Professor Dave here, let’s learn the trigonometric functions.
In the last clip, we introduced radians, and drew some circles.
But trigonometry is about triangles, so where do those come in?
Well let’s go to our coordinate plane and draw a circle.
Now let’s draw a radius and start to move it through quadrant one.
Wherever we stop, we can draw a line from the end of the radius straight down to the
X axis, and we have a triangle.
This will always be a right triangle with an X leg and a Y leg, and the radius as the
hypotenuse.
If these terms are unfamiliar, go back to the geometry tutorials earlier in the series
and catch up, otherwise, let’s observe that as this angle grows, the hypotenuse is always
the same, as it’s the radius of the circle, but the lengths of the legs change in a way
that depends on the angle.
When we get past the vertical, the triangle is now in quadrant two, and as we pass the
horizontal, the triangle appears in quadrant three, and finally quadrant four.
To describe the precise relationship between these side lengths and this angle, we can
use trigonometric functions.
Let’s forget the circle for a second and just look at a right triangle.
Here are the two legs, the hypotenuse, and the angle of interest, theta.
From now on we will differentiate between the two legs by calling this one the adjacent
leg, because it is the leg right next to the angle, and this one will be the opposite leg,
because it is directly opposite the angle.
We will say that the sine of theta, which is abbreviated this way, S, I, N, then the
theta symbol, will be equal to the length of the leg opposite the angle over the length
of the hypotenuse.
The cosine of theta will be equal to the length of the adjacent leg over the length of the
hypotenuse.
And the tangent of theta will be equal to the length of the opposite leg over the length
of the adjacent leg.
We can remember these definitions using the mnemonic SOHCAHTOA.
These letters stand for sin, opposite over hypotenuse, and then cosine, adjacent over
hypotenuse, and lastly tangent, opposite over adjacent.
Say it with me one more time, SOHCAHTOA.
Looking at this triangle with sides of length three, four, and five, sine of theta must
be three over five, cosine theta is four over five, and tangent theta is three over four.
Now we must learn the reciprocals of these functions.
Cosecant is the reciprocal of sine.
If sine is opposite over hypotenuse, cosecant is hypotenuse over opposite.
Secant is the reciprocal of cosine, so instead of adjacent over hypotenuse, it’s hypotenuse
over adjacent.
Cotangent is the reciprocal of tangent, so instead of opposite over adjacent, it’s
adjacent over opposite.
So we definitely have to memorize SOHCAHTOA, and then we just have to memorize which functions
are the reciprocals of these three, since we can just flip these ones to get the others.
So let’s take this triangle with sides of five, twelve, and thirteen, and evaluate all
six trigonometric functions for this angle.
We can place all the letters first to represent the sides involved in each calculation.
Then we fill in the respective numbers, see that there isn’t really any reducing to
be done, so these are the six values end up with.
So now that we know how to evaluate these functions, let’s look at two special triangles
that will come up a lot.
These are the thirty sixty ninety triangle, and the forty-five forty-five ninety triangle.
Let’s find the sine, cosine, and tangent for these three angles.
First for thirty degrees.
As we recall from geometry, if the shorter leg of this triangle has a length of one,
the other sides must be root three, and two.
Therefore, the sin of thirty is one half, the cosine of thirty is root three over two,
and the tangent of thirty is one over root three, but since we don’t like roots in
denominators, we multiply by root three over root three, which gives us root three over three.
We can use the same triangle to evaluate the same functions for the sixty degree angle.
Sine of sixty is root three over two, cosine of sixty is one half, and tangent of sixty
is root three over one, or simply root three.
Now let’s do the same for the forty-five degree angle in this other special triangle.
As we recall, if one leg of this triangle is equal to one, the other sides must be one
and root two.
So the sine of forty-five is one over root two, but we multiply by root two over root
two to get root two over two, and the cosine of forty-five will give us the same value,
since the two legs have the same length.
Then, the tangent of forty-five will be one over one, or simply one.
These triangles will come up again and again, so these values of sine and cosine will be
very common as well.
Ideally we now understand what these six trigonometric functions are and how to calculate them, so
let’s check comprehension.