Trigonometric Identities Made Simple: Mixed Practice, Common Pitfalls, and Mastery Checks

How to Use This Mixed-Practice Chapter

This chapter is a cumulative workout: you will mix reciprocal/quotient, Pythagorean variants, sum–difference, double-angle, half-angle, and algebraic factoring in the same problem sets. The goal is not to remember more identities, but to recognize which family fits the expression’s form and to avoid the most common errors that make correct work collapse at the last line.

A reliable workflow (use it every time)

• Step 1: Scan the form. Look for: (a) mixed sin/cos ratios (suggests convert to tan/sec or common denominator), (b) 1 ± sin or 1 ± cos (suggests multiply by conjugate), (c) squared terms (suggests Pythagorean variants), (d) angle sums like x±y or 2x (suggests sum/difference or double-angle).
• Step 2: Choose a target form. Decide what you want the expression to become (often a single trig function, a constant, or a factored product).
• Step 3: Make one “big move” at a time. Convert, factor, or rationalize; then simplify algebraically.
• Step 4: Guard the domain. Any time you divide by a trig expression, note where it could be zero. If you multiply both sides by an expression, you are implicitly assuming it is nonzero.

Section 1 (Foundation Mix): Simplify by Identity Type

Set 1A: Reciprocal/Quotient + Algebra

Problem 1. Simplify: \frac{\sin x}{1-\cos x}

Step-by-step.

• Recognize 1-\cos x: use the conjugate.
• Multiply by \frac{1+\cos x}{1+\cos x}:

sin x/(1-cos x) * (1+cos x)/(1+cos x) = sin x(1+cos x)/(1-cos^2 x)

• Use 1-\cos^2 x=\sin^2 x:

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

Result: \frac{1+\cos x}{\sin x} (also \csc x+\cot x if you rewrite).

Domain note: original expression requires 1-\cos x\neq 0 (so \cos x\neq 1) and also \sin x can be zero in the simplified form; do not “add” new restrictions incorrectly. Here, when \cos x=1, you also have \sin x=0, so both forms are undefined there. Still, always check.

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Problem 2. Simplify: \frac{\tan x}{\sec x}

tan x/sec x = (sin x/cos x)/(1/cos x) = sin x

Domain note: original requires \cos x\neq 0 (both \tan and \sec undefined). The result \sin x is defined there, but the simplification is only valid where the original is defined.

Mastery Check 1 (Foundation Mix)

• (a) Simplify: \frac{1-\sin x}{\cos x}
• (b) Prove an identity: Show that \frac{\sin x}{1+\cos x}=\csc x-\cot x (state any domain restrictions you used).
• (c) Explain the choice: In one or two sentences, explain why the conjugate method is appropriate for expressions with 1\pm \sin x or 1\pm \cos x.

Section 2 (Intermediate Mix): Pythagorean Variants + Factoring + Rationalization

Set 2A: “Looks like a quadratic” in trig

Problem 1. Simplify: \frac{1-\sin^2 x}{\cos x}

Step-by-step.

• Recognize 1-\sin^2 x as a Pythagorean variant.
• Replace: 1-\sin^2 x=\cos^2 x.

(1-sin^2 x)/cos x = cos^2 x/cos x = cos x

Domain note: you divided by \cos x, so require \cos x\neq 0.

Problem 2. Simplify: \frac{\sec^2 x-1}{\tan x}

• Recognize \sec^2 x-1 as \tan^2 x.

(sec^2 x - 1)/tan x = tan^2 x/tan x = tan x

Domain note: requires \tan x\neq 0 and \cos x\neq 0 (since \sec appears). Don’t forget both.

Set 2B: Rationalizing denominators with trig

Problem 3. Simplify: \frac{\cos x}{1+\sin x}

Step-by-step.

• Use conjugate 1-\sin x:

cos x/(1+sin x) * (1-sin x)/(1-sin x) = cos x(1-sin x)/(1-sin^2 x)

• Use 1-\sin^2 x=\cos^2 x:

= cos x(1-sin x)/cos^2 x = (1-sin x)/cos x

Result: \frac{1-\sin x}{\cos x} (also \sec x-\tan x).

