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Homework Help: Confused with quadrants with Cos Tan Sin

  1. May 9, 2006 #1
    Okay can anyone tell me why is this true for restriction of 0 < x < 360
    sin(x)=0.15, x = 8.6 and 171.4
    cos(x)=-.655 x = 130 and 230
    tan(x)=0.75, x = 36.9 and 216.9

    I don't see a pattern or rule...
    please co-ordinate me through the use of graphs if you can my teacher didn't explain it well.
    Last edited: May 9, 2006
  2. jcsd
  3. May 9, 2006 #2


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    Staff: Mentor

    Sure. It's very handy to have the graphs of sin, cos and tan memorized so that you can sketch them in the margin of your work, especially on tests. I use sketches like that even today, 25 years after school, in my R&D EE work.

    Draw the x (or time, whatever) axis horizontally, and the value axis vertically. I like to actually draw three horizontal parallel lines, so that I can space out the plots of the sin, cos and tangent separately. Put 4 ticks to the right along the axis, at Pi/2, Pi, 3Pi/2, 2Pi. Put one tick to the left of the origin at -Pi/2. Now sketch the sin and cos functions. Sin starts at zero at the origin, rises to the right to the max of 1 at Pi/2, comes back down, etc. all the way to 2Pi. Now sketch cos, starting at its max of 1 at x=0, coming down to zero at Pi/2, etc. Finally, sketch tan(x) by drawing the asymptotes at +/- Pi/2 and 3Pi/2, and draw the whatever-you-call-it plot lines.

    So that combination plot should be something you memorize, and something you can visualize if you need it and aren't able to sketch it out for time reasons or whatever. Now in your problem above, just draw horizontal lines on the y axis at 0.15 on the sin plot, -0.655 on the cos plot, and 0.75 on the tan plot. Where those horizontal value lines intersect the function plots, the two are equal, and you can see the values of x where they are true.

    Make sense?

    EDIT -- BTW, I used Pi radians in my explanation, and your problem is stated in terms of degrees. You should be comfortable with converting back and forth.
    Last edited: May 9, 2006
  4. May 9, 2006 #3
    Need more info. How much do you know? Do you just need help crunching numbers or do you need help with the concepts?
  5. May 9, 2006 #4


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    Homework Helper

    The best way to understand *why* the signs (+ or -) of the trig ratios are the way they are in each quadrant is to plot curves like berkeman said.

    But if you want a quick way to just remember it, this is what I use :

    Just visualise the four quadrants : y and x-axes intersecting to give four square areas. The quadrants are labelled in a counter-clockwise fashion starting with the top right which is the first. Then the top left is the second, bottom left the third and the bottom right the fourth.

    Think of the angles that bound each quadrant. Each quadrant spans 90 degrees or pi/2 radians. The angles go counterclockwise, just like the numbering of the quadrants. So the first stretches from zero to ninety degrees, second from 90 to 180 degrees, third from 180 to 270 and the fourth from 270 to 360 (which is also zero, since we've come a full circle).

    Now, in your visualisation of the quadrants, firmly implant these letters in big bold print : A,S,T,C going counterclockwise. You can use some clever mnemonic to remember the order if you like. The A goes in the first quadrant and stands for "all" meaning all the ratios (sin, cos, tan) are positive in that quadrant. Similarly S means sine, telling you that only the sine is positive in the second quadrant and the tan and cos are negative there. You can work out the implications for the third and fourth quadrants.

    So, when you're told :

    "sin(x)=0.15, x = 8.6 and 171.4"

    The sine is positive (0.15), so we're looking for an x in the first and second quadrants (A and S). The first value is the first quadrant value and the second is the 2nd quadrant value. Now, here's another thing to remember : if you treat the value of the trig ratio of an angle [tex]\theta[/tex] as an unsigned number (don't care if it's positive or negative), you'll get the same absolute value (the same unsigned number) at [tex]\theta, 180 - \theta, 180 + \theta, 360 - \theta[/tex] in degrees. This applies to all the ratios, and it's due to the periodicity of the graphs. The trick is to keep in mind which of those values (which lie in different quadrants) give you the correct sign of the ratio you're looking for. Example sin x = 0.5, which gives a first quadrant value for x as 30 degrees. Using this approach, you know you'll get the same value for [tex]|\sin x|[/tex] for x = 30, 150, 210 and 330 degrees, which correspond to the four quadrants. However, since we want only values of x that give a positive sine, we only use x = 30 and x = 150 as solutions for our equation. (Try doing sin 210 and sin 330, you'll get -0.5 instead, see ?)

    Use the same logic to figure these out :

    "cos(x)=-.655 x = 130 and 230"

    Here's a negative cosine value, so you're looking for quadrants where the cosine is NOT positive.

    "tan(x)=0.75, x = 36.9 and 216.9"
    Last edited: May 9, 2006
  6. Sep 29, 2007 #5
    I need help with my maths homework can someone help me?

    tan50 = 70/x

    do I: x = 70/tan 50

    or do I: X = 70 tan 50
  7. Sep 30, 2007 #6

    Gib Z

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    Homework Helper

    :( leannev4 Please read our rules before posting you questions. It's not polite to 'hijack' another persons thread and post your own question. Also, we can not just tell you what the answer is, many people here will tell you that we don't do your work for you, we just help.

    Show us what you have done to get at those 2 answers, it's not a very difficult problem. All i can do at this point is tell you one of them is correct.
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