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Trigonometric Function Toolkit · Tutorial 146 of 1495

Selecting Identities to Simplify Trigonometric Expressions

Practise selecting a useful identity, simplifying without changing an expression’s domain, and checking the result.

Beginner 9 min read

What You'll Learn

  • Identify which trigonometric identities are useful for a given expression.
  • Combine quotient and reciprocal identities to rewrite expressions using sine and cosine.
  • Use Pythagorean and cofunction identities to simplify expressions.
  • Recognize when angle-combination or sum-to-product formulas create a useful factor.
  • Track the domain of the original expression through each simplification.
  • Check a simplified result on the original domain.

Introduction

Several trigonometric identities may be relevant to the same expression. The main task is to choose one that makes the expression simpler without losing track of where the original expression is defined.

A useful approach is to look for structure: a quotient and a reciprocal function may become easier to handle when both are rewritten using sine and cosine; a squared sine and cosine may suggest a Pythagorean identity; complementary angles may suggest a cofunction identity; and a sum or difference of angles may suggest an angle-combination formula.

In every case, simplification must respect the original domain. An expression involving division is not defined when its denominator is zero, even if algebra later produces a formula that is defined there.

The Domain Comes First

The domain of an expression is the set of input values for which every operation in that expression is defined. For trigonometric expressions, denominators and the definitions of tangent, cotangent, secant, and cosecant impose restrictions.

Key principle: A simplified expression must agree with the original expression at every input in the original domain. The simplified formula may be defined at additional inputs, but those inputs do not automatically become part of the original domain.

For example, \(\tan x\) is defined only when \(\cos x\ne 0\), because \(\tan x=\frac{\sin x}{\cos x}\). Likewise, \(\cot x\) requires \(\sin x\ne 0\), \(\sec x\) requires \(\cos x\ne 0\), and \(\csc x\) requires \(\sin x\ne 0\). If one of these expressions appears in a denominator, it must also be nonzero.

It is often helpful to record the restrictions before choosing an identity. Then, whenever a factor is cancelled or an expression is divided by, you can verify that the factor is nonzero on the domain under consideration.

A Practical Selection Strategy

The identities from earlier material are tools for changing the form of an expression. They are not all equally useful in every situation. The aim is to choose a form that reduces the number of different trigonometric functions, exposes a familiar identity, or creates a factor that can be cancelled safely.

1
Record restrictions.
Find values excluded by denominators and by any tangent, cotangent, secant, or cosecant terms.
2
Look for a visible pattern.
Squared sine and cosine terms suggest a Pythagorean identity; complementary angles suggest cofunction identities; sums or differences of angles suggest angle-combination or sum-to-product formulas.
3
Choose a useful form.
Quotient and reciprocal identities can rewrite the expression in terms of sine and cosine, making common factors or Pythagorean relationships easier to see.
4
Simplify only where each step is valid.
Before cancelling or dividing by a quantity, check that it is nonzero on the original domain.
5
Check the result.
Confirm that the transformations reverse on the original domain, and test a permitted input when an exact numerical check is convenient.

Worked Example: Combining Reciprocal, Quotient, and Pythagorean Identities

Worked Example: Combining Reciprocal, Quotient, and Pythagorean Identities

Simplify \(\frac{\sec x-\cos x}{\tan x}\), stating its original domain.

Step 1: Determine the domain. The expression contains \(\sec x\) and \(\tan x\), so \(\cos x\ne 0\). Also, the denominator \(\tan x\) must be nonzero. Since \(\tan x=\frac{\sin x}{\cos x}\) and \(\cos x\ne 0\), this requires \(\sin x\ne 0\). Thus the original domain is the set of \(x\) for which \(\sin x\ne 0\) and \(\cos x\ne 0\).

