No, you won't stumble over co-functions if you find yourself in Dysfunction Junction. They're deeply embedded in trigonometric functions. Did you notice the "co" in cosine, cosecant, and cotangent?
"Co" is short for complementary.
Co + function = complementary functions.
Does the word complementary ring a bell? In geometry, we learned that complementary angles combine to form a 90° angle.
We also know that the sum of the angles in a triangle must be 180°. So if angle C is a right angle (90°), the other two angles must add up to the other 90°.
Since A + B = 90°, A and B are complementary angles.
And by rearranging things a little, we can see that:
B = 90° – A
and
A = 90° – B
Now let's apply our "co" to the trig functions.
Cosine is the complementary function of sine.
Cosecant is the complementary function of secant.
Cotangent is the complementary function of tangent.
What's this all about? Here's the deal. Take another peek at our triangle:
In this triangle, the cosine of angle A is the same thing as the sine of angle B.
cos A = sin B
Go ahead and let that sink in for a sec. Cosine is the adjacent side over the hypotenuse (CAH), and sine is the opposite side over the hypotenuse (SOH). But the side adjacent to angle A is the same thing as the side opposite angle B. That's why cos A = sin B. They give us the exact same ratio.
Here's where the co-function stuff comes in. Substituting in B = 90° – A gives us:
cos A = sin(90° – A)
See what we did there? The cosine of A is the same as the sine of A's complementary angle.
Similarly:
csc A = sec B
csc A = sec(90° – A)
And:
cot A = tan B
cot A = tan(90° – A)
It works the other way, too:
sin A = cos B = cos(90° – A)
sec A = csc B = csc(90° – A)
tan A = cot B = cot(90° – A)
Sample Problem
Write sin(30°) in terms of its co-function.
This is way more chill than it might sound. The sine's co-function is just the cosine. Using sin A = cos(90° – A), we get:
sin(30°) = cos(90° – 30°)
sin(30°) = cos(60°)
Yep, that's it. The sine of 30° is the exact same thing as the cosine of 60°. Plop 'em into a calculator if you don't believe us.
Sample Problem
Write csc(70°) in terms of its co-function.
Cosecant goes hand-in-hand with secant, so we'll use complementary angles again. The formula this time is csc A = sec(90° – A).
csc(70°) = sec(90° – 70°)
csc(70°) = sec(20°)
Sample Problem
Write tan(10°) in terms of its co-function.
We wanna turn that tangent into cotangent, so we'll use the formula tan A = cot(90°– A).
tan(10°) = cot(90° – 10°)
tan(10°) = cot(80°)
Special Triangles
In trig and—eventually—calculus, you'll find that we use certain angles over and over, ad infinitum, ad nauseam. These angles make up two special triangles. We'll take a look at special triangles in the next section.