Deriving the Pythagorean Identities

Using the definitions of sine and cosine, we will learn how they relate to each other and the unit circle. For any point on the unit circle, 

 $x^2 + y^2 = 1$

We can see how this relates to the Pythagorean theorem, 

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

For a triangle drawn inside a unit circle, the length of the hypotenuse of the triangle is equal to the radius of the circle, which is $1$. The lengths of the sides of the triangle are $x$ and $y$. 

The Pythagorean identity on a unit circle

For a triangle drawn inside a unit circle, the length of the hypotenuse is equal to the radius of the circle. The triangle sides have lengths $x$ and $y$.

Because $x = \cos t$ and $y= \sin t$ on the unit circle, we can substitute for $x$ and $y$ to get the Pythagorean identity:

 $\cos^2 t + \sin^2 t = 1$

which is true for any real number $t$.

We can use the Pythagorean identity to find the cosine of an angle if we know the sine, or vice versa. However, because the equation yields two solutions, we need additional knowledge of the angle to choose the solution with the correct sign. If we know the quadrant where the angle is, we can easily choose the correct solution.

Additional identities can be derived from the Pythagorean identity $\cos^2 t + \sin^2 t = 1$. For example, if we divide the identity by $\cos^2 t$, we have the following identity: 

$\displaystyle{ \begin{aligned} \cos^2 t + \sin^2 t &= 1 \\ \left(\frac{\cos^2 t}{\cos^2 t}\right) + \left(\frac{\sin^2 t}{\cos^2 t}\right) &= \left(\frac{1}{\cos^2 t}\right) \\ 1 + \tan^2 t &= \sec^2 t \end{aligned} }$

Likewise, if we divide the identity $\cos^2 t + \sin^2 t = 1$ by $\sin^2 t$, we have the following:

$\displaystyle{ \begin{aligned} \cos^2 t + \sin^2 t &= 1 \\ \left(\frac{\cos^2 t}{\sin^2 t}\right) + \left(\frac{\sin^2 t}{\sin^2 t}\right) &= \left(\frac{1}{\sin^2 t}\right) \\ 1 + \cot^2 t &= \csc^2 t \end{aligned} } $

These equations are also called Pythagorean trigonometric identities.

Summary

We have derived three Pythagorean identities:

• $\cos^2 t + \sin^2 t = 1$
• $1 + \tan^2 t = \sec^2 t$
• $1 + \cot^2 t = \csc^2 t$

Applications 

The Pythagorean identities can be used to simplify problems by transforming trigonometric expressions. In expressions with multiple trigonometric functions, the Pythagorean identities can be used to substitute and simplify the expression.  

For example, consider the following:

$(1 - \cos^2 t) \csc^2 t$

Let's try to simplify this. We know that cosecant is the reciprocal function of sine.  In other words, we can say 

$\displaystyle{\csc^2 t = \frac{1}{\sin^2 t}}$

We can also recognize that $\sin^2 t = 1 - \cos^2 t $, which is simply the Pythagorean identity rearranged.

Therefore, we can rewrite the expression in terms of sine: 

$\displaystyle{ \left(1 - \cos^2 t\right) \csc^2 t = \left(\sin^2 t\right)\left(\frac{1}{\sin^2 t}\right) }$

The sine functions cancel and this simplifies to $1$, so we have:

 $\displaystyle{ \begin{aligned} \left(1 - \cos^2 t\right) \csc^2 t &= \left(\sin^2 t\right)\left(\frac{1}{\sin^2 t}\right) \\ &= 1 \end{aligned} }$

When simplifying expressions with trigonometric functions, it is helpful to look for ways to use the Pythagorean identities to cancel terms. The problem below provides another helpful example.

Example

Simplify the following expression: $5\sin^2 t + \sec^2 t + 5\cos^2 t - 1 - \tan^2 t$

This looks like a very complicated problem, but let's look for things we can cancel. First, notice that we have both $\sin^2 t$$$ and $\cos^2 t$, and group them together:

$5 \sin^2 t + 5 \cos^2 t+ \sec^2 t - 1 - \tan^2 t$

We can factor out the $5$:

$5(\sin^2 t + \cos^2 t)+ \sec^2 t - 1 - \tan^2 t$

We can now substitute $1$ for $\sin^2 t + \cos^2 t$, applying the Pythagorean identity: 

$5(1) + \sec^2 t - 1 - \tan^2 t$

Look at the remaining terms in the expression. Recall that one of the Pythagorean identities states $1 + \tan^2 t = \sec^2 t$. This can be rearranged to $\sec^2 x - 1 = \tan^2 x$. Substituting this into the expression, we have:

$\displaystyle{ \begin{aligned} 5 + \sec^2 t - 1 - \tan^2 t &= 5 + \left(1 + \tan^2 t\right) - 1 - \tan^2 t \\ &= 5 + 1 - 1 + \tan^2 t - \tan^2 t \\ &= 5 + \tan^2 t - \tan^2 t \\ &= 5 \end{aligned} }$

The expression $5\sin^2 t + \sec^2 t + 5\cos^2 t - 1 - \tan^2 t$ simplifies to $5$.