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Lesson 3 of 5

Trigonometric Functions · Lesson 3 of 5

Angle Combinations and Trigonometric Identities

“Derive a connected family of identities and use their structure to evaluate exact values and prove relationships.”

Learning Objectives

• Explain the geometric and algebraic reasoning behind the sine and cosine addition identities. • Derive difference, complementary-angle and related-angle identities from the addition formulas. • Obtain tangent and cotangent formulas while retaining the conditions needed for division. • Derive double-angle and triple-angle identities and select a useful equivalent form. • Convert sums into products and products into sums using the half-sum and half-difference of angles. • Construct step-by-step identity proofs and check that cancellations are valid.

Why combining angles needs a new rule

Knowing sin x and sin y does not mean that sin(x + y) equals their sum. The input to sine is an angle, so adding angles changes the terminal point in a geometric way. We need relationships that describe this change rather than treating a trigonometric function like ordinary multiplication.

For instance, sin30° + sin30° = 1, but sin60° = √3/2. The mismatch shows why “distributing” sine over a sum is incorrect. The identities in this lesson will express functions of combined angles in terms of functions of the original angles. Their derivations also show which formulas are always valid and which require nonzero denominators.

Definition
Trigonometric identity

A trigonometric identity is an equality that holds at every input where the displayed expressions are defined. It differs from an equation that is true only for selected input values.

A geometric proof of the cosine addition identity

The starting identity can be proved using distances between points on a unit circle. We will compare the lengths of two chords that a rotation carries onto one another. The geometry establishes equality of lengths, and the coordinate distance formula converts that equality into an algebraic relationship.

Choose P₁ = (cos x, sin x), P₂ = (cos(x + y), sin(x + y)), P₃ = (cos(−y), sin(−y)), and P₄ = (1, 0). Rotating the whole circle anticlockwise through y carries P₃ to P₄ and P₁ to P₂. A rotation preserves lengths, so chord P₁P₃ and chord P₂P₄ are equal. Equivalently, their triangles with O have unit radii and equal included angle, so they are congruent.

OP₁: angle xP₂: angle x + yP₃: angle −yP₄: angle 0Rotation by y maps P₁P₃ onto P₂P₄, preserving chord length.
Equal chords for the cosine addition proof— Use the coordinates specified in the text. The two coloured chords have the same length because one is a rotation of the other.

For two points (a, b) and (c, d), the squared distance is (a − c)² + (b − d)². We use squared lengths because they avoid unnecessary square roots. Reflection gives cos(−y) = cos y and sin(−y) = −sin y, so the coordinate differences for P₁P₃ are cos x − cos y and sin x + sin y.

Example — Comparing the two squared distances

Problem
Use the equal chords to derive a formula for cos(x + y).

  1. 1.P₁P₃² = (cos x − cos y)² + (sin x + sin y)². Expanding gives cos²x + cos²y − 2cos x cos y + sin²x + sin²y + 2sin x sin y.
  2. 2.Group cos²x + sin²x and cos²y + sin²y. Each equals 1, so P₁P₃² = 2 − 2cos x cos y + 2sin x sin y = 2 − 2(cos x cos y − sin x sin y).
  3. 3.For P₂P₄, the squared distance is (cos(x + y) − 1)² + sin²(x + y). Expanding and grouping the squared sine and cosine gives P₂P₄² = 2 − 2cos(x + y).
  4. 4.Equal squared lengths give 2 − 2(cos x cos y − sin x sin y) = 2 − 2cos(x + y). Subtract 2 from both sides and divide by −2.
  5. 5.Therefore cos(x + y) = cos x cos y − sin x sin y. The proof used only circle coordinates, reflection, geometry and the fundamental identity. Rotations work for arbitrary real angles, not just the positive angles drawn in one illustration.
Cosine of a sumLaTeX
This identity is valid for all real x and y.

Difference and complementary angles

Once the sum identity is known, a difference can be treated as a sum with a negative angle. Reflection then supplies the sign change. This small substitution avoids learning the difference formula as an unrelated fact.

