100 Trig Identities: The Ultimate 5-Hour Math Marathon

Overview

blackpenredpen works through 100 trigonometric identity problems in a five-hour, single-take math marathon, moving from basic reciprocal and Pythagorean identities to angle-sum, double-angle, half-angle, inverse-trig, product-to-sum, and power-reduction techniques. The final 10 problems emphasize rewriting expressions into integration-ready forms, with explicit attention to substitution choices, domain restrictions, and the unit-circle reasoning behind identities.

Key takeaways

The Pythagorean identity sin²x + cos²x = 1 is most useful when rearranged: 1 − sin²x = cos²x, 1 − cos²x = sin²x, 1 + tan²x = sec²x, and 1 + cot²x = csc²x.
Unit-circle coordinates explain angle transformations geometrically: adding π negates both coordinates, reflecting π − x negates cosine but preserves sine, and adding π/2 rotates the coordinates.
For inverse-trig compositions, set the inverse function equal to an angle and use a right triangle; this makes expressions such as sin(arccos x) and sec(arctan x) concrete while exposing domain and sign constraints.
The final 10 exercises are organized around substitution: retain sin x when using u = cos x, retain cos x when using u = sin x, retain sec x tan x for u = sec x, and retain csc²x for u = cot x.
Square roots of squared trigonometric expressions require absolute values: √(cos²x) = |cos x| and √((sin x − cos x)²) = |sin x − cos x| unless domain restrictions determine the sign.
Angle-sum identities generate both product-to-sum and sum-to-product formulas; these transformations can collapse multi-angle expressions, as in (sin x + sin 5x + sin 3x)/(cos x + cos 5x + cos 3x) = tan 3x.

Chapters

0:00 The 100-Problem Plan and the Integration-Ready Final Question

blackpenredpen previews 100 multiple-choice identity problems, progressing from easy exercises to harder pre-calculus and calculus material.
The planned coverage includes double-, half-, and triple-angle identities, angle sums and differences, and 10 final forms designed to prepare expressions for integration.
The marathon opens with sin x · sec x = tan x, then jumps to question 100: csc⁴x = (1 + cot²x)csc²x, a form suited to u = cot x.

5:53 Question 2: Factoring sin²x with the Pythagorean Identity

Rewrite sin²x as 1 − cos²x, factor it as (1 − cos x)(1 + cos x), and cancel the shared factor in sin²x/(1 − cos x).
The result is 1 + cos x; the example highlights using a rearranged Pythagorean identity rather than stopping at the original form.

11:46 Questions 4–6: Factoring and Simplifying Reciprocal Fractions

Factor cos x from cos x − sin²x cos x, then use 1 − sin²x = cos²x to obtain cos³x.
For (sec x − tan x)sin x, rewrite secant and tangent using sine and cosine; the numerator becomes cos²x, leaving cos x.
For a complex fraction involving tan x, cot x, and sec x, multiply numerator and denominator by the common denominator factors to reduce it to csc x.

15:22 Questions 7–8: Unit-Circle Symmetry and Adding Tangent with Cotangent

The unit circle shows cos(π − x) = −cos x: reflecting an angle across the vertical axis reverses its x-coordinate.
The same reflection preserves the y-coordinate, giving sin(π − x) = sin x.
Combining tan x and cot x over a common denominator gives (sin²x + cos²x)/(sin x cos x) = sec x + csc x.

20:48 Questions 9–10: Quotient Simplification and Complementary Angles

Simplify (1 − sin²x)/cot²x by replacing 1 − sin²x with cos²x and cot²x with cos²x/sin²x; the result is sin²x.
Use the cofunction identity cot(π/2 − x) = tan x, so 1/cot(π/2 − x) simplifies to tan x.
blackpenredpen explains that “co” refers to complementary angles, reducing reliance on memorized isolated formulas.

25:51 Questions 11–12: Odd Functions and the Three Pythagorean Identities

Since sine is odd, sin(−x) = −sin x; multiplying by csc x yields −1.
The unit circle explains oddness: changing θ to −θ preserves the x-coordinate but reverses the y-coordinate.
Using sin²x + cos²x = 1 and 1 + tan²x = sec²x reduces sin²x + cos²x + tan²x to sec²x.

30:49 Questions 13–14: Even Functions and Squared Odd Functions

Cosine is even, so sec(−x) = sec x.
Although sine is odd, sin²(−x) = sin²x because the negative sign disappears when squared.
The distinction depends on whether the negative input is inside the function and whether the function’s output is subsequently squared.

