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    Hi guys does anybody know what is the period of cosine x when you trying to find out solutions in degree interval?i've found my first solution but apparently you need to add 180 to your first solution to get the next one in a defined interval which is between 0 and 360
    attached you the question here
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    apparently this question has nothing to do with what i asked you only need to know that cos^-1(0) is 90 and 270 to find the value o beta but the problem is if you only write 90 and forget about 270 will you lose mark for it?
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    (Original post by Alen.m)
    apparently this question has nothing to do with what i asked you only need to know that cos^-1(0) is 90 and 270 to find the value o beta but the problem is if you only write 90 and forget about 270 will you lose mark for it?
    Yes, the whole purpose of it stating theta is between zero and 360 is so you know how many values it has.
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    (Original post by Alen.m)
    apparently this question has nothing to do with what i asked you only need to know that cos^-1(0) is 90 and 270 to find the value o beta but the problem is if you only write 90 and forget about 270 will you lose mark for it?
    You will. For cosine, besides for the inverse cosine solution, x, there is the solution 360 - x, and then each of these solution generates more with period 360. You'll have to give all the solutions in the range given.
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    am i too late?
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    (Original post by Alen.m)
    apparently this question has nothing to do with what i asked you only need to know that cos^-1(0) is 90 and 270 to find the value o beta but the problem is if you only write 90 and forget about 270 will you lose mark for it?
    You are missing a solution if you don't find the value of theta that gives you 7theta/6 = 270 so you will lose some marks for it.

    In general, you have to look at the range that theta can take and then look for x values such that cosx = .. whatever you're looking for.

    Eg if you're looking for values of theta in the range 0 to 180 degrees such that cos2theta = 0, then you need to look for x values (where x = 2theta) between 0 and 360.
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    (Original post by TeeEm)
    am i too late?
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    Everyone wants to hear it in your own words
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    (Original post by EricPiphany)
    You will. For cosine, besides for the inverse cosine solution, x, there is the solution 360 - x, and then each of these solution generates more with period 360. You'll have to give all the solutions in the range given.
    does this mean that cos and sin have periodic of 360 in degree and 2pi in radian?for example if i find first solution in a certain interval, to get the next one i should add 360 to the first one?
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    (Original post by SeanFM)
    You are missing a solution if you don't find the value of theta that gives you 7theta/6 = 270 so you will lose some marks for it.

    In general, you have to look at the range that theta can take and then look for x values such that cosx = .. whatever you're looking for.

    Eg if you're looking for values of theta in the range 0 to 180 degrees such that cos2theta = 0, then you need to look for x values (where x = 2theta) between 0 and 360.
    seems like i should memorise all the values of cos, sin and tan for example in the picture i attached you here how am i supposed to know that cos of all these values will give -1/2 ?memorising is the only way or am i missing something?
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    (Original post by Alen.m)
    seems like i should memorise all the values of cos, sin and tan for example in the picture i attached you here how am i supposed to know that cos of all these values will give -1/2 ?memorising is the only way or am i missing something?
    You can use a calculator to find one value, then use a CAST diagram or the graphs to find the other values.
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    (Original post by EricPiphany)
    You will. For cosine, besides for the inverse cosine solution, x, there is the solution 360 - x, and then each of these solution generates more with period 360. You'll have to give all the solutions in the range given.
    so basically the periodic of sin and cos is 360?
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    (Original post by Alen.m)
    so basically the periodic of sin and cos is 360?
    Yes, sine and cosine are periodic functions with period 360 degrees.

    Tangent has a period of 180 degrees.
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    (Original post by SeanFM)
    You can use a calculator to find one value, then use a CAST diagram or the graphs to find the other values.
    to be honest I'm still confused how you can find 5 angles using cast diagram found the first value of x using calculator to be 2pi/3 , according to cast diagram i only need to subtract pi from the angle i just found to get the next one between the interval of the question which is 0<x<pi but there are like 5 angles
    here's the full question it's in the factor formula chapter btw
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    (Original post by Zacken)
    Yes, sine and cosine are periodic functions with period 360 degrees.

    Tangent has a period of 180 degrees.
    thanks
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    (Original post by Alen.m)
    found the first value of x using calculator to be 2pi/15 , according to cast diagram i only need to subtract pi from the angle i just found to get the next one between the interval of the question which is 0<x<pi but there are like 5 angles
    here's the full question it's in the factor formula chapter btw
    What range of x values do you think you should be looking for?
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    (Original post by SeanFM)
    What range of x values do you think you should be looking for?
    between zero and pi?
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    (Original post by Alen.m)
    between zero and pi?
    Oh sorry I didn't mean x. What function did you have before finding 2pi/3?
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    cos(5x)= -1/2, 0<x<pi (the intervals signs have equal on each of them as well but couldn't type it) . found 5x=2pi/3 using calculator so x= 2pi/15 but don't know how to find the rest of angles even using cast diagram wouldn't make sense
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    0<x<pi -> 0<5x<5pi, so you need to look for angles with a cos of -1/2 between 0 and 5pi.
    You found 2pi/3, so another solution is 2pi-2pi/3=4pi/3. Now keep adding 2pi onto each of these two solutions to get more solutions, until you go above 5pi.
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    (Original post by constellarknight)
    0<x<pi -> 0<5x<5pi, so you need to look for angles with a cos of -1/2 between 0 and 5pi.
    You found 2pi/3, so another solution is 2pi-2pi/3=4pi/3. Now keep adding 2pi onto each of these two solutions to get more solutions, until you go above 5pi.
    so basically i've got my first answer as 2pi/3 , substrate this from 2pi will give me the second angle which is 4pi/3 now keep adding 2pi to 2pi/3 or 4pi/3?does it really matter?
 
 
 
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