula book or page 9 of the old ones.


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    #3
    gaeilge-abú
    Closed Accounts Posts: 79 ✭✭gaeilge-abú


    yee tried them just cant figure it out!


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    #4
    jumpguy
    Registered Users Posts: 7,962 ✭✭✭jumpguy


    This is an trigonometric identities question and is therefore rather difficult to do sitting on your computer. I'd advise you to ask your maths teacher and do your best. First step anyway would be to multiply out your brackets.


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    #5
    **Timbuk2**
    Moderators, Education Moderators, Motoring & Transport Moderators Posts: 7,395 Mod ✭✭✭✭**Timbuk2**


    gaeilge-abú wrote: »
    any1 able to help me out with this problem?

    show that 2sin(135degree's + A)sin(45degree's + A)=cos2A

    thanks in advance!:D

    [latex]
    2\sin(135 + A)\sin(45+A)
    [/latex]

    From the log tables: [latex]2\sin{A}\sin{B} = \cos(A-B) - \cos(A+B)[/latex]

    Therefore
    [latex]\cos(135+A-45-A) - \cos(135 + A + 45 + A)[/latex]
    [latex]\cos90 - \cos(180 + 2A)[/latex]

    Now, you have to decide what to do next. The cos90 is straight forward, but something needs to be done about the cos(180+2A). If you are stuck, have a look on pg 14 of the log tables. You will find this:
    [latex]\cos(A+B) = \cos{A}\cos{B} - \sin{A}\sin{B}[/latex]

    Therefore
    [latex]\cos90 - (\cos{180}\cos{2A} - \sin{180}\sin{2A})[/latex]
    [latex]0 - [(-1)(\cos{2A}) - 0][/latex]
    [latex]-(-\cos{2A})[/latex]
    [latex]\cos{2A}[/latex]

    I hope this helps! Remember, with trig identities, look in the log tables if you are stuck. There is nearly always a formula that help you - there are usually multiple ways of proving trig identities.


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    #6
    dustintheturkey
    Closed Accounts Posts: 13 ✭dustintheturkey


    You can use the trig pages of the maths book (tables) on page 14.

    Use these two...

    Cos(a+b)=CosaCosb-SinaSinb and
    Cos(a-b)=CosACosB+SinASinB

    Subtract the first one from the second to get

    Cos(a-b)-Cos(a+b)=2SinaSinb, so in your example, we get

    Cos90-Cos(180+2A)=2Sin(135+A)Sin(45+A) which is 0-Cos(180+2A).

    Remember you can't multiply out brackets here.
    The multiplication is the product of two Sines.
    It's the angles that are in brackets,

    Now, remembering that in the circle of radius 1 unit, Cos(angle) is the x co-ordinate
    and Sin(angle) is the y co-ordinate,
    therefore if we increase the angle by 180 degrees, we change the sign of the co-ordinate.
    So, -Cos(180+2A) is Cos2A.

    Or use Cos(180+2A)=Cos180Cos2A-Sin180Sin2A=-Cos2A from Cos(A+B) = CosACosB-SinASinB


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