Equivalent Representations Of Trigonometric Functions

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Trigonometric functions can be written in many different, yet equivalent, ways. Understanding these equivalent representations, or identities, is a crucial skill for simplifying complex expressions, solving equations, and preparing for higher-level mathematics like calculus. Let's explore how these powerful connections work.

What Are Equivalent Trigonometric Representations?

Equivalent representations of trigonometric functions, more commonly known as trigonometric identities, are equations involving trigonometric functions that are true for every value of the variable for which both sides of the equation are defined. Think of it like this: the expressions $\frac{1}{2}$ , $0.5$ , and $50 %$ all look different, but they represent the exact same numerical value. Trigonometric identities work the same way; they are different ways of writing the same trigonometric relationship.

For example, the expression $\sin^{2} (x) + \cos^{2} (x)$ might look complicated, but it is always equal to $1$ , no matter what angle $x$ you choose. This fundamental relationship, $\sin^{2} (x) + \cos^{2} (x) = 1$ , is one of the most important identities in all of mathematics. Mastering these equivalences is not just an exercise in memorization; it is about understanding the deep, geometric relationships between the different trigonometric functions. These identities are the essential tools you will use to:

Simplify complex trigonometric expressions.
Solve trigonometric equations.
Prove other mathematical statements.
Rewrite functions into forms that are easier to work with in calculus.

In this lesson, we will explore three foundational categories of trigonometric identities: the Cofunction Identities, the Negative Angle Identities, and the powerful Pythagorean Identities.

How Do Cofunction Identities Relate Sine and Cosine?

The names of the trigonometric functions themselves hint at a special relationship. The prefix "co-" in cosine, cotangent, and cosecant stands for complementary. Recall that two angles are complementary if they add up to $90^{\circ}$ (or $\frac{π}{2}$ radians). Cofunction identities state that the value of a trigonometric function of an angle is equal to the value of the cofunction of its complementary angle.

Let's see why this works with a right-angled triangle. Consider a triangle with angles $θ$ , $90^{\circ} - θ$ , and $90^{\circ}$ . Let the side opposite $θ$ be $a$ , the side adjacent to $θ$ be $b$ , and the hypotenuse be $c$ .

For angle $θ$ : $\sin (θ) = \frac{opposite}{hypotenuse} = \frac{a}{c}$
For the complementary angle $90^{\circ} - θ$ : The side that was opposite $θ$ (side $a$ ) is now adjacent to this angle. Therefore, $\cos (90^{\circ} - θ) = \frac{adjacent}{hypotenuse} = \frac{a}{c}$ .

As you can see, $\sin (θ) = \cos (90^{\circ} - θ)$ . This relationship holds for all the cofunction pairs.

\sin (θ) = \cos (\frac{π}{2} - θ)

\cos (θ) = \sin (\frac{π}{2} - θ)

Here is a complete list of the cofunction identities:

In Radians	In Degrees
$\sin (θ) = \cos (\frac{π}{2} - θ)$	$\sin (θ) = \cos (90^{\circ} - θ)$
$\cos (θ) = \sin (\frac{π}{2} - θ)$	$\cos (θ) = \sin (90^{\circ} - θ)$
$\tan (θ) = \cot (\frac{π}{2} - θ)$	$\tan (θ) = \cot (90^{\circ} - θ)$
$\cot (θ) = \tan (\frac{π}{2} - θ)$	$\cot (θ) = \tan (90^{\circ} - θ)$
$\sec (θ) = \csc (\frac{π}{2} - θ)$	$\sec (θ) = \csc (90^{\circ} - θ)$
$\csc (θ) = \sec (\frac{π}{2} - θ)$	$\csc (θ) = \sec (90^{\circ} - θ)$

Example 1

If you know that $\cos (40^{\circ}) \approx 0.7660$ , what is the value of $\sin (50^{\circ})$ ?

Solution:
We need to find $\sin (50^{\circ})$ . Notice that $40^{\circ}$ and $50^{\circ}$ are complementary angles because $40^{\circ} + 50^{\circ} = 90^{\circ}$ .
Using the cofunction identity $\sin (θ) = \cos (90^{\circ} - θ)$ , we can set $θ = 50^{\circ}$ .
$\sin (50^{\circ}) = \cos (90^{\circ} - 50^{\circ})$
$\sin (50^{\circ}) = \cos (40^{\circ})$
Since we are given that $\cos (40^{\circ}) \approx 0.7660$ , it follows that $\sin (50^{\circ}) \approx 0.7660$ .

What Happens When the Angle is Negative?