Common Pitfall Spotlight: Incorrect “cancellation” and combining like terms

Trig expressions behave like algebraic expressions, but only when you apply algebra correctly.

• Wrong: \frac{\sin x+\sin x}{\sin x}=\frac{\sin x}{\sin x}+\frac{\sin x}{\sin x} is fine, but concluding \sin x+\sin x=\sin^2 x is not.
• Correct: \sin x+\sin x=2\sin x because they are like terms.
• Wrong: \frac{\sin x+\cos x}{\sin x}=1+1 (you cannot cancel across addition).
• Correct approach: split only when each term is divisible: \frac{\sin x+\cos x}{\sin x}=\frac{\sin x}{\sin x}+\frac{\cos x}{\sin x}=1+\cot x (valid when \sin x\neq 0).

Mastery Check 2 (Intermediate Mix)

• (a) Simplify: \frac{1-\cos^2 x}{\sin x}
• (b) Prove an identity: Prove \frac{1}{1-\sin x}=\frac{1+\sin x}{\cos^2 x} and state the domain restrictions.
• (c) Explain the choice: Explain why Pythagorean variants are the “right tool” when you see 1-\sin^2, 1-\cos^2, or \sec^2-1.

Section 3 (Advanced Mix): Angle Changes (Sum/Difference, Double, Half) in Proofs and Simplification

Set 3A: Expressions that want a double-angle

Problem 1. Simplify: \frac{2\sin x\cos x}{\sin^2 x-\cos^2 x}

Step-by-step.

• Recognize numerator 2\sin x\cos x as \sin 2x.
• Recognize denominator \sin^2 x-\cos^2 x as -\cos 2x (since \cos 2x=\cos^2 x-\sin^2 x).

(2 sin x cos x)/(sin^2 x - cos^2 x) = (sin 2x)/(-cos 2x) = -tan 2x

Common pitfall: sign errors when matching \sin^2-\cos^2 to \cos 2x. Write the identity and compare term order.

Problem 2. Prove: \sin(x+y)\sin(x-y)=\sin^2 x-\sin^2 y

Step-by-step (product-to-sum via expansion).

• Expand each sine using sum/difference formulas:

sin(x+y)=sin x cos y + cos x sin y

sin(x-y)=sin x cos y - cos x sin y

• Multiply as a difference of squares:

sin(x+y)sin(x-y) = (sin x cos y)^2 - (cos x sin y)^2

• Rewrite and group:

= sin^2 x cos^2 y - cos^2 x sin^2 y

• Convert cos^2 terms using cos^2=1-sin^2:

= sin^2 x(1-sin^2 y) - (1-sin^2 x)sin^2 y

= sin^2 x - sin^2 x sin^2 y - sin^2 y + sin^2 x sin^2 y

= sin^2 x - sin^2 y

Common Pitfall Spotlight: Losing domain restrictions (division by zero)

Identity proofs often multiply or divide by expressions like \sin x, \cos x, or 1\pm\sin x. That can silently exclude values where those expressions are zero.

• If you divide by \sin x, you must assume \sin x\neq 0.
• If you multiply both sides by \cos x, you must remember the step is only reversible when \cos x\neq 0.
• Best practice: write a short note like \cos x\neq 0 next to the step where it matters.

Set 3B: Half-angle sign errors and square roots

Half-angle identities can produce square roots. The sign is not “whatever looks nice”; it depends on the quadrant of the half-angle.

Example. Evaluate \sin\left(\frac{5\pi}{6}\right) using a half-angle idea from \sin\left(\frac{\theta}{2}\right) with \theta=\frac{5\pi}{3}.

