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    Ok, im really stuck on a certain question, and have been through the whole book and internet and cannot find anything to help me, so hopefully you can?

    The question is:

    Show that the equation
    Sin( x + 30° ) = 2 cos ( x + 60° )
    Can be written in the form 3(root)3 sin x = 1 cos x

    Thanks
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     \cos \theta \equiv \sin (90-\theta ) .
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    (Original post by B_9710)
     \cos \theta \equiv \sin (90-\theta ) .
    But don't you have to keep both the cos and the sin as they are both in the final answer?
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    Just expand and rearrange to get the final expression.
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    (Original post by BAMItzCallum)
    But don't you have to keep both the cos and the sin as they are both in the final answer?
    I will look at it properly. Give me a minute.
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    Thanks
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    (Original post by BAMItzCallum)
    Ok, im really stuck on a certain question, and have been through the whole book and internet and cannot find anything to help me, so hopefully you can?

    The question is:

    Show that the equation
    Sin( x + 30° ) = 2 cos ( x + 60° )
    Can be written in the form 3(root)3 sin x = 1 cos x

    Thanks
    Expand Sin (A+B) = sin A cos B + cos A sin B and cos (A+B) = cosA cos B - sin A sin B and use sin 60 = root3/2 = cos 30 and sin 30= 1/2 = cos 60. Then rearrange to get final expression.
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    (Original post by tangotangopapa2)
    Just expand and rearrange to get the final expression.
    I tried that, and came out with (root3)/2 sinx + 1/2 cosx = 1 cox - (root3) sinx
    Which is completely wrong
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    That is not wrong take cos x terms to one side and sin x terms to other side. you get 3 root 3 sinx /2 = cos x /2. Cancel 2 to get final expression

    (Original post by BAMItzCallum)
    I tried that, and came out with (root3)/2 sinx + 1/2 cosx = 1 cox - (root3) sinx
    Which is completely wrong
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    (Original post by tangotangopapa2)
    That is not wrong take cos x terms to one side and sin x terms to other side. you get 3 root 3 sinx /2 = cos x /2. Cancel 2 to get final expression
    if i put them on the same side though, i get:
    (root3)/2 sinx + (root3)sinx = 1cosx + 1/2 cosx
    :/
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    root3/2 + root3 is 3root3/2
    You should get 1 cos x - 1/2 cos x on RHS.

    (Original post by BAMItzCallum)
    if i put them on the same side though, i get:
    (root3)/2 sinx + (root3)sinx = 1cosx + 1/2 cosx
    :/
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    (Original post by tangotangopapa2)
    root3/2 + root3 is 3root3/2
    You should get 1 cos x - 1/2 cos x on RHS.
    Derp. Thanks so much!
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    (Original post by BAMItzCallum)
    Derp. Thanks so much!
    You're very welcome.
 
 
 
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