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Easiest way to learn exact values for trig functions?

  1. Oct 30, 2015 #1
    I'm realizing now how much I need to know the exact values of various trigonometric functions, as shown in various trig tables. Memorizing is pretty arduous, and I'd prefer to understand it, so how can I learn all of these?
     
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  3. Oct 30, 2015 #2

    DrClaude

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    A long long time ago, my math teacher made us build our own trigonometric circles. Having a geometrical representation really helped me, and the values of the functions have stuck with me ever since.
     
  4. Oct 30, 2015 #3
    Thanks! I guess I'd forgotten that these circles existed. Does this mean that if I memorize the important angles/values from the first quadrant I can use the identities of -sin(θ) = sin(-θ) and cos(-θ) = cos(θ) to figure out the values in each other quadrant?
     
  5. Oct 30, 2015 #4

    DrClaude

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    It's even more than that. Consider for instance π/6 (30°): you get halfway up on the y axis, so obviously sin π/6 = 1/2. Likewise, for π/4, you can see it as building a square of side 1/2 (along x and y), so the diagonal is (according to Pythagoras) ##\sqrt{(1/2)^2 + (1/2)^2} = 1/\sqrt{2} = \sqrt{2}/{2}##. All of these geometric equivalents have really helped me.
     
  6. Oct 30, 2015 #5

    symbolipoint

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    That graphical circle representation is typically called, the Unit Circle, and is a very important part of Trigonometry instruction. Is is used frequently. The picture can often help because of how the circle is symmetric. Full ray rotation, 2*pi radians. Same function values as for zero degrees rotation.
     
  7. Oct 30, 2015 #6
    What would you guys say are the most important angles to learn? I've seen some charts that go from 0, adding π/12 each segment, and that seems like an awful lot to go through in order to find these angles, are the three, π/6, π/4, and π/3 and their counterparts typically sufficient?
     
  8. Oct 30, 2015 #7

    symbolipoint

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    THE MOST COMMON you must know are 0, 2*pi, pi/2, pi/4, pi/3, pi/6.
     
  9. Oct 30, 2015 #8

    DrClaude

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    I realize that this might not be as clear as it can be. Rather, take the radius as the hypotenuse of a right triangle with two equal sides of length ##a## (along the x and y axes), so by Pythagoras ##2a^2 = 1 \Rightarrow a = 1/\sqrt{2}##, thus ##\cos \pi/4 = \sin \pi/4 = \sqrt{2}/2##.
     
  10. Oct 30, 2015 #9

    Mark44

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    πTo expand on what symbolipoint said, the important angles in the first quadrant, and their sines and cosines are:
    ##\sin(0) = \frac {\sqrt{0}} 2 =\cos(\pi/2)##
    ##\sin(\pi/6) = \frac {\sqrt{1}} 2 =\cos(\pi/3)##
    ##\sin(\pi/4) = \frac {\sqrt{2}} 2 =\cos(\pi/4)##
    ##\sin(\pi/3) = \frac {\sqrt{3}} 2 =\cos(\pi/6)##
    ##\sin(\pi/2) = \frac {\sqrt{4}} 2 =\cos(0)##
    These values should be memorized, but you can use symmetry to figure out the sines, cosines, tangents, etc. of the counterparts of these angles in the other three quadrants.
     
  11. Oct 30, 2015 #10

    Vanadium 50

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    There are 7 special angles worth memorizing:0, 15, 30, 45, 60, 75 and 90. Sines going up are cosines coming down. Tangent=sine/cosine may be a bit slow for calculating in your head for 15 and 75, and arguably 30 and 60, but the other three are trivial. So you have at most 10 - possibly 8 - facts to remember. That shouldn't be too arduous.
     
  12. Oct 30, 2015 #11

    Student100

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    Another method that may help you should you forget is to draw a picture and use geometry/trig:

    Snapshot.jpg

    From my poorly drawn circle you can see you want to find the point, (x, y) on the unit circle that intersects with a line of the radial length drawn 30 degrees from the origin. The right triangle formed by hypotenuse (which is our radius of 1) forms a special 30, 60, 90 triangle. A property of such a triangle is that the hypotenuse is twice the length of the shorter leg, and the longer leg is ##\sqrt{3}## times the length of the shorter leg. Since the hypotenuse is one, the length of the shorter leg, ## y=\frac{1}{2}## and the longer leg, x is ##\sqrt{3}## times more so, ##x=\frac{\sqrt{3}}{2}##. So the ##sin(30)=\frac{1}{2}## while the ##cosine(30)=\frac{\sqrt{3}}{2}##

    You can repeat the same logic for 60 degrees, or just realize that the x and y's are swapped. For a 45-45-90 triangle, the legs are congruent, so the hypotenuse is ##\sqrt{2}## the length of either leg. So lets say the hypotenuse has length ##x\sqrt{2}## in this case, since we know the length to be one, we can simply say ##x\sqrt{2}=1## or ##x=\frac{1}{\sqrt{2}}##. The legs are congruent so ##y=\frac{1}{\sqrt{2}}##.

    For 15 and 75 degrees you can use ##sin(45-30)## and solve using trig rules. For 0 and 90 you can again draw a line from the origin to the corresponding point on the circle and see either, x=1 in the case of an angle of 0, or x=0 in the case of an angle of 90 or vice versus.

    Not sure if this will help or not, hopefully it'll allow you to re-derive these values should you forget them on a test.
     
  13. Oct 30, 2015 #12

    symbolipoint

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    The common reference angles that are learned on the Unit Circle come from a couple of Special Triangles.
     
  14. Nov 4, 2015 #13
    If you look at the picture here: http://etc.usf.edu/clipart/43200/43216/unit-circle8_43216_lg.gif you will see that only the first quadrant is needed (and even then, only the angles between 0 and 45 degrees are needed (everything else can be derived by symmetry of the circle. As for the coordinates, I find it easier to remember them as:

    0: [itex]\left( \sqrt{\frac{4}{4}}, \sqrt{\frac{0}{4}}\right) [/itex]
    pi/6: [itex]\left( \sqrt{\frac{3}{4}}, \sqrt{\frac{1}{4}}\right) [/itex]
    pi/4: [itex]\left( \sqrt{\frac{2}{4}}, \sqrt{\frac{2}{4}}\right) [/itex]
    pi/3: [itex]\left( \sqrt{\frac{1}{4}}, \sqrt{\frac{3}{4}}\right) [/itex]
    pi/2: [itex]\left( \sqrt{\frac{0}{4}}, \sqrt{\frac{4}{4}}\right) [/itex]
     
  15. Nov 5, 2015 #14

    Erland

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    Easiest way is to use to draw a half square and a half equiliteral triangle, and use the Pythagorean theorem to obtaim sin, cos and tan for the angles in tjose triangles, i.e. 45, 30 and 60 degrees.
     
  16. Nov 5, 2015 #15

    Fredrik

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    And once you have those, you can use them to find the values for some other angles. For example,
    \begin{align*}
    &\sin 15=\sin(45-30)=\sin 45\cos 30-\cos 45\sin 30\\
    &\sin 120=\sin(2\cdot 60)=2\sin 60\cos 60
    \end{align*} The only thing worth committing to memory in my opinion, is (what Erland said) that you start with half a square and half an equilateral triangle.
     
  17. Nov 5, 2015 #16

    symbolipoint

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    The unit circle and its typically shown values can be easily memorized.

    THIS here is essential:
     
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