Replace y by −y: cos(x − y) = cos x cos(−y) − sin x sin(−y). Cosine keeps its value under reflection, while sine reverses sign. Thus the subtraction between the products becomes addition.

Cosine of a differenceLaTeX

Now substitute x = π/2 into that difference identity, keeping the other angle as a variable t. Since cos(π/2) = 0 and sin(π/2) = 1, cos(π/2 − t) = sin t. Replace t by π/2 − x to obtain sin(π/2 − x) = cos x as well. These relationships exchange horizontal and vertical roles.

Complementary-angle identitiesLaTeX

Deriving the sine formulas

The complementary-angle relationship turns a sine into a cosine whose input is shifted by a quarter turn. Applying the cosine difference identity then produces the sine sum formula. The pattern of products is similar, but its signs differ.

Example — Sine of a sum and difference

Problem
Derive the sine angle-combination identities from the cosine results.

  1. 1.Start with sin(x + y) = cos(π/2 − (x + y)) = cos((π/2 − x) − y).
  2. 2.Apply the cosine difference formula: cos(π/2 − x)cos y + sin(π/2 − x)sin y.
  3. 3.Use the complementary identities to replace these by sin x cos y + cos x sin y. Thus sin(x + y) = sin x cos y + cos x sin y.
  4. 4.Replace y by −y. The term cos(−y) remains cos y and sin(−y) becomes −sin y, giving sin(x − y) = sin x cos y − cos x sin y.
Sine of a sum and differenceLaTeX

In the sine formulas the products are mixed: one sine and one cosine in each term. In the cosine formulas the products are like with like: cosine with cosine and sine with sine. For cosine the sign reverses relative to the angle sign; for sine it follows the angle sign. These patterns are useful checks, but the derivations explain why they hold.

Example — An exact value at 15 degrees

Problem
Find sin15° without a decimal approximation.

  1. 1.Write 15° = 45° − 30°. Apply the sine difference identity.
  2. 2.sin15° = sin45°cos30° − cos45°sin30° = (1/√2)(√3/2) − (1/√2)(1/2).
  3. 3.Combining over the common denominator gives (√3 − 1)/(2√2), equivalently (√6 − √2)/4.
  4. 4.It is positive, as expected in quadrant I. The value is not sin45° − sin30°, because the subtraction acts on the angle before evaluation.

Related-angle formulas from the same identities

Angles involving a half turn, a full turn or a quarter turn occur repeatedly. Rather than applying a memorised sign rule without explanation, substitute the known axis values into the addition and difference formulas. The unit-circle quadrant provides an independent check.

InputSine valueCosine value
−x−sin xcos x
π/2 − xcos xsin x
π/2 + xcos x−sin x
π − xsin x−cos x
π + x−sin x−cos x
2π − x−sin xcos x
2π + xsin xcos x
3π/2 + x−cos xsin x
3π/2 − x−cos x−sin x

For example, sin(π + x) = sinπ cos x + cosπ sin x = 0 − sin x. Cos(π/2 + x) = cos(π/2)cos x − sin(π/2)sin x = −sin x. Tangent and cotangent relationships follow by taking the corresponding sine–cosine quotient, while secant and cosecant follow by taking reciprocals wherever defined.

Example — Evaluating a mixed standard-angle expression

Problem
Show that 3sin(π/6)sec(π/3) − 4sin(5π/6)cot(π/4) equals 1.

  1. 1.The standard values are sin(π/6) = 1/2, sec(π/3) = 2 and cot(π/4) = 1.
  2. 2.Since 5π/6 = π − π/6, the related-angle identity gives sin(5π/6) = sin(π/6) = 1/2.
  3. 3.The expression becomes 3(1/2)(2) − 4(1/2)(1) = 3 − 2 = 1.
  4. 4.All reciprocal values used have nonzero denominators at the stated angles, so the evaluation is valid.
A shifted angle may change the function name

A shift by π/2 exchanges sine and cosine, often with a sign change. A shift by π changes signs without exchanging them. Derive unfamiliar shifts from the addition formulas rather than applying one sign rule to every shift.

