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    Solve equation root2sin(theta+45) = 3costheta, 0<theta<360.

    Okay where do I even start!!! I'm so STUCK, I tried the addition formula but I don't think I'm getting anywhere ...
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    (Original post by plower)
    Solve equation root2sin(theta+45) = 3costheta, 0<theta<360.

    Okay where do I even start!!! I'm so STUCK, I tried the addition formula but I don't think I'm getting anywhere ...
     \displaystyle \sin (a+b)  = ( \sin a)( \cos b) + ( \cos a)( \sin b)

    try using that on  \displaystyle \sin ( \theta + 45)
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    (Original post by plower)
    Solve equation root2sin(theta+45) = 3costheta, 0<theta<360.

    Okay where do I even start!!! I'm so STUCK, I tried the addition formula but I don't think I'm getting anywhere ...
    Your approach is correct - please post your working.
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    (Original post by DylanJ42)
     \displaystyle \sin (a+b)  = ( \sin a)( \cos b) + ( \cos a)( \sin b)

    try using that on  \displaystyle \sin ( \theta + 45)

    I did and then I got root2sinthetacos45 + root2costhetasin45 = 3costheta . So do I now divide by cos fit all of them?
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    (Original post by plower)
    I did and then I got root2sinthetacos45 + root2costhetasin45 = 3costheta . So do I now divide by cos fit all of them?
    If you evaluate all the constants then it should make things simpler i.e. what is cos(45), sin(45) ?

    Then you should be able to simplify to just have a single sine term and a single cosine term and it will look like a C2 question.
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    here is a theta for you : Θ

    and a square root: √

    and a degree symbol: °




    you cannot "divide by cos"....

    √2sinΘcos45° + √2cosΘsin45° = 3cosΘ

    instead join the two terms with cosΘ together... also put in fractions for cos45° and sin45°
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    (Original post by notnek)
    If you evaluate all the constants then it should make things simpler i.e. what is cos(45), sin(45) ?

    Then you should be able to simplify to just have a single sine term and a single cosine term and it will look like a C2 question.
    (Original post by the bear)
    here is a theta for you : Θ

    and a square root: √

    and a degree symbol: °




    you cannot "divide by cos"....

    √2sinΘcos45° + √2cosΘsin45° = 3cosΘ

    instead join the two terms with cosΘ together... also put in fractions for cos45° and sin45°
    (Original post by notnek)
    Your approach is correct - please post your working.
    Oooh GOT IT !! Thankyou so much guys for correcting me <3
 
 
 
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