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    Express  7\sin \theta + 3 \cos \theta into Rsin(\theta + \alpha) where R > 0 and \alpha is an acute angle.

    I have no idea what to do.... I got 3 = Rcosx and 7 = Rsinx, but I'm not sure If i'm doing the right direction to get the answer R = \sqrt 58 and \alpha = sin^-1 (\frac {3}{R})
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    tan (alpha) = (7/3) -------------(from Rsin(alpha)/Rcos(alpha)=7/3)
    Then sub in to get R to one of the previous two equations
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    (Original post by KingPin)
    tan (alpha) = (7/3) -------------(from Rsin(alpha)/Rcos(alpha)=7/3)
    Then sub in to get R to one of the previous two equations
    correct, pretty simple rlly.
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    (Original post by piccolo12345)
    Express  7\sin \theta + 3 \cos \theta into Rsin(\theta + \alpha) where R > 0 and \alpha is an acute angle.

    I have no idea what to do.... I got 3 = Rcosx and 7 = Rsinx, but I'm not sure If i'm doing the right direction to get the answer R = \sqrt 58 and \alpha = sin^-1 (\frac {3}{R})
    That's right. Time to get the calc out to get alpha.

    EDIT: I meant the textbook was right.
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    use R^2(sin^2a+cos^2a) to find R
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    (Original post by KingPin)
    tan (alpha) = (7/3) -------------(from Rsin(alpha)/Rcos(alpha)=7/3)
    Then sub in to get R to one of the previous two equations
    How do the answers get alpha as 23.2? I'm supposed to do this by putting tan^-1(7/3) in the calculator arent I? which gets 66.8
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    (Original post by piccolo12345)
    How do the answers get alpha as 23.2? I'm supposed to do this by putting tan^-1(7/3) in the calculator arent I? which gets 66.8
    tanx = 3/7 in this case, he did it wrong. And it was a pointless post, because you already had R = sqrt58 and an expression for alpha.
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    (Original post by Physics Enemy)
    tanx = 3/7 in this case, he did it wrong. And it was a pointless post, because you already had R = sqrt58 and an expression for alpha.
    Sorry, I went from his results...lazy I know
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    (Original post by Physics Enemy)
    tanx = 3/7 in this case, he did it wrong. And it was a pointless post, because you already had R = sqrt58 and an expression for alpha.
    Oh did I get the 3 and 7 mixed up at the start?
    And so now I've got alpha as 23.2

    So now to find R, I substitute it into an equation, so R = 3/cos(23.2), which gets 3.26??
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    (Original post by piccolo12345)
    Oh did I get the 3 and 7 mixed up at the start?
    And so now I've got alpha as 23.2

    So now to find R, I substitute it into an equation, so R = 3/cos(23.2), which gets 3.26??
    Rsin(x + a) = (Rcosa)sinx + (Rsina)cosx
    Rcosa = 7, Rsina = 3
    R^2 = 58 => R = Sqrt(58)
    a = sin^-1[3/Sqrt(58)]

    Yes you got them mixed up. And no I wouldn't use the angle to get R.
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    I always find it easy to remember the proof (then again, I know a lot of people who can't). But remember your double angle formula:
    RSin(\theta + \alpha) = RCos(\alpha)Sin(\theta) + RSin(\alpha)Cos(\theta)
    So looking at the question:
    RCos(\alpha) = 7
    RSin(\alpha) = 3
    So Tan(\alpha) = ?
    Therefore:
    R^2 Sin^2(\alpha) + R^2 Cos^2(\alpha) = \sqrt(7^2 + 3^2)
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    (Original post by Physics Enemy)
    Rsin(x + a) = (Rcosa)sinx + (Rsina)cosx
    Rcosa = 7, Rsina = 3
    R^2 = 58 => R = Sqrt(58)
    a = sin^-1[3/Sqrt(58)]