38:33 Questions 15–16: Inverse Cosine Reflection and Inverse Secant

Let θ = arccos x; for x in [−1, 1], arccos(−x) = π − θ, so arccos x + arccos(−x) = π.
The inverse-function notation cos⁻¹x means arccos x, not 1/cos x; blackpenredpen stresses interpreting inverse trig outputs as angles.
Deriving from sec θ = x gives arcsec x = arccos(1/x), with domain restrictions requiring care, especially at x = 0.

43:39 Question 17: Double-Angle Expansion Produces tan x

Expand sin 2x as 2 sin x cos x and choose cos 2x = 2cos²x − 1 to match the denominator’s cosine terms.
After factoring sin x from the numerator and cos x from the denominator, the quotient reduces to tan x.
The example warns that trigonometric functions are not linear: sin x + sin 2x is not sin 3x.

50:44 Questions 18–20: Double-Angle Factoring and sec x − tan x

Factor cos 2x = cos²x − sin²x as (cos x − sin x)(cos x + sin x); the denominator cancels one factor.
Rewrite (1 − tan²x)/(1 + tan²x) in sine and cosine form to identify the double-angle identity cos 2x.
Separate (1 − sin x)/cos x into sec x − tan x.

55:13 Question 21: Adding Conjugate Sine Denominators

Add 1/(1 − sin x) and 1/(1 + sin x) using a common denominator.
The denominator becomes 1 − sin²x = cos²x, while the numerator becomes 2.
The result is 2sec²x, illustrating how conjugate factors reveal a Pythagorean identity.

1:02:19 Questions 24–26: Complex Fractions and Factoring Shared Terms

Rewrite (sec x − cos x)/tan x using sine and cosine; simplifying the numerator gives sin²x, and division by tan x leaves sin x.
Distribute csc x across (csc x − sin x) to obtain csc²x − 1 = cot²x.
Adding sin x/(1 + cos x) and (1 + cos x)/sin x produces 2csc x after combining terms and factoring 1 + cos x.

1:07:13 Question 27: Difference of Squares Gives −1

Use (tan x + sec x)(tan x − sec x) = tan²x − sec²x.
Since sec²x = 1 + tan²x, the expression equals −1.
The solution demonstrates both direct use of the Pythagorean identity and conversion to sine and cosine.

1:10:15 Question 28: Deriving the Cotangent Addition Formula

Start with cot(x + y) = cos(x + y)/sin(x + y), then expand both functions with angle-sum identities.
Divide through by sin x sin y to express the result using cotangents.
The derived identity is cot(x + y) = (cot x cot y − 1)/(cot x + cot y), with the denominator’s plus sign arising from the sine addition formula.

1:16:33 Question 30: Why cos(π/2 + x) = −sin x

The angle-sum formula gives cos(π/2 + x) = cos(π/2)cos x − sin(π/2)sin x = −sin x.
On the unit circle, rotating a point by 90° maps its coordinates so the new x-coordinate is the negative of the original y-coordinate.
The coordinate argument also establishes sin(π/2 + x) = cos x and connects the rotation to perpendicular-line slopes.

1:22:59 Questions 31–34: Products of Shifted Angles and π-Shift Identities

Expand cos(x + y)cos(x − y) and simplify to cos²x − sin²y.
Expanding sin(x + y)sin(x − y) similarly gives sin²x − sin²y.
Unit-circle shifts show that adding π negates both sine and cosine, while tan x · tan(π/2 − x) = tan x · cot x = 1.

1:27:19 Questions 35–36: Right Triangles for Inverse-Trig Compositions

For x ≥ 1, set θ = arcsec x; a triangle with hypotenuse x and adjacent side 1 gives sin(arcsec x) = √(x² − 1)/x.
For x in [−1, 1], set θ = arccos x; adjacent side x and hypotenuse 1 give tan(arccos x) = √(1 − x²)/x.
The examples distinguish the radicands x² − 1 and 1 − x², which come from different triangle side assignments.

1:37:01 Questions 37–38: Evaluating Secant and Cosine of Inverse Tangent

A triangle for tan θ = x gives sec²(arctan x) = 1 + x²; the Pythagorean identity provides a shorter derivation.
For tan θ = x/3, use opposite side x and adjacent side 3, making the hypotenuse √(x² + 9).
Thus cos(arctan(x/3)) = 3/√(x² + 9), with the triangle method tracking which side belongs to each ratio.

