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Math Help - Inverse Trig functions - due tomorrow!

  1. #1
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    Exclamation Inverse Trig functions - due tomorrow!

    Hey guys, I'm stuck on these 3 probs. I'll be trying to work on them for an hour or so, but I don't have any more time for them tonight. Any help MUCH appreciated. I tried the third one but got stuck, you would think the hint would help me out(d'oh!) Thanks for any help.

    Evaluate the following without a calculator.
    1. Sin^-1(sin11(pi)/14)
    2. Cos^-1[Cos(7(pi)/5)]
    3. Sin[2csc^-1(5/3)] Hint: Let csc^-1(5/3) = theta, then use the appropriate double angle formula.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by leviathanwave View Post
    Hey guys, I'm stuck on these 3 probs. I'll be trying to work on them for an hour or so, but I don't have any more time for them tonight. Any help MUCH appreciated. I tried the third one but got stuck, you would think the hint would help me out(d'oh!) Thanks for any help.

    Evaluate the following without a calculator.
    1. Sin^-1(sin11(pi)/14)
    2. Cos^-1[Cos(7(pi)/5)]
    3. Sin[2csc^-1(5/3)] Hint: Let csc^-1(5/3) = theta, then use the appropriate double angle formula.
    Okay, so all these questions are trick questions, especially the first two. the third is the same trick, but it takes some manipulation to see it. here goes.

    Sin^-1(sin11(pi)/14)
    i presume this should be Sin^-1[sin(11pi/14)]

    short answer, if you sign something, and then take the sin^-1 of the answer, you just get back the original something you started with. obvious huh?

    so Sin^-1[sin(11pi/14)] = 11pi/14 ........without working anything out! trick question, boy they are sly.

    if you don't what i said before, i'll prove it

    let's say we had sin^-1(sinx) = y
    => sinx = siny
    => x = y

    so the original thing we applied sine to is the answer


    Cos^-1[Cos(7(pi)/5)]
    same thing: Cos^-1[Cos(7(pi)/5)] = 7pi/5


    Sin[2csc^-1(5/3)] Hint: Let csc^-1(5/3) = theta, then use the appropriate double angle formula.
    this question is a little more interesting. let's start by following the hint.

    i don't know how to type theta, so let's just call it x

    let csc^-1(5/3) = x

    so Sin[2csc^-1(5/3)] = sin(2x) = sinxcosx + sinxcosx = 2sinxcosx

    and they expected this to make it easier?! oh, but it is easier. let's examine csc^-1(5/3) a bit more

    consider csc^-1(5/3)

    let csc^-1(5/3) = y
    => 5/3 = csc(y)
    => 5/3 = 1/sin(y) ..........this is a trig identity
    => 3/5 = sin(y) .............i took the inverse of both sides
    => sin^-1(3/5) = y

    so we see that csc^-1(5/3) = sin^-1(3/5)

    Aha. we have a sin^-1 translation of our guy, so we know what sinx is right off the bat (see the trick in the previous questions). now all that's left is to find cosx.

    we had x = csc^-1(5/3) = sin^-1(3/5)

    so cosx = cos[sin^-1(3/5)], now let's examine sin^-1(3/5)

    let sin^-1(3/5) = z
    => 3/5 = sin(z)

    now recall that sin^2(z) + cos^2(z) = 1
    => (3/5)^2 + cos^2(z) = 1
    => cos^2(z) = 1 - (3/5)^2 = 16/25
    => cos(z) = sqrt(16/25) = 4/5

    so we have cos(z) = 4/5
    => cos^-1(4/5) = z ..............but wait! sin^-1(3/5) = z

    so sin^-1(3/5) = cos^-1(4/5)

    so cos[sin^-1(3/5)] = cos(cos^-1(4/5)) = 4/5 ....remember the trick

    (note, we could also find out what cosx is by using a right-triangle, see diagram below)

    so finally:

    sin(2x) = 2sinxcosx

    = 2*sin(sin^-1(3/5))*cos(cos^-1(4/5))
    = 2*(3/5)*(4/5)
    = 24/25

    this problem was very involved, if you didnt get it the first time around, go over it, it's really simple when you see how to use the trick.
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    thanks!