Tangent and cotangent of combined angles

Tangent is sine divided by cosine, so its addition formula comes from dividing the corresponding sine and cosine formulas. Cotangent follows by reversing that quotient. This division introduces conditions that did not occur in the sine and cosine identities.

Example — Deriving tangent addition

Problem
Derive tan(x + y) in terms of tan x and tan y.

  1. 1.Assume cos x, cos y and cos(x + y) are nonzero. Then tan(x + y) = [sin x cos y + cos x sin y]/[cos x cos y − sin x sin y].
  2. 2.Divide every term in numerator and denominator by the nonzero product cos x cos y.
  3. 3.The numerator becomes tan x + tan y; the denominator becomes 1 − tan x tan y.
  4. 4.Therefore tan(x + y) = (tan x + tan y)/(1 − tan x tan y). The denominator must remain nonzero because the combined-angle tangent must exist.
  5. 5.Replacing y by −y gives tan(x − y) = (tan x − tan y)/(1 + tan x tan y), with analogous restrictions.
Tangent sum and differenceLaTeX
Require tan x and tan y to exist, and the relevant denominator to be nonzero. Equivalently, the tangent at the combined angle must also exist.
Example — Deriving cotangent addition

Problem
Derive the cotangent sum formula and obtain the difference formula.

  1. 1.Assume sin x, sin y and sin(x + y) are nonzero. Form cos(x + y)/sin(x + y) using the cosine and sine sum identities.
  2. 2.This quotient is [cos x cos y − sin x sin y]/[sin x cos y + cos x sin y]. Divide all terms by sin x sin y.
  3. 3.The numerator becomes cot x cot y − 1 and the denominator becomes cot y + cot x.
  4. 4.Thus cot(x + y) = (cot x cot y − 1)/(cot y + cot x).
  5. 5.For the difference substitute −y: cot(−y) = −cot y. Multiplying numerator and denominator of the resulting quotient by −1 gives cot(x − y) = (cot x cot y + 1)/(cot y − cot x).
Cotangent sum and differenceLaTeX
Require cot x, cot y and the relevant combined-angle cotangent to exist, so every displayed denominator is nonzero.
Example — A periodic angle and an exact tangent

Problem
Find tan(13π/12).

  1. 1.13π/12 = π + π/12. Tangent has period π, so the value equals tan(π/12).
  2. 2.Write π/12 = π/4 − π/6. The tangent difference formula gives (1 − 1/√3)/(1 + 1/√3) = (√3 − 1)/(√3 + 1).
  3. 3.Multiply numerator and denominator by √3 − 1: (√3 − 1)²/(3 − 1) = (4 − 2√3)/2 = 2 − √3.
  4. 4.The positive result matches the quadrant-I angle π/12. The relevant denominator is nonzero, so the formula is applicable.
Example — A quotient of two sines

Problem
Prove sin(x + y)/sin(x − y) = (tan x + tan y)/(tan x − tan y) where all displayed expressions are defined.

  1. 1.Assume sin(x − y) ≠ 0 and cos x cos y ≠ 0. Expand both sine expressions using their sum and difference identities.
  2. 2.The quotient becomes [sin x cos y + cos x sin y]/[sin x cos y − cos x sin y].
  3. 3.Divide all four terms by cos x cos y. This produces (tan x + tan y)/(tan x − tan y).
  4. 4.The condition sin(x − y) ≠ 0 ensures the original denominator is nonzero. With cos x cos y ≠ 0, it also ensures tan x − tan y ≠ 0, because sin(x − y) = cos x cos y(tan x − tan y).

Double angles as sums of equal inputs

A double angle is just x + x. Substituting equal angles into the addition formulas gives identities for sine, cosine and tangent of 2x. The fundamental identity lets us rewrite the cosine result in several useful forms, each suited to different information.