Step 2: Use the reciprocal identity. Replace \(\sec x\) by \(\frac{1}{\cos x}\):

$$ \frac{\sec x-\cos x}{\tan x} = \frac{\frac{1}{\cos x}-\cos x}{\tan x}. $$

Step 3: Combine the terms in the numerator.

$$ \frac{\frac{1}{\cos x}-\cos x}{\tan x} = \frac{\frac{1-\cos^2 x}{\cos x}}{\tan x}. $$

Step 4: Use the Pythagorean and quotient identities. Since \(1-\cos^2 x=\sin^2 x\) and \(\tan x=\frac{\sin x}{\cos x}\),

$$ \frac{\frac{1-\cos^2 x}{\cos x}}{\tan x} = \frac{\frac{\sin^2 x}{\cos x}}{\frac{\sin x}{\cos x}} = \frac{\sin^2 x}{\cos x}\cdot\frac{\cos x}{\sin x} = \sin x. $$

The cancellation is valid on the original domain: both \(\cos x\) and \(\sin x\) are nonzero there. The simplified expression is \(\sin x\), with the understanding that it agrees with the original only where the original expression is defined.

Check: At \(x=\frac{\pi}{6}\), both \(\sin x\) and \(\cos x\) are nonzero. The original expression has value

$$ \frac{2-\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{3}} = \frac{\frac{4-\sqrt{3}}{2}}{\frac{\sqrt{3}}{3}} = \frac{3(4-\sqrt{3})}{2\sqrt{3}}. $$

This does not equal \(\sin\frac{\pi}{6}=\frac12\), so it exposes an arithmetic error in that check: \(\sec\frac{\pi}{6}=\frac{2}{\sqrt{3}}\), not \(2\). Using the correct value gives

$$ \frac{\frac{2}{\sqrt{3}}-\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{3}} = \frac{\frac{4-3}{2\sqrt{3}}}{\frac{\sqrt{3}}{3}} = \frac{1}{2\sqrt{3}}\cdot\frac{3}{\sqrt{3}} = \frac12, $$

which matches the simplified value. The step-by-step identity chain also verifies the simplification throughout the original domain.

Why the Transformations Work

An identity can replace one expression with another equal expression wherever both are defined. For instance, \(\sec x=\frac{1}{\cos x}\) is valid when \(\cos x\ne 0\), precisely the restriction already required for \(\sec x\). Similarly, the quotient identity for tangent is valid wherever \(\tan x\) is defined.

Cancellation needs an additional check. The rule \(\frac{a}{b}=c\) or cancelling a common factor is valid only when the relevant divisor or factor is nonzero. In the first worked example, the original denominator requires \(\tan x\ne 0\); together with \(\cos x\ne 0\), this implies \(\sin x\ne 0\). Therefore cancelling \(\sin x\) does not remove any points from the original domain.

Domain-preserving rule: If each identity used is valid and every cancellation is by a quantity known to be nonzero, the resulting formula agrees with the original expression on its original domain. Do not infer that the original expression is defined at any additional inputs where the new formula happens to exist.

Worked Example: Using Cofunction and Pythagorean Identities

Worked Example: Using Cofunction and Pythagorean Identities

Simplify \(\sin\left(\frac{\pi}{2}-x\right)\cos x+\cos\left(\frac{\pi}{2}-x\right)\sin x\).

There are no denominators or restricted trigonometric functions here, so the original expression is defined for every real \(x\). The angles \(x\) and \(\frac{\pi}{2}-x\) are complementary. Apply the cofunction identities \(\sin(\frac{\pi}{2}-x)=\cos x\) and \(\cos(\frac{\pi}{2}-x)=\sin x\):

$$ \begin{aligned} \sin\left(\frac{\pi}{2}-x\right)\cos x +\cos\left(\frac{\pi}{2}-x\right)\sin x &=\cos x\cos x+\sin x\sin x\\ &=\cos^2 x+\sin^2 x\\ &=1. \end{aligned} $$

The last step uses the Pythagorean identity. Thus the expression simplifies to \(1\), for every real \(x\).

Check: Take \(x=\frac{\pi}{6}\). The original expression is

$$ \sin\left(\frac{\pi}{3}\right)\cos\left(\frac{\pi}{6}\right) +\cos\left(\frac{\pi}{3}\right)\sin\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{3}}{2} +\frac12\cdot\frac12 = \frac34+\frac14 = 1. $$

This agrees with the simplified result. No domain restrictions were introduced because the identities used are defined for every real \(x\).

Worked Example: Choosing a Sum-to-Product Formula

Worked Example: Choosing a Sum-to-Product Formula

Simplify \(\frac{\sin(x+y)-\sin(x-y)}{2\cos x}\), and state its domain.

