Archives for posts with tag: Special Right Triangles

Trig students are expected to be able to compute things like the sine and cosine of 5π/6, 3π/4, and the like. How do we locate points on the unit circle corresponding to these angles, and how do we find their coordinates? This post explains how.

[AP students: the course description for AP Calculus has this sentence under the “Prerequisites” heading: “Students must…know the values of the trigonometric functions at the numbers 0, π/6, π/4, π/3, π/2, and their multiples.”  If you need a brush-up on trig, this post is for you!]

Let’s begin at the beginning: we can define the radian measure of an angle in terms of arc lengths in a unit circle (that is, a circle with radius r = 1).  How much arc is in a full circle?  Since the circumference of a circle is 2πr, we get 2π(1) = 2π radians for the full unit circle.  So one complete rotation encompasses 2π radians.  That means a half-rotation is π radians.


Consequence #1: If a “straight angle” is 1π radians, then a right angle is (1/2)π radians.

Proof: 1/2 + 1/2 = 1.


Consequence #2: If you bisect a right angle, you get (1/4)π radians.

Proof: 1/4 + 1/4 = 2/4 = 1/2.


Consequence #3: If you trisect a right angle, you get (1/6)π radians.

Proof: 1/6 + 1/6 + 1/6 = 3/6 = 1/2.


Now for some theorems about triangles (or trigons!), the figure of interest in trigonometry.

Theorem #1: Since the angles in a triangle can be arranged in a straight line, they must make π radians altogether.


Theorem #2: Since the angles in an equilateral triangle are the same, each angle must be (1/3)π radians.

Proof: 1/3 + 1/3 + 1/3 = 1.


Theorem #3: If you bisect an angle in an equilateral triangle, you get two angles with measure (1/6)π radians.  Proof: 1/6 + 1/6 = 2/6 = 1/3.


Theorem #4: The acute angles in an isosceles right triangle are both (1/4)π radians.

Proof: 1/4 + 1/4 + 1/2 = 2/4 + 1/2 = 1/2 + 1/2 = 1.


Now let’s actually answer some trig questions: what are the sine and cosine associated with an angle of π/4 radians?


Per Theorem #4 above, we are dealing with an isosceles right triangle.


The vertical and horizontal legs must both be sqrt(1/2).  Proof: if you square each side in the triangle, you get (1/2) + (1/2) = 1, which is obviously correct.  A more detailed treatment of this type of triangle is found here.

Note: many folks write the sqrt(1/2) as sqrt(2)/2, which is the same thing (how so?).

Thus sine(π/4) is sqrt(1/2) and cosine(π/4) is sqrt(1/2).  From this one can easily work out the sine and cosine of (3/4)π, (5/4)π, (7/4)π, etc.  We just have to pay attention to whether the triangle is “pointing up/down/left/right” to get the sign correct.

On to the next major question: what are the sine and cosine associated with an angle of π/6 radians?


Theorem #3 above tells us we’re really dealing with half of an equilateral triangle, so the answer can be found via symmetry:


Reflecting the triangle across the x-axis gives an equilateral triangle with side length 1.  So clearly the vertical leg is 1/2.  This means the horizontal leg must be sqrt(3/4).  Proof: squaring each side gives 3/4 + 1/4 = 1, which is obviously true.

A more detailed treatment of this type of triangle can be found here.

Note: many folks write sqrt(3/4) as sqrt(3)/2, which is clearly the same thing, since both quantities are equal to sqrt(3)/sqrt(4).

Thus sine(π/6) is 1/2 and cosine(π/6) is sqrt(3/4).  From this it’s easy to work out the sine and cosine of (5/6)π, (7/6)π, (11/6)π, etc.  Again we just have to look to see which way the triangle is pointing to know whether to assign a positive or negative value to these lengths.


That’s pretty much it!  The only thing left to do is to write down all possible ratios of the sides of a right triangle.  We have 3 sides to work with, one of which is to be called “top” and the other “bottom,” so there are 6 ratios altogether.

Ratio #1: a/b.  We call this ratio the “tangent” of x.

Ratio #2: b/a.  This one is known as the “cotangent” of x.

Ratio #3: a/c.  The sine of x!

Ratio #4: c/a.  The “cosecant” of x.

Ratio #5: b/c.  The cosine of x!

Ratio #6: c/b.  The “secant” of x.

Learning the names of the ratios above is truly a task for flash cards.  If there is any intuition to the names, it has surely been lost on the likes of me!  Someone out there can probably unpack the Latin origins of these terms, but that someone is not me!

As an example, let’s work out the secant of π/4 radians.  The relevant triangle has legs of length sqrt(1/2) and hypotenuse 1, so the ratio must be 1/(sqrt(1/2)).  So we just need to find the reciprocal of sqrt(1/2).  But this is easy: sqrt(1/2) = sqrt(1)/sqrt(2), which is 1/sqrt(2), so the reciprocal must simply be sqrt(2).

Exercise 1: work out all 6 of the fundamental trig ratios for π/6.  Repeat for each of the first 12 multiples of π/6 (i.e. through one full rotation).

Exercise 2: work out all 6 of the fundamental trig ratios for π/4.  Repeat for each of the first 8 multiples of π/4 (which is, again, one full rotation).

Calculus students: if you can do the above 2 exercise sets fluidly, confidently, and correctly, you’ve officially mastered half of one sentence from the “Prerequisites” section of the AP Course Description!  Congrats.


Here’s my take on how to “see” the sides of a 30-60-90 triangle, without “memorizing” anything.  Students learning trigonometry are expected to know the side ratios for “special angles,” and so it’s useful to know how to work them out quickly.

photo 1

First, students should learn to recognize that a 30-60-90 triangle is half of an equilateral triangle.  Thus, if the hypotenuse is 1 (thinking of a unit circle here), then the shorter leg must be 1/2.

photo 2

Now let’s deal with that pesky longer leg.  Assuming we’ve worked out this length using the Pythagorean theorem before, it won’t come as a surprise that it’s “the square root of something.”  But what is that “something?”

photo 3

Applying the Pythagorean theorem, we get this:

photo 4

Clearly there is only room for one quantity here: if “something” plus 1/4 is 1, then that something had better be 3/4!

photo 5

Thus the long leg is the square root of 3/4.  The advantage of writing the leg in this way is that we can do a quick mental check: the square root of 3/4 squared, plus 1/2 squared, is indeed equal to 1 squared, as 3/4 + 1/4 = 1.  Done!