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Math Help - Just a little clarification...

  1. #1
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    Just a little clarification...

    Is it true that to find the basic angle, we have to always use the positive number?

    for example, cosx = -1
    so to find the basic angle = cos^-1 (1)?
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  2. #2
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    Quote Originally Posted by Punch View Post
    Is it true that to find the basic angle, we have to always use the positive number?

    for example, cosx = -1
    so to find the basic angle = cos^-1 (1)?
    what do you mean by basic angel?
    if \cos x = -1, then x = \cos^{-1} (-1)
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  3. #3
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    Basic angle is the acute angle of a general angle
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  4. #4
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    Hello Punch
    Quote Originally Posted by Punch View Post
    Is it true that to find the basic angle, we have to always use the positive number?

    for example, cosx = -1
    so to find the basic angle = cos^-1 (1)?
    Yes; because it's only acute angles that have their sine, cosine and tangent all positive.

    Grandad
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  5. #5
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    How about this,

    taking the cosine inverse of -1 will get me 180, that is obtuse. so i take the acute angle of it that is 0. And it is equals to cosine inverse of 1.

    So this shows that whenever u sin cos tan something and get the obtuse angle, just use the acute angle? am i right?
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  6. #6
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    Hello Punch
    Quote Originally Posted by Punch View Post
    How about this,

    taking the cosine inverse of -1 will get me 180, that is obtuse. so i take the acute angle of it that is 0. And it is equals to cosine inverse of 1.

    So this shows that whenever u sin cos tan something and get the obtuse angle, just use the acute angle? am i right?
    I think you need to be very careful to say exactly what you mean. There are lots of things to learn about trig functions, and you can't condense them into a single sentence! Let me try to clarify a bit:

    • To find the angle x between 0^o and 180^o if \cos x is negative (say, \cos x = -k), find the acute angle whose cosine is +k and subtract it from 180^o.


    • This works in exactly the same with with \tan x. If \tan x is negative, find the angle whose tangent is the same positive number, and subtract from 180^o.


    • With \sin x, there are no negative values between 0^o and 180^o: \sin x and \sin(180^o-x) have exactly the same positive value. You need angles outside this range if you want negative values of sine. For instance, if \sin x = -k, then you can find the acute angle whose sine is +k and either add this to 180^o or subtract it from 360^o.

    I told you that, sadly, it wasn't as simple as you were trying to make it!

    Grandad
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