The denominator requires \(\cos x\ne 0\). The numerator is defined for all real \(x\) and \(y\), so the original domain consists of all pairs \((x,y)\) with \(\cos x\ne 0\).

The numerator is a difference of sines. The sum-to-product formula \(\sin A-\sin B=2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)\) is useful here. Set \(A=x+y\) and \(B=x-y\). Then

$$ \frac{A+B}{2} = \frac{(x+y)+(x-y)}{2} = x, \qquad \frac{A-B}{2} = \frac{(x+y)-(x-y)}{2} = y. $$

Therefore,

$$ \frac{\sin(x+y)-\sin(x-y)}{2\cos x} = \frac{2\cos x\sin y}{2\cos x} = \sin y. $$

Because the original domain requires \(\cos x\ne 0\), cancelling \(\cos x\) is valid. The simplified result is \(\sin y\) on that domain; the formula does not extend the domain of the original expression.

Check: Choose \(x=\frac{\pi}{6}\) and \(y=\frac{\pi}{4}\). Then \(\cos x=\frac{\sqrt{3}}{2}\ne 0\), and the original expression becomes

$$ \frac{\sin\left(\frac{5\pi}{12}\right)-\sin\left(-\frac{\pi}{12}\right)} {2\cos\left(\frac{\pi}{6}\right)} = \frac{\frac{\sqrt{6}}{2}}{\sqrt{3}} = \frac{\sqrt{2}}{2}. $$

The simplified value is \(\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}\), so the numerical check agrees.

A Common Domain Mistake

Consider the expression \(\frac{1-\cos^2 x}{\sin x}\). Using the Pythagorean identity gives \(\frac{\sin^2 x}{\sin x}=\sin x\), but the original expression requires \(\sin x\ne 0\). The simplification is therefore valid only on that original domain.

A common incorrect conclusion is that the original expression now has value \(0\) at \(x=0\), because the simplified formula \(\sin 0=0\). But at \(x=0\), the original expression is \(\frac{1-1}{0}=\frac00\), which is undefined. The correct statement is that the original expression equals \(\sin x\) whenever \(\sin x\ne 0\); its domain has not been enlarged.

Takeaway: Simplifying changes the form of an expression, not the domain of the original expression. Keep the original restrictions with the final answer.

Choosing Among Identity Families

When several identities seem relevant, prefer the one that exposes the expression’s structure with the fewest unnecessary steps.

Visible featureUseful identity choice
\(\tan x\), \(\cot x\), \(\sec x\), or \(\csc x\)Quotient or reciprocal identities can rewrite the expression in sine and cosine.
A pair such as \(\sin^2 x\) and \(\cos^2 x\)A Pythagorean identity may combine or replace the terms.
An angle of the form \(\frac{\pi}{2}-x\)Cofunction identities can replace the function by one of the complementary angle.
Sines or cosines of sums or differences of anglesAngle-combination or sum-to-product formulas may reveal a common factor.

These are clues, not automatic instructions. One expression may call for more than one identity, as in the worked examples. After each transformation, check whether the new form is genuinely simpler and whether the restrictions still justify the operations.

Summary

Effective simplification begins with the domain, then matches visible features of the expression to a useful identity. Quotient and reciprocal identities rewrite function types; Pythagorean identities handle squared sine and cosine terms; cofunction identities handle complementary angles; and angle-combination or sum-to-product formulas can reorganize sums and differences.

The central discipline is to preserve the original restrictions. Every equality in a simplification must hold at the inputs being considered, and cancellation is valid only when the cancelled quantity is nonzero. A final formula should always be read together with the domain of the original expression.

Check Your Understanding

For each question, attend both to the identity choice and to the domain.

  1. What values of \(x\) must be excluded from the domain of \(\frac{\csc x}{\cot x}\)?
  2. Which identity is especially useful for simplifying \(\sin^2 x+\cos^2 x\), and what is the result?
  3. Simplify \(\frac{\sin(\frac{\pi}{2}-x)}{\cos x}\), and state the restriction inherited from the original expression.
  4. Why is cancelling a factor valid only after checking that the factor is nonzero?
  5. If a simplification produces a formula defined at an input where the original expression was undefined, what should you report about the original domain?

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