Negative angle identities describe the relationship between the trigonometric value of an angle $θ$ and its opposite, $- θ$ . These are best understood by visualizing the unit circle. An angle $θ$ in standard position corresponds to a point $(x, y)$ on the unit circle. A negative angle $- θ$ is measured clockwise from the positive x-axis by the same amount, corresponding to the point $(x, - y)$ .

Remember that on the unit circle, $\cos (θ) = x$ and $\sin (θ) = y$ . Let's see how this applies:

For Cosine: The x-coordinate is the same for both $θ$ and $- θ$ . Therefore, $\cos (- θ) = x = \cos (θ)$ . Because $f (- x) = f (x)$ , cosine is an even function.
For Sine: The y-coordinate for $- θ$ is the negative of the y-coordinate for $θ$ . Therefore, $\sin (- θ) = - y = - \sin (θ)$ . Because $f (- x) = - f (x)$ , sine is an odd function.

We can find the identity for tangent by using the quotient identity $\tan (θ) = \frac{\sin (θ)}{\cos (θ)}$ :
$\tan (- θ) = \frac{\sin (- θ)}{\cos (- θ)} = \frac{- \sin (θ)}{\cos (θ)} = - \tan (θ)$ . So, tangent is also an odd function.

\sin (- θ) = - \sin (θ)

\cos (- θ) = \cos (θ)

\tan (- θ) = - \tan (θ)

The identities for the reciprocal functions follow the same even/odd pattern: secant is even, while cosecant and cotangent are odd.

Example 2

Simplify the expression $\sec (- x) - \sin (- x) \cot (- x)$ .

Solution:
We will apply the negative angle identities to each part of the expression.
First, recall the identities: $\sec (- θ) = \sec (θ)$ (even), $\sin (- θ) = - \sin (θ)$ (odd), and $\cot (- θ) = - \cot (θ)$ (odd).
Substitute these into the expression:
$\sec (x) - (- \sin (x)) (- \cot (x))$ The two negative signs in the second term multiply to become a positive:
$\sec (x) - (\sin (x)) (\cot (x))$ Now, rewrite $\cot (x)$ in terms of sine and cosine: $\cot (x) = \frac{\cos (x)}{\sin (x)}$ .
$\sec (x) - \sin (x) (\frac{\cos (x)}{\sin (x)})$ The $\sin (x)$ terms cancel out, provided $\sin (x) \neq 0$ .
$\sec (x) - \cos (x)$ This is the simplified form. We could also write it as $\frac{1}{\cos (x)} - \cos (x)$ if needed.

Why Are Pythagorean Identities So Important?

The Pythagorean identities are the cornerstone of simplifying trigonometric expressions. They are derived directly from the Pythagorean theorem applied to the unit circle. The equation of a unit circle is $x^{2} + y^{2} = 1$ . For any angle $θ$ on the unit circle, the coordinates of the corresponding point are $(x, y) = (\cos (θ), \sin (θ))$ . Substituting these into the circle's equation gives us the fundamental Pythagorean identity:

\sin^{2} (θ) + \cos^{2} (θ) = 1

This single identity is incredibly powerful. It provides a direct link between the sine and cosine of any angle. From this one equation, we can derive two other important Pythagorean identities.

Deriving the second identity:
Start with $\sin^{2} (θ) + \cos^{2} (θ) = 1$ and divide every term by $\cos^{2} (θ)$ (assuming $\cos (θ) \neq 0$ ):
$\frac{\sin^{2} (θ)}{\cos^{2} (θ)} + \frac{\cos^{2} (θ)}{\cos^{2} (θ)} = \frac{1}{\cos^{2} (θ)}$ Using the quotient and reciprocal identities, this simplifies to:
$\tan^{2} (θ) + 1 = \sec^{2} (θ)$

Deriving the third identity:
Start with $\sin^{2} (θ) + \cos^{2} (θ) = 1$ and divide every term by $\sin^{2} (θ)$ (assuming $\sin (θ) \neq 0$ ):
$\frac{\sin^{2} (θ)}{\sin^{2} (θ)} + \frac{\cos^{2} (θ)}{\sin^{2} (θ)} = \frac{1}{\sin^{2} (θ)}$ This simplifies to:
$1 + \cot^{2} (θ) = \csc^{2} (θ)$

These three identities are essential for converting between different trigonometric functions and simplifying expressions that involve squared terms.

Example 3

Given that $\cos (θ) = - \frac{12}{13}$ and angle $θ$ is in Quadrant III, find the values of $\sin (θ)$ and $\tan (θ)$ .