Example — Deriving the double-angle forms

Problem
Obtain the sine and cosine formulas for 2x, including versions involving tan x.

  1. 1.Put y = x in the sine sum identity: sin2x = sin x cos x + cos x sin x = 2sin x cos x.
  2. 2.Put y = x in the cosine sum identity: cos2x = cos²x − sin²x. Replacing sin²x by 1 − cos²x gives cos2x = 2cos²x − 1. Replacing cos²x by 1 − sin²x gives cos2x = 1 − 2sin²x.
  3. 3.Since cos²x + sin²x = 1, write cos2x = (cos²x − sin²x)/(cos²x + sin²x). If cos x ≠ 0, divide both numerator and denominator by cos²x to get (1 − tan²x)/(1 + tan²x).
  4. 4.Likewise sin2x = 2sin x cos x/(cos²x + sin²x). Dividing by cos²x when permitted gives 2tan x/(1 + tan²x).
  5. 5.The sine–cosine forms hold for all real x. The tangent-based forms additionally require cos x ≠ 0; the denominator 1 + tan²x is then always positive.
Double-angle sine and cosineLaTeX
Double-angle formulas using tangentLaTeX
These two forms require cos x ≠ 0.

For tangent, substitute y = x into its sum formula. The numerator becomes 2tan x and the denominator becomes 1 − tan²x. Both tan x and tan2x must exist: cos x ≠ 0, and the denominator must not vanish.

Double-angle tangentLaTeX
Require cos x ≠ 0 and cos2x ≠ 0. At x = π/4 the denominator vanishes and tan2x is undefined. At x = π/2 the left side exists but the tangent-based right side does not.
Example — Choosing a double-angle form

Problem
If sin x = 3/5 and x is in quadrant I, find sin2x and cos2x.

  1. 1.The identity gives cos²x = 1 − 9/25 = 16/25. Quadrant I gives cos x = 4/5.
  2. 2.sin2x = 2(3/5)(4/5) = 24/25.
  3. 3.For cosine, choose the form that uses the given sine: cos2x = 1 − 2(9/25) = 7/25.
  4. 4.The pair (cos2x, sin2x) is (7/25, 24/25), and its squared coordinates sum to (49 + 576)/625 = 1. The check is consistent with the unit circle.
Distinguish a doubled angle from a squared value

sin2x means sin(2x); sin²x means (sin x)². They are usually different. In particular, sin2x is not 2sin x. Keep parentheses or superscripts visible when writing intermediate steps.

Triple angles from a double angle and one more angle

To triple an angle, combine 2x and x. This makes the triple-angle formulas consequences of the addition and double-angle identities. The derivation shows why cubic powers appear rather than leaving the formulas as unexplained patterns.

Example — Deriving triple-angle sine

Problem
Show that sin3x = 3sin x − 4sin³x.

  1. 1.Start with sin3x = sin(2x + x) = sin2x cos x + cos2x sin x.
  2. 2.Use sin2x = 2sin x cos x and cos2x = 1 − 2sin²x. This gives 2sin x cos²x + sin x − 2sin³x.
  3. 3.Replace cos²x by 1 − sin²x: 2sin x(1 − sin²x) + sin x − 2sin³x.
  4. 4.Expand to get 2sin x − 2sin³x + sin x − 2sin³x = 3sin x − 4sin³x. No divisions occurred, so the result holds for every real x.
Example — Deriving triple-angle cosine

Problem
Show that cos3x = 4cos³x − 3cos x.

  1. 1.Start with cos3x = cos(2x + x) = cos2x cos x − sin2x sin x.
  2. 2.Use cos2x = 2cos²x − 1 and sin2x = 2sin x cos x. This gives (2cos²x − 1)cos x − 2sin²x cos x.
  3. 3.Replace sin²x by 1 − cos²x: 2cos³x − cos x − 2cos x(1 − cos²x).
  4. 4.Expanding and collecting gives 2cos³x − cos x − 2cos x + 2cos³x = 4cos³x − 3cos x. This is also valid for every real x.
Triple-angle sine and cosineLaTeX

For triple-angle tangent, divide sin3x by cos3x where the denominator is nonzero. To express the result entirely in tan x, assume cos x ≠ 0 and rewrite the sine numerator as 3sin x cos²x − sin³x. This is equal to 3sin x − 4sin³x because cos²x = 1 − sin²x. Rewrite the cosine denominator as cos³x − 3cos x sin²x for the same reason. Dividing both by cos³x gives the next formula.