• Compute \cos\theta: \cos\left(\frac{5\pi}{3}\right)=\frac{1}{2}.
• Use the half-angle magnitude: \sin\left(\frac{\theta}{2}\right)=\pm\sqrt{\frac{1-\cos\theta}{2}}.

sin(5π/6) = ± sqrt((1 - 1/2)/2) = ± sqrt((1/2)/2) = ± sqrt(1/4) = ± 1/2

• Decide the sign: 5\pi/6 is in Quadrant II, where sine is positive.

sin(5π/6) = 1/2

Common pitfall: taking the principal square root automatically. The square root gives a nonnegative number, but the trig value may be negative depending on the angle.

Mastery Check 3 (Advanced Mix)

• (a) Simplify: \frac{\sin(x+y)+\sin(x-y)}{\cos(x+y)+\cos(x-y)}
• (b) Prove an identity: Prove \frac{1-\cos 2x}{\sin 2x}=\tan x and list the values of x excluded by any division steps.
• (c) Explain the choice: Explain why spotting 2\sin x\cos x, \sin^2 x-\cos^2 x, or 1-\cos 2x suggests using double-angle relationships.

Section 4 (Mixed Cumulative Sets): Organized by Difficulty and Identity Type

Level 1: Quick Simplifications (single dominant idea)

Level 2: Two-step Mix (convert + factor or rationalize + substitute)

Level 3: Proof-Style Mix (structure recognition matters)

Section 5: Common Pitfalls (Explicit Fixes You Should Practice)

Pitfall A: Incorrect cancellations across addition

Bad move: \frac{\sin x+\sin x}{\sin x}=\frac{\sin x}{\sin x}=1

Fix: Combine like terms first, or distribute division correctly:

(sin x + sin x)/sin x = (2 sin x)/sin x = 2   (sin x ≠ 0)

Pitfall B: Losing domain restrictions when dividing

Bad move: Starting with \sin x\cos x=0 and dividing by \sin x to get \cos x=0 loses the solutions where \sin x=0.

Fix: Factor and use the zero-product property, or split cases:

• \sin x=0 or \cos x=0.

Pitfall C: Sign errors from square roots in half-angle contexts

Bad move: \sin(\theta/2)=\sqrt{\frac{1-\cos\theta}{2}} (missing the \pm).

Fix: Use \pm and decide sign from the quadrant of \theta/2.

Pitfall D: Misapplied Pythagorean variants

Common confusion: mixing up which identity matches which function.

• 1-\sin^2 x=\cos^2 x and 1-\cos^2 x=\sin^2 x
• 1+\tan^2 x=\sec^2 x and 1+\cot^2 x=\csc^2 x
• Not true: 1+\sin^2 x=\csc^2 x (wrong family)

Section 6: Mastery Checks (Cumulative, Understanding-Based)

Mastery Check 4 (Cumulative Simplify)

• (a) Simplify: \frac{(1-\cos x)(1+\cos x)}{\sin x}
• (b) Prove an identity: Prove \frac{\sec x-\tan x}{\sec x+\tan x}=\frac{1-\sin x}{1+\sin x} (track restrictions).
• (c) Explain the choice: Explain why multiplying by a conjugate is appropriate in part (b), and identify exactly which subexpression is being rationalized.

Mastery Check 5 (Cumulative Proof + Domain)

• (a) Simplify: \frac{\sin 2x}{\cos^2 x}
• (b) Prove an identity: Prove \frac{\tan x}{1+\sec x}=\frac{\sin x}{1+\cos x}.
• (c) Explain the choice: Explain why converting everything to \sin and \cos is a strong strategy for ratios involving \tan and \sec, and list the domain restrictions that come from the original denominators.

Mastery Check 6 (Half-Angle Sign Reasoning)

• (a) Simplify: \frac{1-\cos 2x}{\sin 2x} (give a simplified trig function of x).
• (b) Prove an identity: Prove \sin^2\left(\frac{x}{2}\right)=\frac{1-\cos x}{2} and state when each side is defined.
• (c) Explain the choice: Explain how you would determine the correct sign if you instead solved for \sin(x/2) (not squared) using a square root.