    Cool, i understand the last one fine, but i don't know about the first 2. Theres a lot of "work space" on the paper with the problems on it, so I dunno why they would be that simple. I know that inverse functions "undo" whatever the normal functions do, but I could swear there was something special that had to do with what quadrant the angles are in, so I'm not sure if it's just that simple. If it is, awesome
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by leviathanwave View Post
    Cool, i understand the last one fine, but i don't know about the first 2. Theres a lot of "work space" on the paper with the problems on it, so I dunno why they would be that simple. I know that inverse functions "undo" whatever the normal functions do, but I could swear there was something special that had to do with what quadrant the angles are in, so I'm not sure if it's just that simple. If it is, awesome
    it is that simple. there are questions that you have to worry about what quadrants the angles are in, but these don't seem to be such questions. like i said, they're trick questions, they put large work spaces to trick you! the scoundrels!
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by leviathanwave View Post
    Cool, i understand the last one fine, but i don't know about the first 2. Theres a lot of "work space" on the paper with the problems on it, so I dunno why they would be that simple. I know that inverse functions "undo" whatever the normal functions do, but I could swear there was something special that had to do with what quadrant the angles are in, so I'm not sure if it's just that simple. If it is, awesome
    if you want to show "work", just do it in the way i did the proof.

    example, for the first one

    Sin^-1[sin(11pi/14)]

    let Sin^-1[sin(11pi/14)] = x
    => sin(11pi/14) = sinx
    => x = 11pi/14

    therefore, Sin^-1[sin(11pi/14)] = 11pi/14

    sorry to disappoint you, but it doesn't get much more complicated than that
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  6. #6
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by Jhevon View Post
    Okay, so all these questions are trick questions, especially the first two. the third is the same trick, but it takes some manipulation to see it. here goes.

    i presume this should be Sin^-1[sin(11pi/14)]

    short answer, if you sign something, and then take the sin^-1 of the answer, you just get back the original something you started with. obvious huh?

    so Sin^-1[sin(11pi/14)] = 11pi/14 ........without working anything out! trick question, boy they are sly.

    if you don't what i said before, i'll prove it

    let's say we had sin^-1(sinx) = y
    => sinx = siny
    => x = y

    so the original thing we applied sine to is the answer


    same thing: Cos^-1[Cos(7(pi)/5)] = 7pi/5


    this question is a little more interesting. let's start by following the hint.

    i don't know how to type theta, so let's just call it x

    let csc^-1(5/3) = x

    so Sin[2csc^-1(5/3)] = sin(2x) = sinxcosx + sinxcosx = 2sinxcosx

    and they expected this to make it easier?! oh, but it is easier. let's examine csc^-1(5/3) a bit more

    consider csc^-1(5/3)

    let csc^-1(5/3) = y
    => 5/3 = csc(y)
    => 5/3 = 1/sin(y) ..........this is a trig identity
    => 3/5 = sin(y) .............i took the inverse of both sides
    => sin^-1(3/5) = y

    so we see that csc^-1(5/3) = sin^-1(3/5)

    Aha. we have a sin^-1 translation of our guy, so we know what sinx is right off the bat (see the trick in the previous questions). now all that's left is to find cosx.

    we had x = csc^-1(5/3) = sin^-1(3/5)

    so cosx = cos[sin^-1(3/5)], now let's examine sin^-1(3/5)

    let sin^-1(3/5) = z
    => 3/5 = sin(z)

    now recall that sin^2(z) + cos^2(z) = 1
    => (3/5)^2 + cos^2(z) = 1
    => cos^2(z) = 1 - (3/5)^2 = 16/25
    => cos(z) = sqrt(16/25) = 4/5

    so we have cos(z) = 4/5
    => cos^-1(4/5) = z ..............but wait! sin^-1(3/5) = z

    so sin^-1(3/5) = cos^-1(4/5)

    so cos[sin^-1(3/5)] = cos(cos^-1(4/5)) = 4/5 ....remember the trick

    (note, we could also find out what cosx is by using a right-triangle, see diagram below)

    so finally:

    sin(2x) = 2sinxcosx

    = 2*sin(sin^-1(3/5))*cos(cos^-1(4/5))
    = 2*(3/5)*(4/5)
    = 24/25

    this problem was very involved, if you didnt get it the first time around, go over it, it's really simple when you see how to use the trick.
    There's a slight problem with this:

