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Help with Maths Homework! Compound and Double angle formula !! watch

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    Find the value of tanx and hence solve the equation for 0<x<360
    a) sin(x-30)=cosx
    For this one I got x=30, 210 but I'm not sure if that's correct!
    b) cos(x+60) = 2cos(x+30)
    I am COMPLETELY stuck ...
    Any help would be appreciated ! Thanks!
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    expand sin( x - 30° ) using sin ( A - B ) ≡ sinAcosB - cosAsinB
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    (Original post by the bear)
    expand sin( x - 30° ) using sin ( A - B ) ≡ sinAcosB - cosAsinB
    This.

    And I'll also add that you should always just draw out sine and cosine curves if you're in any doubt. For example, if x was 30 in the first question (sin(x-30) = cos(x)) you'd get sin(0) = cos(30) which can't ever be true.
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    (Original post by the bear)
    expand sin( x - 30° ) using sin ( A - B ) ≡ sinAcosB - cosAsinB
    Which i've done and got : sin(x)cos(-30) - cos(x)sin(-30) = cosx Then I divided it by (cosx)(cos(-30)) Should I have done that?!
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    here A is x

    B is 30° not -30°
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    (Original post by the bear)
    here A is x

    B is 30° not -30°

    Ok so I've worked it out and I got: x=60,240

    Part b)??
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    I wouldn't use addition formula unless you're specifically asked to here.
    Note that  \cos x\equiv \sin (90-x) . Then you get  \sin (x-30)=\sin (90-x) .
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    (Original post by Sophie_Banwell)
    Ok so I've worked it out and I got: x=60,240

    Part b)??
    those look correct.

    b) i would use the expansion for cos ( D + E )....
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    (Original post by the bear)
    those look correct.

    b) i would use the expansion for cos ( D + E )....

    The cos(A+B)=cosAcosB+sinAsinB identity? So I would then get cosxcos60 + sinxsin60 = 2(cosxcos30 + sinxsin30) ??
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    (Original post by Sophie_Banwell)
    The cos(A+B)=cosAcosB+sinAsinB identity? So I would then get cosxcos60 + sinxsin60 ??
    there is a slight mistake in your version of the identity.
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    (Original post by the bear)
    there is a slight mistake in your version of the identity.
    Sorry a typo it should be : cos(A+B) = cosAcosB - sinAsinB So I would get: cosxcos60 - sinxsin60 = 2(cosxcos30 - sinxsin30) ??
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    (Original post by Sophie_Banwell)
    Sorry a typo it should be : cos(A+B) = cosAcosB - sinAsinB So I would get: cosxcos60 - sinxsin60 = 2(cosxcos30 - sinxsin30) ??
    yes
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    (Original post by the bear)
    yes
    Now what do I do? Do I divide it all by ((cosx)(cos30)) ?
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    (Original post by Sophie_Banwell)
    Now what do I do? Do I divide it all by ((cosx)(cos30)) ?
    you could divide each term by cosx ...
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    But if I divide by cos(x) I get :
    tanx(2sin30-sin60) = 2cos30-cos60

    Is this even right?
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    (Original post by the bear)
    you could divide each term by cosx ...
    But if I divide by cos(x) I get :
    tanx(2sin30-sin60) = 2cos30-cos60

    Is this even right?
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    (Original post by Sophie_Banwell)
    But if I divide by cos(x) I get :
    tanx(2sin30° -sin60° ) = 2cos30° -cos60°

    Is this even right?
    that looks fine...

    now using your knowledge of special angles, replace the sin30° etc with fractions.
 
 
 
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