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    hello guys,
    i need a little explanation finishing off this question which i attached the picture of it here basically i've managed to find the value of beta which is 63.43 degrees but at the end the text book add 180 to the answer which i don't really understand why any explanation would be helpful thanks
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    There is another solution, 180 degrees further. If you draw your solution on a tan graph between 0-360 degrees, you will see that solution is there too (243.43 degrees).
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    oh ok so basically to find the next solution you need to add 180 to the first solution, does this always make sense for tan questions tho?
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    (Original post by Alen.m)
    oh ok so basically to find the next solution you need to add 180 to the first solution, does this always make sense for tan questions tho?
    Yep.

    tan(x - 180o) = tan(x) = tan(x + 180o)... and so forth.
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    too late
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    (Original post by Alexion)
    Yep.

    tan(x - 180o) = tan(x) = tan(x + 180o)... and so forth.
    but why for the question which i attached you here the answer has been added two times once by pi and once by 2pi?
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    (Original post by Alen.m)
    but why for the question which i attached you here the answer has been added two times once by pi and once by 2pi?
    Presumably the calculation gives you a negative answer then? For a tan(x) question, there will always be two solutions in the range (0<x<2pi).
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    (Original post by Alexion)
    Presumably the calculation gives you a negative answer then? For a tan(x) question, there will always be two solutions in the range (0<x<2pi).
    that's right i had a negative answer of -0.296 which has been added by pi to give me 2.84 and by 2pi to give me 5.99.
    do you suggest that everytime i get a negative answer for tan x i should add the answer once by pi and once by 2pi?
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    (Original post by Alen.m)
    that's right i had a negative answer of -0.296 which has been added by pi to give me 2.84 and by 2pi to give me 5.99.
    do you suggest that overtime i get a negative answer for tan x i should add the answer once by pi and once by 2pi?
    Yeah, like I say, you're looking for all the possible solutions in that range - so add any multiples of pi that'll get you those solutions.
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    (Original post by Alexion)
    Yeah, like I say, you're looking for all the possible solutions in that range - so add any multiples of pi that'll get you those solutions.
    can't add by 3pi this because of the range is that right?
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    (Original post by Alen.m)
    can't add by 3pi this because of the range is that right?
    Usually adding 3pi to the answer you get would take it above 2pi, so yeah.
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    (Original post by Alexion)
    Usually adding 3pi to the answer you get would take it above 2pi, so yeah.
    appreciate your help mate
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    (Original post by Alen.m)
    appreciate your help mate
    No problem :cute:
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    (Original post by Alen.m)
    appreciate your help mate
    Feel free to PM me or tag me if you need any more help with stuff...
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    (Original post by Alexion)
    No problem :cute:
    oh forgot to ask does this rule apply for cos and sin as well?
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    \tan is periodic with period \pi, so if x is a solution to \tan x = k, then so is x + \pi.

    \sin is periodic with period 2\pi, so if x is a solution to \sin x = k_s or \cos x = k_c, then so is x+2\pi.
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    (Original post by Alen.m)
    oh forgot to ask does this rule apply for cos and sin as well?
    ...

    \tan is periodic with period \pi, so if x is a solution to \tan x = k, then so is x + \pi.

    \sin is periodic with period 2\pi, so if x is a solution to \sin x = k_s or \cos x = k_c, then so is x+2\pi.
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    (Original post by Zacken)
    ...
    does the sin periodic will be the same as cos periodic?
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    (Original post by Alen.m)
    does the sin periodic will be the same as cos periodic?
    Yes, I forgot to say that sorry. \sin and \cos have the same period. They are just shifts of each other, afterall.
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    (Original post by Zacken)
    Yes, I forgot to say that sorry. \sin and \cos have the same period. They are just shifts of each other, afterall.
    thanks mate
 
 
 
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