    The arcsin (inverse sin) function has a limited domain: (-pi/2,pi/2) and so any solution given by the arcsin function must appear within that domain.
    a) 11pi/14 is in the second quadrant and is > pi/2, therefore arcsin(sin(11pi/14)) does not equal 11pi/14, but rather equals its first quadrant equivolent*: pi - 11pi/14 = 3pi/14.
    *Note: the reason the second quadrant angle becomes its first quadrant equivolent is because sin is positive in the first and second quadrant while negative in the thrid and forth. If the angle was in the third quadrant then the necessary equivolent angle would have to come from the negative forth quadrant.

    The arccos (inverse cosine) function is limited to the interval (0,pi).
    b) 7pi/5 is in the third quadrant > pi, therefore arccos(cos(7pi/5)) equals cosine's second quadrant equivolent*: 2pi - 7pi/5 = 3pi/5
    *Note: the reason the third quadrant angle becomes its second quadrant equivlent is because cosine is negative is the third and second quadrant. If the angle were in the forth quadrant, then the first quadrant equivolent would be needed since cosine is positive in both quadrants.

    Notes:
    To change an angle from one quadrant to its equivolent in another, use the following rules:

    Let x be the given angle, then:

    To change an angle from fourth quadrant to:
    First: 2pi - x
    Second: x - pi
    Third: 3pi - x

    To change an angle from third quadrant to:
    First: x - pi
    Second: 2pi - x
    Forth: 3pi - x

    To change an angle from second quadrant to:
    First: pi - x
    Third: 2pi - x
    Forth: x + pi

    To change an angle from first quadrant to:
    Second: pi - x
    Third: x + pi
    Forth: 2pi - x
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ecMathGeek View Post
    There's a slight problem with this:

    The arcsin (inverse sin) function has a limited domain: (-pi/2,pi/2) and so any solution given by the arcsin function must appear within that domain.
    a) 11pi/14 is in the second quadrant and is > pi/2, therefore arcsin(sin(11pi/14)) does not equal 11pi/14, but rather equals its first quadrant equivolent*: pi - 11pi/14 = 3pi/14.
    *Note: the reason the second quadrant angle becomes its first quadrant equivolent is because sin is positive in the first and second quadrant while negative in the thrid and forth. If the angle was in the third quadrant then the necessary equivolent angle would have to come from the negative forth quadrant.

    The arccos (inverse cosine) function is limited to the interval (0,pi).
    b) 7pi/5 is in the third quadrant > pi, therefore arccos(cos(7pi/5)) equals cosine's second quadrant equivolent*: 2pi - 7pi/5 = 3pi/5
    *Note: the reason the third quadrant angle becomes its second quadrant equivlent is because cosine is negative is the third and second quadrant. If the angle were in the forth quadrant, then the first quadrant equivolent would be needed since cosine is positive in both quadrants.

    Notes:
    To change an angle from one quadrant to its equivolent in another, use the following rules:

    Let x be the given angle, then:

    To change an angle from fourth quadrant to:
    First: 2pi - x
    Second: x - pi
    Third: 3pi - x

    To change an angle from third quadrant to:
    First: x - pi
    Second: 2pi - x
    Forth: 3pi - x

    To change an angle from second quadrant to:
    First: pi - x
    Third: 2pi - x
    Forth: x + pi

    To change an angle from first quadrant to:
    Second: pi - x
    Third: x + pi
    Forth: 2pi - x


    ok, thanks ecMathGeek, leviathanwave will be happy. he's upset that the problem wasn't complicated enough.

    thanks for looking out, you're a good guy to have around ecMathGeek
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  8. #8
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by Jhevon View Post
    ok, thanks ecMathGeek, leviathanwave will be happy. he's upset that the problem wasn't complicated enough.

    thanks for looking out, you're a good guy to have around ecMathGeek
    Aw, shucks

    I'm glad to help. And thanks, Jhevon. I have a lot of respect for you and all the other people on this site who freely give their time to help students in need. I'm just glad that I can do the same .
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