## The Complements Theorem

In this post, we discuss the proof behind one of the most commonly used identities in trigonometry. We examine the equations below and  show why the relationships always hold.

$\sin \theta = \cos (90^{\circ} - \theta)$
$\cos \theta = \sin(90^{\circ} - \theta)$

To students who have taken trigonometry, I’m sure that you have met these equation before.  The proof of these equations are as follows.

Consider triangle $ABC$ right angled at $C$. From the definitions, we know that

$\sin A = \displaystyle\frac{a}{c}$
$\cos A = \displaystyle\frac{b}{c}$
$\sin B = \displaystyle\frac{b}{c}$
$\cos B = \displaystyle\frac{a}{c}$

Therefore, (1) $\sin A = \cos B$ and (2) $\cos A = \sin B$.

Now, If we let $A = \theta$, then $B = 90^{\circ} - \theta$, then substituting the values of $A$ and $B$ in (1) and (2), we have

$\sin \theta = \cos (90^{\circ} - \theta)$
$\cos \theta = \sin (90^{\circ} - \theta)$

and these are what we want to show.

As exercises, use the strategy above, or any strategy you want to prove the following identities.

1.) $\cot\theta = \tan (90^{\circ} - \theta)$
2.) $\sec \theta = \csc(90^{\circ} - \theta)$

## Pythagorean Theorem, Distance Formula, and Equation of a Circle

In my Algebraic and Geometric Proof of the Pythagorean Theorem post, we have learned that a right triangle with side lengths $a$ and $b$ and hypotenuse length $c$, the sum of the squares of $a$ and $b$ is equal to the square of $c$. Placing it in equation form we have $c^2 = a^2 + b^2$.

If we place the triangle in the coordinate plane, having $A$ and $B$ coordinates of $(x_1,y_1)$ and $(x_2,y_2)$ respectively, it is clear that the length of $AC$ is $|x_2 - x_1|$ and the length of $BC$ is $|x_2 - x_1|$.  We are finding the length, which means that we want a positive value; the absolute value signs guarantee that the result of the operation is always positive. But in the final equation,$c^2 = |x_2 - x_1|^2 + |y_2-y_1|^2$, the absolute value sign is not needed since we squared all the terms, and squared numbers are always positive. Getting the square root of both sides we have,

$c = \sqrt{|x_2 - x_1|^2 + |y_2-y_1|^2}$

We say that $c$ is the distance between $A$ and $B$, and we call the formula above, the distance formula. » Read more

## The Algebraic and Geometric Proofs of Pythagorean Theorem

The Pythagorean Theorem states that if a right triangle has side lengths $a, b$ and $c$, where $c$ is the hypotenuse, then the sum of the squares of the two shorter lengths is equal to the square of the length of the hypotenuse.

Figure 1 – A right triangle with side lengths a, b and c.

Putting it in equation form, we have

$a^2 + b^2 = c^2$.

For example, if a right triangle has side lengths $5$ and $12$, then the length of its hypotenuse is $13$, since $c^2 = 5^2 + 12^2 \Rightarrow c = 13$.

Exercise 1: What is the hypotenuse of the triangle with sides $1$ and $\sqrt{3}$?

The converse of the theorem is also true. If the side lengths of the triangle satisfy the equation $a^2 + b^2 = c^2$, then the triangle is right. For instance, a triangle with side lengths $(3, 4, 5)$ satisfies the equation $3^2 + 4^2 = 5^2$, therefore, it is a right triangle.

Geometrically, the Pythagorean theorem states that in a right triangle with sides $a, b$ and $c$ where $c$ is the hypotenuse, if three squares are constructed whose one of the sides are the sides of the triangle as shown in Figure 2, then the area of the two smaller squares when added equals the area of the largest square.

Figure 2 – The geometric interpretation of the Pythagorean theorem states that the area of the green square plus the area of the red square is equal to the area of the blue square.

One specific case is shown in Figure 3: the areas of the two smaller squares are $9$ and $16$ square units, and the area of the largest square is $25$ square units.

Exercise 2: Verify that the area of the largest square in Figure 3 is 25 square units by using the unit squares.

Figure 3 – A right triangle with side lengths 3, 4 and 5.

Similarly, triangles with side lengths $(7, 24, 25)$ and  $(8, 15, 17)$ are right triangles. If the side lengths of a right triangle are all integers, we call them Pythagorean triples. Hence, $(7, 24, 25)$ and  $(8, 15, 17)$ are Pythagorean triples.

Exercise 3: Give other examples of Pythagorean triples.

Exercise 4: Prove that there are infinitely many Pythagorean triples.

Proofs of the Pythagorean Theorem

There are more than 300 proofs of the Pythagorean theorem. More than 70 proofs are shown in tje Cut-The-Knot website. Shown below are two of the proofs.  Note that in proving the Pythagorean theorem, we want to show that for any right triangle with hypotenuse $c$, and sides $a$, and $b$, the following relationship holds: $a^2 + b^2 = c^2$.

Geometric Proof

First, we draw a triangle with side lengths $a, b$ and $c$ as shown in Figure 1. Next, we create 4 triangles identical to it and using the triangles form a square with side lengths $a + b$ as shown in Figure 4-A. Notice that the area of the white square in Figure 4-A is $c^2$.

Figure 4 – The Geometric proof of the Pythagorean theorem.

Rearranging the triangles, we can also form another square with the same side length as shown in Figure 4-B.This means that the area of the white square in the Figure 4-A is equal to the sum of the areas of the white squares in Figure 4-B (Why?). That is, $c^2 = a^2 + b^2$ which is exactly what we want to show. *And since we can always form a (big) square using four right triangles with any dimension (in higher mathematics, we say that we can choose arbitrary $a$ and $b$ as side lengths of a right triangle), this implies that the equation $a^2 + b^2 = c^2$ stated above is always true regardless of the size of the triangle.

Exercise 5: Prove that the quadrilateral with side length C in Figure 4-A is a square.

Algebraic Proof

In the second proof, we will now look at the yellow triangles instead of the squares.  Consider Figure 4-A. We can compute the area of a square with side lengths $a + b$ using two methods: (1) we can square the side lengths and (2) we can add the area of the 4 congruent triangles and then add them to the area of the white square which is $c^2$.  If we let $A$ be the area of the square with side $b + a$, then calculating we have

Method 1: $A = (b + a)^2 = b^2 + 2ab +a^2$

Method 2:  $A = 4(1/2ab) + c^2 = 2ab + c^2$

Methods 1 and 2 calculated the area of the same square, therefore they must be equal. This means that we can equate both expressions.  Equating we have,

$b^2 + 2ab + a^2 = 2ab + c^2 \Rightarrow a^2 + b^2 = c^2$

which is exactly what we want to show.