Solution:
We are given $\cos (θ)$ and need to find $\sin (θ)$ . The Pythagorean identity $\sin^{2} (θ) + \cos^{2} (θ) = 1$ is the perfect tool for this.
Substitute the given value of $\cos (θ)$ :
$\sin^{2} (θ) + {(- \frac{12}{13})}^{2} = 1$ $\sin^{2} (θ) + \frac{144}{169} = 1$ Subtract $\frac{144}{169}$ from both sides:
$\sin^{2} (θ) = 1 - \frac{144}{169} = \frac{169}{169} - \frac{144}{169} = \frac{25}{169}$ Now, take the square root of both sides. Remember to include both positive and negative possibilities:
$\sin (θ) = \pm \sqrt{\frac{25}{169}} = \pm \frac{5}{13}$ To choose the correct sign, we use the quadrant information. In Quadrant III, the y-values are negative, so sine is negative. Therefore, we must choose the negative root.
$\sin (θ) = - \frac{5}{13}$ .

Now we can find $\tan (θ)$ using the quotient identity:
$\tan (θ) = \frac{\sin (θ)}{\cos (θ)} = \frac{- 5 / 13}{- 12 / 13}$ The denominators and the negative signs cancel out:
$\tan (θ) = \frac{5}{12}$ This makes sense, as tangent is positive in Quadrant III.

How Can You Combine Identities to Simplify Expressions?

Simplifying trigonometric expressions or proving new identities is like solving a puzzle. You have a set of tools—the identities—and you need to apply them strategically to transform a complicated expression into a simpler one. There isn't always a single correct path, but some strategies are more effective than others.

Here are some general tips to guide your work:

Start with the more complex side: If you are trying to prove an identity like $A = B$ , pick the side that looks more complicated and try to manipulate it until it looks like the other side.
Convert to sine and cosine: This is often the most reliable strategy. When in doubt, rewrite all functions ( $\tan, \cot, \sec, \csc$ ) in terms of $\sin$ and $\cos$ . This often reveals hidden cancellations or simplifications.
Look for Pythagorean forms: Be on the lookout for squared terms like $\sin^{2} (x)$ or $\tan^{2} (x)$ . These are strong clues that a Pythagorean identity might be useful. Sometimes you might see $1 - \cos^{2} (x)$ , which can be immediately replaced with $\sin^{2} (x)$ .
Use algebra: Don't forget your fundamental algebra skills! You may need to:
- Find a common denominator to add or subtract fractions.
- Factor expressions (e.g., difference of squares).
- Multiply by a clever form of 1, such as multiplying the numerator and denominator by a conjugate.

Let's walk through an example that requires proving an identity by manipulating one side to match the other, a common task in trigonometry.

Example 4

Prove the identity: $\frac{\sin (x)}{1 + \cos (x)} = \frac{1 - \cos (x)}{\sin (x)}$ .

Solution:
We will start with the left-hand side (LHS) and try to transform it into the right-hand side (RHS).
LHS = $\frac{\sin (x)}{1 + \cos (x)}$
The expression on the right has $1 - \cos (x)$ in the numerator. This suggests that multiplying by the conjugate of the denominator, which is $1 - \cos (x)$ , might be a good strategy. We will multiply the LHS by $\frac{1 - \cos (x)}{1 - \cos (x)}$ , which is a form of $1$ .
$\frac{\sin (x)}{1 + \cos (x)} \cdot \frac{1 - \cos (x)}{1 - \cos (x)}$ Multiply the numerators and the denominators:
$\frac{\sin (x) (1 - \cos (x))}{(1 + \cos (x)) (1 - \cos (x))}$ The denominator is in the form $(a + b) (a - b) = a^{2} - b^{2}$ . So, $(1 + \cos (x)) (1 - \cos (x)) = 1^{2} - \cos^{2} (x) = 1 - \cos^{2} (x)$ .
$\frac{\sin (x) (1 - \cos (x))}{1 - \cos^{2} (x)}$ Now, we use the Pythagorean identity $\sin^{2} (x) + \cos^{2} (x) = 1$ . Rearranging it gives $\sin^{2} (x) = 1 - \cos^{2} (x)$ . We can substitute this into the denominator.
$\frac{\sin (x) (1 - \cos (x))}{\sin^{2} (x)}$ Finally, we can cancel one factor of $\sin (x)$ from the numerator and the denominator.
$\frac{1 - \cos (x)}{\sin (x)}$ This is exactly the right-hand side of the original equation. We have successfully proven the identity.

What Are Common Mistakes to Avoid?

Working with trigonometric identities can be tricky, and a few common errors often trip students up. Being aware of these pitfalls is the first step to avoiding them.

