Trigonometric Identities

# What is a Trigonometric Identity?

A trigonometric identity is essentially another form of a trigonometric expression that is equivalent to the original. For example, we have already learned two very important trigonometric identities. One for the tangent function in relation to sine and cosine:

(1)
\begin{align} tanx = \frac{sinx}{cosx} \end{align}

And of course, the unit circle:

(2)
$$cos^2x + sin^2x = 1$$

We will now extend our knowledge on trigonometric identities for the purpose of simplifying seemingly difficult expressions for future applications.

## Example 1

Simplify the following expression:

(3)
\begin{align} [cos^4x + 2cos^2xsin^2x + sin^4x] \frac{cotx}{cosx} \end{align}

This expression may first appear very daunting to simplify, however, we can certainly complete it without further knowledge into trigonometric identities. Let's first simplify the part of the expression contained within the square brackets. By factorization, we should notice that we can simplify it to obtain:

(4)
\begin{align} [(cos^2x + sin^2x)^2] \frac{cotx}{cosx} \end{align}

But we know the following identity from which we can substitute for to obtain a modified yet equivalent expression:

(5)
$$cos^2x + sin^2x = 1$$

Thus:

(6)
\begin{align} [(1)^2] \frac{cotx}{cosx} \end{align}
(7)
\begin{align} \frac{cotx}{cosx} \end{align}

Now we know that cotangent is simply the reciprocal trigonometric function of tangent, so we can rewrite our expression further to obtain:

(8)
\begin{align} \frac{\frac{cosx}{sinx}}{cosx} \end{align}

Which when simplified we obtain:

(9)
\begin{align} \frac{1}{sinx} \end{align}

Or simply:

(10)
$$cscx$$

As you can see, the use of trigonometric identities for simplifying trigonometric expressions is very useful.

# Reciprocal Trigonometric Identities

The reciprocal trigonometric identities are straightforward and are exactly the same as our definitions for the reciprocal trigonometric identities. That is:

(11)
\begin{align} secx = \frac{1}{cosx} \quad cscx = \frac{1}{sinx} \quad cotx = \frac{cosx}{sinx} \end{align}

# Pythagorean Trigonometric Identities

We already are aware of one of the pythagorean trigonometric identities which is the definition of the unit circle such that:

(12)
$$cos^2x + sin^2x = 1$$

However, there are other pythagorean trigonometric identities such as the following:

(13)
\begin{align} 1^2 + tan^2x = sec^2x \quad 1^2 + cot^2x = csc^2x \end{align}

## Proof of Pythagorean Trigonometric Identity Equivalencies

We will not prove the unit circle trigonometric identity because it is already geometrically proven on the unit circle page. Nevertheless we will begin the following trigonometric identity and prove that the left hand side is equivalent to the righthand side with the trigonometric identities we already know are true.

(14)
$$1^2 + tan^2x = sec^2x$$
(15)
\begin{align} (cos^2x + sin^2x) + \frac{sin^2x}{cos^2x} = sec^2x \end{align}
(16)
\begin{align} \frac{cos^2x(cos^2x + sin^2x)}{cos^2x} + \frac{sin^2x}{cos^2x} = sec^2x \end{align}
(17)
\begin{align} \frac{cos^4x + cos^2xsin^2x + sin^2x}{cos^2x} = sec^2x \end{align}
(18)
\begin{align} \frac{cos^4x + cos^2x(1 - cos^2x) + (1 - cos^2x)}{cos^2x} = sec^2x \end{align}
(19)
\begin{align} \frac{cos^4x + cos^2x - cos^4x + 1 - cos^2x}{cos^2x} = sec^2x \end{align}
(20)
\begin{align} \frac{1 }{cos^2x} = sec^2x \end{align}
(21)
$$sec^2x = sec^2x$$

We will now prove the other Pythagorean identity as being equivalent:

(22)
$$1^2 + cot^2x = csc^2x$$
(23)
\begin{align} (cos^2x + sin^2x) + \frac{cos^2x}{sin^2x} = csc^2x \end{align}
(24)
\begin{align} \frac{sin^2x(cos^2x + sin^2x)}{sin^2x} + \frac{cos^2x}{sin^2x} = csc^2x \end{align}
(25)
\begin{align} \frac{sin^2x(cos^2x + sin^2x) + cos^2x}{sin^2x} = csc^2x \end{align}
(26)
\begin{align} \frac{sin^2xcos^2x + sin^4x + cos^2x}{sin^2x} = csc^2x \end{align}
(27)
\begin{align} \frac{sin^2x(1-sin^2x) + sin^4x + (1 - sin^2x)}{sin^2x} = csc^2x \end{align}
(28)
\begin{align} \frac{sin^2x -sin^4x + sin^4x + 1 - sin^2x}{sin^2x} = csc^2x \end{align}
(29)
\begin{align} \frac{1}{sin^2x} = csc^2x \end{align}
(30)
$$csc^2x = csc^2x$$

# Odd and Even Trigonometric Identities.

The odd and even trigonometric identities are more difficult to prove. They are:

(31)
\begin{align} -sin(x) = sin(-x) \quad cos(x) = cos(-x) \quad -tan(x) = tan(-x) \end{align}

## Odd Functions

First let's acknowledge what an odd and an even function is. An odd function is a function that satisfies the condition such that:

(32)
$$f(-x) = -f(x)$$

Additionally, functions that are odd can be rotated 180 degrees about the origin to obtain the same graph (look at the graph of cosx for reference).

So let's say that we have a function such that f(x) = cosx. Thus, -f(x) = -cosx, and f(-x) = cos(-x).

We should acknowledge thatn f(x) and f(-x) are horizontal reflections about the y-axis, and f(x) and f(-x) are vertical reflections about the x-axis. Hence, a function f(x) that is odd has a function g(x) that is equivalent to f(x) being reflected along the y-axis and the x-axis.

We can thus determine that the cosine function is an odd function because cosx can be rotated 180 degrees about the origin to obtain cosx and is a reflection of itself across the y-axis and x-axis.

Therefore:

(33)
$$cos(-x) = -cos(x)$$

## Even Functions

An even function on the other hand is a function that satisfies the condition:

(34)
$$f(x) = f(-x)$$

This transformation states that a function, f(x) is symmetric about the y-axis as a horizontal reflection. (Look at the graph of sinx as an example). Hence, a function is even if it satisfies only this property. A function that satisfies f(x) = f(-x) and f(-x) = -f(x) is no longer even because it satisfies both conditions and hence, is an odd function. Thus we can geometrically see that sinx is an even function. Therefore:

(35)
$$sinx = sin(-x)$$

### Example 2

Determine the most simplified trigonometric expression for:

(36)
\begin{align} cos^2(-x) + sin^2(-x) + cot^2x \frac{sin^2x}{cos^3x} - 1 + tan^2x \end{align}

The simplification using the trigonometric identities explained on this page are all that is necessary to simplify this. The simplification is as follows:

(37)
\begin{align} (-cosx)^2 + (-sinx)^2 + cot^2x + \frac{sin^2x}{cos^3x} - 1 + tan^2x \end{align}
(38)
\begin{align} 1 + cot^2x + \frac{sin^2x}{cos^3x} - 1 + tan^2x \end{align}
(39)
\begin{align} cot^2x + \frac{sin^2x}{cos^3x} + tan^2x \end{align}
(40)
\begin{align} \frac{cos^2x}{sin^2x} + \frac{sin^2x}{cos^3x} + tan^2x \end{align}
(41)
\begin{align} \frac{cos^2x sin^2x}{sin^2x cos^3x} + tan^2x \end{align}
(42)
\begin{align} \frac{1}{ cosx} + tan^2x \end{align}
(43)
$$secx + tan^2x$$

We could stop here for simplification, however, we can express this expression in terms of tangent only if we decide since we know that:

(44)
$$sec^2x = 1 + tan^2x$$

Thus we obtain that:

(45)
\begin{align} secx = \pm \sqrt{1 + tan^2x} \end{align}

Thus we obtain:

(46)
\begin{align} \pm \sqrt{1 + tan^2x} + tan^2x \end{align}