1:41:36 Questions 39–40: Double-Angle Fractions and csc x sec x

Simplifying 2tan x/(1 + tan²x) in sine and cosine form yields sin 2x.
Rewrite csc x sec x as 1/(sin x cos x), then use sin 2x = 2sin x cos x.
The product becomes 2csc 2x, a rewrite that also prepares the expression for later integration techniques.

1:48:39 Questions 41–43: Angle Addition and Double-Angle Tangent

Expand sin(x + π/3) using sin π/3 = √3/2 and cos π/3 = 1/2, obtaining (1/2)sin x + (√3/2)cos x.
Simplify tan²x/(csc x sec³x) by converting to sine and cosine and canceling powers, leaving sin³x cos x.
Recognize tan x/(1 − tan²x) as part of tan 2x; the full double-angle identity is tan 2x = 2tan x/(1 − tan²x).

1:54:51 Questions 44–45: Double-Angle Identities with Inverse Trig Inputs

For tan(2 arctan x), apply the tangent double-angle formula before canceling tan(arctan x); the result is 2x/(1 − x²).
For cos(2 arcsin x), choose cos 2θ = 1 − 2sin²θ so the inverse sine composition cancels directly.
The second expression simplifies to 1 − 2x² without constructing a triangle.

1:59:15 Question 46: Complementary Angles Turn Two Sine Squares into 1

Rewrite x + π/4 as π/2 + (x − π/4), then use sin(π/2 + θ) = cos θ.
The expression becomes cos²(x − π/4) + sin²(x − π/4), which equals 1.
Recognizing complementary-angle structure avoids expanding and squaring two separate angle-sum expressions.

2:10:18 Question 47: Secant of an Arctangent of sin x

Set θ = arctan(sin x), so tan θ = sin x = sin x/1.
A right triangle has opposite side sin x, adjacent side 1, and hypotenuse √(1 + sin²x).
Therefore sec(arctan(sin x)) = √(1 + sin²x); the square root does not cancel because its radicand is a sum.

2:16:08 Questions 48–49: Combining Sine and Cosine into One Sine

For sin x + cos x, factor out √2 and identify the coefficients 1/√2 as sin(π/4) and cos(π/4).
The angle-addition formula gives sin x + cos x = √2 sin(x + π/4).
Likewise, √3 sin x − cos x = 2sin(x − π/6), using the 30°–60°–90° triangle.

2:19:05 Questions 50–51: A Unit-Circle View of a Half-Angle Tangent

The ratio (sin x + sin y)/(cos x + cos y) can be interpreted by adding the two unit-circle coordinate vectors.
The resulting vector points at the average angle, giving tan((x + y)/2), when the denominator is nonzero.
For tan 3x − tan 2x − tan x, expand tan(x + 2x) and rearrange to obtain tan x · tan 2x · tan 3x.

2:28:41 Questions 52–53: Factoring Powers and Shifting Cosine by π

Factor cos x − cos³x as cos x(1 − cos²x); since 1 − cos²x = sin²x, division by sin²x leaves cos x.
The angle-addition formula and the unit-circle point at π establish cos(x + π) = −cos x.
The same π shift also gives sin(x + π) = −sin x.

2:35:48 Questions 55–57: Triple-Angle Cosine and Product-to-Sum Formulas

The cosine triple-angle identity is cos 3x = 4cos³x − 3cos x, paralleling sin 3x = 3sin x − 4sin³x.
Subtracting the cosine angle-sum and angle-difference formulas yields sin x sin y = ½[cos(x − y) − cos(x + y)].
Adding those cosine formulas gives cos x cos y = ½[cos(x − y) + cos(x + y)].

2:41:01 Questions 58–60: Sum-to-Product and a Cosine Double-Angle Rearrangement

Adding the sine sum and difference formulas produces sin x + sin y = 2sin((x + y)/2)cos((x − y)/2).
The corresponding cosine identity is cos x + cos y = 2cos((x + y)/2)cos((x − y)/2).
Use cos 2x = 2cos²x − 1 to reduce 2cos²x − cos 2x to 1.

2:48:01 Questions 61–63: Cosine of 4x and High-Power Pythagorean Factoring

Apply cos 2θ = 2cos²θ − 1 twice to derive cos 4x = 8cos⁴x − 8cos²x + 1.
Factor sin⁶x + cos⁶x as (sin²x + cos²x)(sin⁴x − sin²x cos²x + cos⁴x).
Using sin²x + cos²x = 1 and completing a square further rewrites the sixth-power sum as 1 − (3/4)sin²2x.