Incorrectly Applying Functions to Sums: A very common mistake is to think that a trig function can be distributed over a sum or difference. For example, $\sin (A + B)$ is not equal to $\sin (A) + \sin (B)$ . These require special sum and difference identities, which are a separate topic.
Confusing $\sin^{2} (x)$ and $\sin (x^{2})$ : The notation $\sin^{2} (x)$ is shorthand for $(\sin (x))^{2}$ , which means you find the sine of the angle first, then square the result. This is completely different from $\sin (x^{2})$ , where you square the angle first, then find the sine of that new angle.
Forgetting the Sign ( $\pm$ ): When using a Pythagorean identity to solve for a function value, you often have to take a square root. For instance, if $\sin^{2} (x) = \frac{1}{4}$ , then $\sin (x) = \pm \frac{1}{2}$ . You must use the quadrant information given in the problem to determine whether the positive or negative value is correct. Forgetting this step leads to an incomplete or incorrect answer.
Mixing Degrees and Radians: The cofunction identities can be written with $90^{\circ}$ or $\frac{π}{2}$ . Be sure you are using the correct form based on the units of your angle. Don't mix them, for example by writing $\cos (π - θ)$ when you mean $\cos (180^{\circ} - θ)$ or vice versa if the context is specific.
Algebraic Errors: Sometimes the mistake isn't with the trigonometry but with the underlying algebra. Double-check your factoring, fraction manipulation, and simplification steps. It's easy to make a small error that derails the entire problem.

Quick Reference: Key Trigonometric Identities

This table summarizes the fundamental identities discussed in this lesson. They are essential tools for your mathematical toolkit and are worth committing to memory or keeping as a handy reference.

Identity Type	Key Formulas	Core Concept
Cofunction Identities	$\sin (θ) = \cos (\frac{π}{2} - θ)$ $\tan (θ) = \cot (\frac{π}{2} - θ)$ $\sec (θ) = \csc (\frac{π}{2} - θ)$	A function of an angle equals the cofunction of its complement.
Negative Angle Identities (Even/Odd)	$\sin (- θ) = - \sin (θ)$ (Odd) $\cos (- θ) = \cos (θ)$ (Even) $\tan (- θ) = - \tan (θ)$ (Odd)	Describes how a function behaves with a negative input angle.
Pythagorean Identities	$\sin^{2} (θ) + \cos^{2} (θ) = 1$ $1 + \tan^{2} (θ) = \sec^{2} (θ)$ $1 + \cot^{2} (θ) = \csc^{2} (θ)$	Relates the squares of different trigonometric functions based on the Pythagorean theorem.

Frequently Asked Questions

Why are these called 'identities'?

They are called identities because they are true for all possible values of the input variable (the angle), whereas a regular equation is only true for specific values. For example, $\sin (x) = 1$ is an equation true only for certain $x$ , but $\sin^{2} (x) + \cos^{2} (x) = 1$ is an identity because it's true for any $x$ .

Do I need to memorize all of these identities?

It is highly recommended to memorize the fundamental Pythagorean identity $\sin^{2} (θ) + \cos^{2} (θ) = 1$ and the negative angle (even/odd) properties. The other two Pythagorean identities can be quickly derived from the first one, and the cofunction identities can be understood conceptually from a right triangle.

What's the difference between an identity and an equation?

An identity is a statement of equivalence that holds true for any value of the variable for which the expressions are defined. An equation is a statement that is only true for a specific set of values, called the solutions. The goal with an identity is to prove it's always true, while the goal with an equation is to find its solutions.

How do I know which identity to use when simplifying an expression?

Practice is key. Look for clues in the expression: squared terms suggest a Pythagorean identity, negative angles suggest a negative angle identity, and a mix of functions might be simplified by converting everything to sine and cosine. Over time, you will develop an intuition for which strategy to try first.

Are there more identities than the ones listed here?

Yes, many more! This lesson covers the foundational identities. You will also learn about sum and difference identities (like for $\sin (A + B)$ ), double-angle identities (for $\sin (2 A)$ ), half-angle identities, and product-to-sum identities. These all build upon the fundamental ones covered here.

Can I use these identities with radians instead of degrees?

Absolutely. All of these identities are true regardless of whether the angle is measured in degrees or radians. The only place the unit matters is in the cofunction identities, where the complement is either $90^{\circ}$ or $\frac{π}{2}$ radians.

Why is cos(x) an even function but sin(x) is an odd function?

This comes from the symmetries of the unit circle. For any angle $x$ , the angle $- x$ has the same x-coordinate, making $\cos (x) = \cos (- x)$ (even). However, the angle $- x$ has the opposite y-coordinate, making $\sin (- x) = - \sin (x)$ (odd).