Triple-angle tangentLaTeX
Require cos x ≠ 0 and cos3x ≠ 0. With tan x defined, the latter means 1 − 3tan²x ≠ 0.

Another route begins with tan(2x + x), as in the source development. That intermediate route also requires tan2x to exist during the calculation. The sine–cosine derivation above proves the final formula without adding that unnecessary intermediate restriction; for example, x = π/4 is allowed in the final formula even though tan2x is undefined.

Example — An identity linking three tangent values

Problem
Prove tan3x tan2x tan x = tan3x − tan2x − tan x where all three tangent values are defined.

  1. 1.Since 3x = 2x + x, the addition formula gives tan3x = (tan2x + tan x)/(1 − tan2x tan x).
  2. 2.Multiply by the nonzero denominator: tan3x − tan3x tan2x tan x = tan2x + tan x.
  3. 3.Move the product term to the right and the two single tangent terms to the left. This gives tan3x − tan2x − tan x = tan3x tan2x tan x.
  4. 4.Reversing the equality gives the required statement. The multiplication was legitimate because the tangent addition denominator is nonzero on the stated common domain.

Turning sums into products

Sometimes a sum of two trigonometric terms is difficult to simplify directly, while a product exposes common factors. Pairing the addition and difference formulas lets us move between these forms. The relevant angles become the half-sum and half-difference of the original inputs.

Example — Deriving all four sum-to-product identities

Problem
Derive formulas for cos A ± cos B and sin A ± sin B.

  1. 1.Add cos(u + v) = cos u cos v − sin u sin v and cos(u − v) = cos u cos v + sin u sin v. The sine products cancel, giving cos(u + v) + cos(u − v) = 2cos u cos v.
  2. 2.Subtract the second cosine equation from the first. The cosine products cancel, giving cos(u + v) − cos(u − v) = −2sin u sin v.
  3. 3.Add the two sine equations to get sin(u + v) + sin(u − v) = 2sin u cos v. Subtract them to get sin(u + v) − sin(u − v) = 2cos u sin v.
  4. 4.Set A = u + v and B = u − v. Adding these gives u = (A + B)/2; subtracting gives v = (A − B)/2.
  5. 5.Substitute these expressions for u and v in all four results. A and B may be any real angles, so the resulting identities hold for all real inputs.
Cosine sum and difference as productsLaTeX
Sine sum and difference as productsLaTeX

The half-sum locates the centre of the two angles, and the half-difference measures their separation. For angles 7x and 5x these become 6x and x. Compute both before substituting; the order of a difference matters because sine reverses sign when its input is negated.

Example — A symmetric cosine sum

Problem
Prove cos(π/4 + x) + cos(π/4 − x) = √2 cos x.

  1. 1.Let A = π/4 + x and B = π/4 − x. Their half-sum is π/4 and their half-difference is x.
  2. 2.Apply cosine sum-to-product: the left side becomes 2cos(π/4)cos x.
  3. 3.Since cos(π/4) = 1/√2, this is (2/√2)cos x = √2 cos x. No cancellation or division by a variable expression was required.
Example — Matching half-sums in a quotient

Problem
Prove (cos7x + cos5x)/(sin7x − sin5x) = cot x on the original quotient’s domain.

