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    a) Find all possible values of sin C for which cos C = 1/2.
    b) Find the values of D for which -180 degrees <D< 180 degrees and tan D = 5 sin D.

    Right so for part a), I did cos-1 (1/2) = 60, and then sin 60 = root 3 / 2. I am told that this is one of the answers, and the other is meant to be - root 3 / 2, how do I get this?

    I don't know how to approach part b), so any help on both would be really appreciated thanks!
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    for c2 you only learn 2 things:

    cos^2 x + sin^2 x =1

    tanx=(sinx)/(cosx)

    sin is postive for angles between 0 and 180 and negative 180< to 360, cos is positive for angles 0 to 90 and 270 to 360 negative 90-270

    use these in a) and b) in a way you think helps you the most.......
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    you get - \frac{\sqrt3}{2} by looking at your graph of y=cosx and figuring out that there is another value C for which cosC=\frac{1}{2}
    I'm assuming somewhere in the question it says -180\le C \le 180

    for b, I'd try to find a value D for which both sind and tand = 0. because then 5sind and tand will = 0.
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    (Original post by mathz)
    for c2 you only learn 2 things:

    cos^2 x + sin^2 x =1

    tanx=(sinx)/(cosx)

    sin is postive for angles between 0 and 180 and negative 180< to 360, cos is positive for angles 0 to 90 and 270 to 360 negative 90-270

    use these in a) and b) in a way you think helps you the most.......
    So for a), to get the other answer you would do 360 - 60 = 300, and then do sin 300 = - root 3/2?
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    (Original post by adil12)
    a) Find all possible values of sin C for which cos C = 1/2.
    b) Find the values of D for which -180 degrees <D< 180 degrees and tan D = 5 sin D.

    Right so for part a), I did cos-1 (1/2) = 60, and then sin 60 = root 3 / 2. I am told that this is one of the answers, and the other is meant to be - root 3 / 2, how do I get this?

    I don't know how to approach part b), so any help on both would be really appreciated thanks!
    awwwwwwwww, its been about 7 months since i didnt do any maths, i miss it lol

    the first part is this:
    cos^2 C + sin^2 C = 1
    cos ^2 C = 1/4
    sin^2 C = 1 - 1/4 = 3/4
    sin C = root(3) / 2 = pi / 3 (Im not sure, double check on calculator)
    and then depending on the range, there are certain other answers, i think you can do it from here.

    second part:
    tan D = 5sin D
    sin D / cos D = 5 sin D
    sinD / sind D = 5 cos D
    cos D = 1/5
    and calculate this on your calculator
    i hope this helps)
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    (Original post by vahik92)
    awwwwwwwww, its been about 7 months since i didnt do any maths, i miss it lol

    the first part is this:
    cos^2 C + sin^2 C = 1
    cos ^2 C = 1/4
    sin^2 C = 1 - 1/4 = 3/4
    sin C = root(3) / 2 = pi / 3 (Im not sure, double check on calculator)
    and then depending on the range, there are certain other answers, i think you can do it from here.

    second part:
    tan D = 5sin D
    sin D / cos D = 5 sin D
    sinD / sind D = 5 cos D
    cos D = 1/5
    and calculate this on your calculator
    i hope this helps)
    I see Thanks for the help For the second part, once you get cos D = 1/5, you calculate D to be 78.46 degrees, but the answers at the back say that you should get 78.46 (which I have got), -78.46 and 0. Any idea how to get the other 2 answers?
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    (Original post by TimmonaPortella)
    you get - \frac{\sqrt3}{2} by looking at your graph of y=cosx and figuring out that there is another value C for which cosC=\frac{1}{2}
    I'm assuming somewhere in the question it says -180\le C \le 180

    for b, I'd try to find a value D for which both sind and tand = 0. because then 5sind and tand will = 0.
    I am asking how to do it without a graph And nope, in the question it just says "all possible values". And oh right you are right there
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    (Original post by adil12)
    I am asking how to do it without a graph And nope, in the question it just says "all possible values". And oh right you are right there
    oh no, actually what am I thinking, C doesn't need confining :rolleyes:
    I apologise.

    and really, you should just be able to sketch out a graph of y=cosx. though you needn't actually use it for that question, you should just know that cos60=0.5 and that cos(-60)= 0.5
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    (Original post by TimmonaPortella)
    oh no, actually what am I thinking, C doesn't need confining :rolleyes:
    I apologise.

    and really, you should just be able to sketch out a graph of y=cosx. though you needn't actually use it for that question, you should just know that cos60=0.5 and that cos(-60)= 0.5
    Oh right I see So for the final value you would do 360 - 60 = 300, and then sin 300 = - root 3/2 ?
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    (Original post by adil12)
    I see Thanks for the help For the second part, once you get cos D = 1/5, you calculate D to be 78.46 degrees, but the answers at the back say that you should get 78.46 (which I have got), -78.46 and 0. Any idea how to get the other 2 answers?
    erm, i know how to get -78.46 , cos X = cos (-X) for any value of X, i.e. cos 78.46 = cos (-78.46).
    but for 0, i dont know ( for 0, cos X = 1, which we didnt get, we got cos X = 1/5.. sorry :confused:
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    (Original post by vahik92)
    erm, i know how to get -78.46 , cos X = cos (-X) for any value of X, i.e. cos 78.46 = cos (-78.46).
    but for 0, i dont know ( for 0, cos X = 1, which we didnt get, we got cos X = 1/5.. sorry :confused:
    Oh right so is cos X = cos (-X) a general rule which only applies for cos?
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    (Original post by adil12)
    Oh right so is cos X = cos (-X) a general rule which only applies for cos?
    yep, cos x = cos -x for all the values of x. these are all basic things which you should really know for C2 and you will have to use for C3 and C4,, make sure u learn these formulaes as there are quite a lot of them, without them you wont be able to get good grades .. just an advice
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    (Original post by mathz)
    for c2 you only learn 2 things:

    cos^2 x + sin^2 x =1

    tanx=(sinx)/(cosx)

    sin is postive for angles between 0 and 180 and negative 180< to 360, cos is positive for angles 0 to 90 and 270 to 360 negative 90-270

    use these in a) and b) in a way you think helps you the most.......
    This Get used to the basics of trig now..it gets harder at A2..once youve got these basics (ie learning how to solve sinx = a, cosx = a, tanx = a (ie sin-1(a) = x, then using a cast diagram) will help a lot..also learn to use sin^2(x)+cos^2(x) = 1 and (sinx)(cosx) = tanx will help you a lot..and once you learn these it isnt that bad, honest!
 
 
 
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