    Yes you got them mixed up. And no I wouldn't use the angle to get R.
    Cheers
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    (Original post by Wednesday Bass)
    I always find it easy to remember the proof (then again, I know a lot of people who can't). But remember your double angle formula:
    RSin(\theta + \alpha) = RCos(\alpha)Sin(\theta) + RSin(\alpha)Cos(\theta)
    So looking at the question:
    RCos(\alpha) = 7
    RSin(\alpha) = 3
    So Tan(\alpha) = ?
    Therefore:
    R^2 Sin^2(\alpha) + R^2 Cos^2(\alpha) = \sqrt(7^2 + 3^2)
    You mind helping me on part c, although i dont think they are related. i got so much to learn for the exam in a few weeks @@@@@@@@@

    Question is

    ci) Given that \beta is an acute angle and tan \beta= 2 \sqrt 2 , show that cos\beta = \frac {1}{3}

    ii) Hence show that sin2\beta = p\sqrt 2 where p is a rational number

    ii uses the double angle formula im guessing but i dont know what to do at all for part i)
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    (Original post by piccolo12345)
    You mind helping me on part c, although i dont think they are related. i got so much to learn for the exam in a few weeks @@@@@@@@@

    Question is

    ci) Given that \beta is an acute angle and tan \beta= 2 \sqrt 2 , show that cos\beta = \frac {1}{3}

    ii) Hence show that sin2\beta = p\sqrt 2 where p is a rational number

    ii uses the double angle formula im guessing but i dont know what to do at all for part i)
    I think I've seen this question before in one of my past papers.

    First Part:
    So you know that Tan\beta = 2\sqrt{2}
    Therefore we can conclude that:
    \displaystyle{\frac{Sin\beta}{Co  s\beta} = 2\sqrt{2}}
    \displaystyle{\frac{\sqrt{1 - Cos^2{\beta}}}{Cos\beta} = 2\sqrt{2}}
    So...
    \sqrt{1 - Cos^2{\beta}} =2\sqrt{2}Cos\beta
    Then square both sides and it should be obvious.

    Second Part:
    We know Tan\beta and Cos\beta. So we can find Sin\beta
    And remember:
    Sin{2\theta} = 2Sin{\theta}Cos{\theta}

    Problem solved
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    (Original post by Wednesday Bass)
    I think I've seen this question before in one of my past papers.

    First Part:
    So you know that Tan\beta = 2\sqrt{2}
    Therefore we can conclude that:
    \displaystyle{\frac{Sin\beta}{Co  s\beta} = 2\sqrt{2}}
    \displaystyle{\frac{\sqrt{1 - Cos^2{\beta}}}{Cos\beta} = 2\sqrt{2}}
    So...
    \sqrt{1 - Cos^2{\beta}} =2\sqrt{2}Cos\beta
    Then square both sides and it should be obvious.

    Second Part:
    We know Tan\beta and Cos\beta. So we can find Sin\beta
    And remember:
    Sin{2\theta} = 2Sin{\theta}Cos{\theta}

    Problem solved
    sorry must sound like an idiot but what do you get if you square  2\sqrt 2 cos \beta

    And just wondering the mark scheme seems to have a "shortcut answer" ... using pythagorous ... h^2 = 1 + (2 \sqrt 2 )^2 , h = 3, therefore cos \beta = \frac {1}{3} ?

    and thanks perfectly understand part ii
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    (Original post by piccolo12345)
    sorry must sound like an idiot but what do you get if you square  2\sqrt 2 cos \beta

    And just wondering the mark scheme seems to have a "shortcut answer" ... using pythagorous ... h^2 = 1 + (2 \sqrt 2 )^2 , h = 3, therefore cos \beta = \frac {1}{3} ?

    and thanks perfectly understand part ii
    (2\sqrt{2}Cos\beta)^2 = (2)^2(\sqrt{2})^2 (Cos\beta)^2 = 8Cos^2\beta
 
 
 
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