Fun with trig

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    OK, the title is a bit of a lie. I don't understand Sin Cos quadrant two angle thing.

    My understanding was that Sin existed in the first and fourth quadrants, and Cos existed in the first and second.

    Now, if I have cosine 40 I would expect the other angle to be 140

    If the Sin angle is 40 I would expect the other angle to be 320.

    If the Sin angle is -40 I would expect the other angle to be 40.

    However, doing these offers the following angle answers and I do not understand it.

    Sin x = 1/4 14.8 and 165.52
    Sin x = -0.4 203.58 and 336.42

    I understand how they're calculating it of course, but I don't understand why it's happening.
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    (Original post by Brainfrozen)
    OK, the title is a bit of a lie. I don't understand Sin Cos quadrant two angle thing.

    My understanding was that Sin existed in the first and fourth quadrants, and Cos existed in the first and second.

    Now, if I have cosine 40 I would expect the other angle to be 140

    If the Sin angle is 40 I would expect the other angle to be 320.

    If the Sin angle is -40 I would expect the other angle to be 40.

    However, doing these offers the following angle answers and I do not understand it.

    Sin x = 1/4 14.8 and 165.52
    Sin x = -0.4 203.58 and 336.42

    I understand how they're calculating it of course, but I don't understand why it's happening.
    The easiest way of course is to just use the general solutions. Just learn those and you'll be sailing smoothly through any of these.
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    (Original post by RDKGames)
    The easiest way of course is to just use the general solutions. Just learn those and you'll be sailing smoothly through any of these.
    The course is asking me to find both angles, and it will be examined as such I guess, so I'll have to do it the way they're wanting it.

    But I'd rather understand what I'm doing wrong regarding my thinking of the quadrants.
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    Sine and cosine exist in all quadrants.
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    Can I have someone explain to me please instead of giving me pointers?

    I've been out of school for more than a decade and I'm coming back to it, this means I've got holes in my understanding due to things not being used. It is not like I've not paid attention in class, I did and now work as an engineer and I'm doing my HNC, but that means I need a helping hand with some of the basics.
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    (Original post by Brainfrozen)
    Can I have someone explain to me please instead of giving me pointers?

    I've been out of school for more than a decade and I'm coming back to it, this means I've got holes in my understanding due to things not being used. It is not like I've not paid attention in class, I did and now work as an engineer and I'm doing my HNC, but that means I need a helping hand with some of the basics.
    It's easiest to conceptualise everything by considering the graph. I presume the question you're doing asks for solutions in the range 0 to 360 degrees. So, the first answer for sinx = 1/4 is easy enough, you just take the inverse sin of 1/4 on a calculator. Then look at the graph. For another answer, you need x to be no more than 180 degrees, because for 180 < x < 360 sinx is non-positive so clearly sinx =/= 1/4. It's quite apparent that, considering only 0 < x < 180, the graph is symmetrical about x = 90 degrees, right? So if you move the same distance backwards from 180 degrees as you did forwards from 0 to get your first solution, you'll get another solution. Hence the answer is 180 - sin^-1(1/4).

    As for sinx = -0.4. First you take the inverse sin. Note by the periodicity of sin, adding 360 degrees gives you an answer, so that's one solution. After that, it's very similar to before; you consider 180 < x < 360 and see symmetry about x = 270, so to get another solution you move the same distance forwards from 180 as you did backwards from 360.
 
 
 
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