2:51:47 Question 64: Pairing x and 5x in a Three-Term Trig Quotient

Apply sum-to-product to sin x + sin 5x and cos x + cos 5x, pairing the outer angles because their averages are 3x.
The numerator becomes 2sin 3x cos 2x + sin 3x; the denominator becomes 2cos 3x cos 2x + cos 3x.
Factoring sin 3x and cos 3x reveals that the quotient simplifies to tan 3x.

3:02:51 Question 65: Sum-to-Product Reduces a 3x-versus-x Quotient

Use sin A − sin B = 2cos((A + B)/2)sin((A − B)/2) for the numerator sin 3x − sin x.
Use the cosine sum-to-product identity for cos 3x + cos x.
Shared factors cancel, leaving tan x.

3:12:10 Questions 66–67: Half-Angle Tangent and Difference of Fourth Powers

Rewrite (1 − cos x)/(1 + cos x) using x = 2(x/2) and double-angle identities to obtain tan²(x/2).
Factor cos⁴x − sin⁴x as (cos²x − sin²x)(cos²x + sin²x).
The second factor is 1 and the first is cos 2x, so the difference of fourth powers is cos 2x.

3:16:16 Questions 68–69: Absolute Values and Combining Sine–Cosine Terms

Recognize 1 − sin 2x as (sin x − cos x)²; its square root is |sin x − cos x|, not automatically the signed difference.
Normalize cos x − sin x by √2 and use the cosine difference identity to get √2 cos(x + π/4).
The absolute-value step matters because sin x − cos x can be negative for some x.

3:21:09 Questions 70–71: Simplifying Powers and Preserving Square-Root Signs

Factor 4 inside the cube in 2sec²x/[4(1 + tan²x)]³ and use 1 + tan²x = sec²x.
After cancellation, the expression reduces to cos⁴x/32.
For √(4 − 4sin²x), factor to 2√(cos²x) = 2|cos x| unless the domain guarantees cos x is nonnegative.

3:25:20 Questions 72–73: Double Angles of Inverse Cosine and Inverse Tangent

For sin(2 arccos x), draw a triangle with adjacent side x and hypotenuse 1; the opposite side is √(1 − x²).
The result is 2x√(1 − x²), with the nonnegative sine sign justified by arccos x ∈ [0, π].
For cos(2 arctan x), a triangle with opposite x and adjacent 1 gives (1 − x²)/(1 + x²).

3:33:21 Questions 74–75: A Reciprocal Pythagorean Identity and Tangent Addition

Combine sec²x + csc²x over a common denominator; the numerator becomes sin²x + cos²x = 1.
Thus sec²x + csc²x = sec²x · csc²x, provided the functions are defined.
Combining tan x + tan y over cos x cos y turns the numerator into sin(x + y), giving sin(x + y)/(cos x cos y).

3:37:20 Questions 76–77: Inverse-Tangent Addition and Cosine of Arcsine Sums

Set α = arctan x and β = arctan y, then apply the tangent addition formula to derive arctan x + arctan y = arctan((x + y)/(1 − xy)) under the stated xy < 1 condition.
For cos(arcsin x + arcsin y), expand cosine of a sum and use cos(arcsin x) = √(1 − x²).
The result is √(1 − x²)√(1 − y²) − xy, with x and y restricted to [−1, 1].

3:43:20 Questions 78–79: Power Reduction and a Half-Angle Quotient

Rewrite cos²x sin²x as ¼sin²2x, then use sin²θ = ½(1 − cos 2θ) to obtain ⅛ − ¼cos 4x.
The power-reduction identity is motivated by calculus: first-power sine and cosine forms are often more useful for integration.
For (1 − cos x + sin x)/(1 + cos x + sin x), rewrite x as 2(x/2) and apply double-angle identities to simplify to tan(x/2).

3:54:43 Question 80: Reducing sin⁴x to First-Power Cosines

Write sin⁴x as [½(1 − cos 2x)]², then reduce cos²2x with cos²θ = ½(1 + cos 2θ).
The resulting first-power expression is 3/8 − 1/2cos 2x + 1/8cos 4x.
This power-reduction sequence illustrates how a single fourth power becomes a sum of cosine terms with different angles.

4:01:25 Questions 81–82: Factoring Sums of Sine and Cosine

Rewrite 1 + sin 2x as (sin x + cos x)² and cos 2x as (cos x − sin x)(cos x + sin x).
Cancel the common factor to simplify (1 + sin 2x)/cos 2x to (sin x + cos x)/(cos x − sin x).
Expanding (sin x + cos x)² + (sin x − cos x)² cancels the cross terms and yields 2.

