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    Using the identity cos(A+B)= cosAcosB - sinAsinB, prove that

    cos2A = 1-2sin^2 A


    So far I've got that cos2A = cos^2 A - sin^2 A

    but I dont know where to go from here.. any help?
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    (Original post by JosephinaMarley)

    Using the identity cos(A+B)= cosAcosB - sinAsinB, prove that

    cos2A = 1-2sin^2 A


    So far I've got that cos2A = cos^2 A - sin^2 A

    but I dont know where to go from here.. any help?
    Use the identity sin^2 A + cos^2 A = 1
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    (Original post by Clarity Incognito)
    Use the identity sin^2 A + cos^2 A = 1
    But its not '+', all I have is a minus...
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    (Original post by JosephinaMarley)

    Using the identity cos(A+B)= cosAcosB - sinAsinB, prove that

    cos2A = 1-2sin^2 A


    So far I've got that cos2A = cos^2 A - sin^2 A

    but I dont know where to go from here.. any help?

    say b=A

    cos(A+A)=cosAcosA-sinASinA

    Cos2A=(cosA)^2-(sinA)^2

    we want sin only so substitue in 1-sin^2A=(cosA)^2

    cos2a=1-sin^2A-sin^2A=1-2sin^2A
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    (Original post by JosephinaMarley)
    But its not '+', all I have is a minus...
    Rearrange and substitute then!
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    Cos2A = Cos^2A - Sin^2A
    =2cos^2A - 1
    =1-2sin^2A
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    cos (A+B) = cosAcosB - sinAsinB
    so
    cos (A+A) = cosAcosA - sinAsinA
    cos (2A) = cos^2 A - sin^2 A
    cos 2A = cos^2 A - (1-cos^2 A) *(using sin^2 A + cos^2 A = 1)
    cos 2A = cos^2 A -1 + cos^2 A
    cos 2A = 2cos^2 A -1

    Hope this helped...
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    (Original post by Remarqable M)
    say b=A

    cos(A+A)=cosAcosA-sinASinA

    Cos2A=(cosA)^2-(sinA)^2

    we want sin only so substitue in 1-sin^2A=(cosA)^2

    cos2a=1-sin^2A-sin^2A=1-2sin^2A
    Thankyou!!! I feel so dull for not seeing that!
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    Ryt, here goes.myt be a bit l8 considering the exam is in 2 hours

    Cos^2 A - Sin^2 A= Cos^2 A -(1- Cos^2 A)

    = 2Cos^2 A -1= 2(1-Sin^2 A)-1

    = 2-2Sin^2 A -1

    =1-2Sin^2 A


    Boom!

    ur welcome
 
 
 
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