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Homework Help: Trigonometry - Associated Acute Angles

  1. Dec 19, 2013 #1

    i'm doing some revision and working through the textbook my course follows and have a small problem.

    The question is an incomplete table of values for sinθ, cosθ, tanθ, [itex]\alpha[/itex] and θ. I have to work out the blanks using what is given.

    I thought I knew the correct method of doing this, and I got all the answers correct except for the last two lines where I got the value correct, but did not see why I needed to include a ± sign.

    I'll show what I did for one of the lines, and hopefully someone can see why i'm not realising the values can be negative or positive.

    1. The problem statement, all variables and given/known data

    sinθ =
    cosθ =
    tanθ =
    [itex]\alpha[/itex] = 45°
    θ =

    3. The attempt at a solution

    I worked out sin, cos and tan for this value of alpha and they were all positive, so I thought I could assume, based on the CAST diagram, that the associated acute angle [itex]\alpha[/itex], must lie in the 1st quadrant. Therefore θ = 45°

    However, my textbook says all the trig values I obtained can be ± and θ = 45° or 135°

    What am I missing?
  2. jcsd
  3. Dec 19, 2013 #2


    Staff: Mentor

    How are α and θ related?
    Are you given a picture of the triangle?
    You haven't provided enough information for us to be able to help you.
  4. Dec 19, 2013 #3
    Sorry, I didn't stop to think whether my explanation involving α would make sense.

    I'll do my best to explain it, but it's easier with a set of axes in front of you.

    Take o to be the origin, and draw a line op at say 45° to the x axis. Directly below p, on the x axis, we label the point q to construct a right angled triangle. The anticlockwise direction is taken as +ve. We can say that α is the associated acute angle for θ here because it lies in the 'first quadrant'. They are numbered from 1-4 anticlockwise.

    Now suppose you rotate the line op to a position of 135°, p lies in the 'second' quadrant and θ = 135°, but the associated acute angle α is 45° (the triangle opq)

    I hope that makes enough sense to see what my question is getting at.
  5. Dec 19, 2013 #4


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    i don't get it :confused:

    if OP is 120°, what is α?​
  6. Dec 19, 2013 #5
    If the line op is rotated 120° anticlockwise, then θ is 120° and α is the acute angle made with the x axis, so it's 60°.


    This is in relation to the CAST diagram, that tells us which trig ratios are positive in which quadrant.

  7. Dec 19, 2013 #6


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    I see.

    And what about 240° and 300° … are they 60° or -60° ?
  8. Dec 19, 2013 #7
    In the first quadrant α = θ - 360°
    second quadrant α = 180° - θ
    Third quadrant α = θ - 180°
    Fourth quadrant α = 360° - θ

    So, if θ = 240°, the line op lies in the third quadrant and α = 60°.

    If θ = 300°, the line op lies in the fourth quadrant and α = 60°

    I'm pretty sure it's setup so that α is always positive.
  9. Dec 19, 2013 #8


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    ok, then if θ = 45° or 135°, then both cosθ and tanθ can be ± (though sinθ can only be +)

    however, on your explanation of α, i don't see why θ can't be 225° or 315° :confused:
  10. Dec 19, 2013 #9
    I'm sorry, this whole question is due to me mis-reading the answer at the back of the book.

    It was written as θ = ±45° or ±135° and I didn't see the signs in front of the values it gave for θ. I have it all making sense now.

    Thanks for your patience.
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