4:04:34 Questions 83–84: Cotangent at 2x and a Reciprocal-Sum Quotient

Rewrite csc 2x + cot 2x using sine and cosine of 2x; applying double-angle identities simplifies the expression to cot x.
For (sin x + cos x)/(csc x + sec x), multiply numerator and denominator by sin x cos x.
The denominator becomes sin x + cos x, which cancels the numerator’s shared factor and leaves sin x cos x.

4:10:37 Questions 85–87: Product-to-Sum and Fourth-Power Reduction

Sum-to-product gives sin 5x + sin 3x = 2sin 4x cos x.
Product-to-sum gives cos 2x cos 5x = ½[cos 7x + cos 3x], using cosine’s evenness for the negative difference angle.
Complete the square in sin⁴x + cos⁴x to obtain 1 − 2sin²x cos²x = 1 − ½sin²2x.

4:16:08 Questions 88–89: Tangent Shifts by π/4 and π/2

The tangent addition formula with tan(π/4) = 1 yields tan(x + π/4) = (tan x + 1)/(1 − tan x).
Do not substitute tan(π/2) into the addition formula: tan(π/2) is undefined because cos(π/2) = 0.
Using the cofunction identity instead gives tan(x + π/2) = −cot x wherever the expressions are defined.

4:24:29 Question 90: Sine of a Sum of Inverse Sine and Cosine Angles

Expand sin(arcsin x + arccos y) using the sine addition formula.
Substitute sin(arcsin x) = x, cos(arcsin x) = √(1 − x²), sin(arccos y) = √(1 − y²), and cos(arccos y) = y.
The result is x y + √(1 − x²)√(1 − y²), for x and y in [−1, 1].

4:27:50 Question 91: Preparing sin⁴x for u = cos x

The target form is a polynomial in cos x multiplied by sin x, so that u = cos x enables substitution.
Write sin⁴x sin x as (1 − cos²x)²sin x and expand to (1 − 2cos²x + cos⁴x)sin x.
This is the first of the final integration-ready exercises, emphasizing the required differential factor rather than merely simplifying the trig expression.

4:30:31 Questions 92–94: Matching Trig Factors to Substitution Differentials

Rewrite tan³x sec x as (sec²x − 1)sec x tan x, using tan²x = sec²x − 1; the outside sec x tan x matches the derivative of sec x.
For tan³x/sec x, retain sin x outside and express the remaining factor as (1 − cos²x)/cos²x.
For sin³x cos³x, reserve one cos x and rewrite cos²x as 1 − sin²x, producing (sin³x − sin⁵x)cos x for u = sin x.

4:35:27 Questions 95–96: Rewriting Cotangent and Cosecant Powers for Integration

Convert cos²x/cot³x to a form with sin x outside and a cosine-only expression inside: [(1 − cos²x)/cos x]sin x.
Rewrite csc⁶x as csc²x(csc²x)² and use csc²x = 1 + cot²x.
The final form is (cot⁴x + 2cot²x + 1)csc²x, suited to u = cot x.

4:40:08 Question 97: Rewriting cot⁵x to Isolate cos x

Express cot⁵x as cos⁵x/sin⁵x and reserve one cos x as the outside factor.
Replace the remaining cos⁴x with (cos²x)² = (1 − sin²x)².
The resulting sine-polynomial factor multiplied by cos x is suited to u = sin x.

4:46:42 Questions 98–99: Producing Secant-Squared and Cosine Factors

For 1/(1 + sin²x), divide numerator and denominator by cos²x to produce sec²x/[sec²x + tan²x].
Use sec²x = 1 + tan²x to rewrite the denominator as 1 + 2tan²x, leaving an expression in tan x multiplied by sec²x.
Rewrite sec x as (1/cos²x)cos x = [1/(1 − sin²x)]cos x, exposing the cosine factor for u = sin x.

4:48:23 Questions 100–101: The Final Cotangent Form and a Three-Angle Sine Sum

Question 100 factors csc⁴x as csc²x · csc²x and rewrites one factor with csc²x = 1 + cot²x, giving (1 + cot²x)csc²x for u = cot x.
For x + y + z = π, blackpenredpen combines sin x + sin y with sum-to-product and uses z = π − x − y to reorganize the remaining sine term.
The final identity is sin x + sin y + sin z = 4cos(x/2)cos(y/2)cos(z/2); the marathon closes by encouraging repeated practice of all 100 problems.

Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, blackpenredpen.