  1. 1.The numerator becomes 2cos6x cos x, using half-sum 6x and half-difference x.
  2. 2.The denominator becomes 2cos6x sin x, using the sine difference identity.
  3. 3.The original denominator is nonzero, so both cos6x and sin x are nonzero. Cancel the common factor 2cos6x to obtain cos x/sin x = cot x.
  4. 4.The equality holds on the original domain. Cot x also exists at some inputs excluded by the original quotient, but the simplification does not restore those inputs.
Example — Grouping terms before transforming

Problem
Prove (sin5x − 2sin3x + sin x)/(cos5x − cos x) = tan x where the original denominator is nonzero.

  1. 1.Group sin5x + sin x together. Their sum is 2sin3x cos2x, so the numerator becomes 2sin3x(cos2x − 1).
  2. 2.The denominator cos5x − cos x becomes −2sin3x sin2x. Its nonzero value requires both sin3x and sin2x to be nonzero.
  3. 3.Cancel 2sin3x to get (1 − cos2x)/sin2x. Use 1 − cos2x = 2sin²x and sin2x = 2sin x cos x.
  4. 4.Because sin2x is nonzero, both sin x and cos x are nonzero. Cancel 2sin x to get sin x/cos x = tan x.
  5. 5.The grouping exposed a common angle 3x. Beginning with that structure is more effective than expanding every term independently.

Turning products into sums

The paired addition equations can also be read in the reverse direction. A product becomes a sum or difference at the combined angles x + y and x − y. This is useful when different products can produce terms that cancel.

Product-to-sum identitiesLaTeX
These are the same addition/subtraction relationships used in the preceding derivation, rearranged. They hold for every real x and y.

The sine–sine formula is also written −2sin x sin y = cos(x + y) − cos(x − y). Both versions say the same thing. For the last two mixed products, keep track of which factor carries the sine: interchanging x and y changes the sign of the sine of their difference.

Example — A difference of squared sine values

Problem
Show that sin²6x − sin²4x = sin2x sin10x.

  1. 1.Factor the left side as (sin6x − sin4x)(sin6x + sin4x).
  2. 2.The difference is 2cos5x sin x, while the sum is 2sin5x cos x.
  3. 3.Their product is 4sin5x cos5x sin x cos x. Group it as (2sin5x cos5x)(2sin x cos x).
  4. 4.The double-angle identity turns these two factors into sin10x and sin2x, giving the required result. No denominator restrictions are added.

Applying the identities to larger multiples

A larger multiple can be handled by repeating a known doubling operation. The chapter’s applications at 4x and 6x do not require a separate family of starting identities. They are opportunities to choose substitutions carefully and simplify with a plan.

Example — A fourth-angle tangent expression

Problem
Derive tan4x in terms of t = tan x where the relevant expressions are defined.

  1. 1.Apply the double-angle formula twice: tan4x = 2tan2x/(1 − tan²2x), and tan2x = 2t/(1 − t²). For this route, assume tan x, tan2x and tan4x exist.
  2. 2.Substitute: tan4x = [4t/(1 − t²)]/[1 − 4t²/(1 − t²)²].
  3. 3.Combine the denominator over (1 − t²)² and multiply numerator and denominator by that nonzero square. The result is 4t(1 − t²)/[(1 − t²)² − 4t²].
  4. 4.Expand the denominator: (1 − 2t² + t⁴) − 4t² = 1 − 6t² + t⁴. Thus tan4x = 4t(1 − t²)/(1 − 6t² + t⁴).
  5. 5.The final expression remains valid when tan x and tan4x exist even if tan2x does not. To verify without that intermediate restriction, use sin4x = 4sin x cos x(cos²x − sin²x) and cos4x = cos⁴x − 6sin²x cos²x + sin⁴x, then divide by cos⁴x. Require cos x ≠ 0 and the final denominator ≠ 0.
Example — Cosine at four and six times an angle

Problem
Derive cos4x = 1 − 8sin²x cos²x and cos6x as a polynomial in cos x.

  1. 1.For the first result, apply cos2u = 1 − 2sin²u with u = 2x. Then cos4x = 1 − 2sin²2x.
  2. 2.Substitute sin2x = 2sin x cos x and square it: cos4x = 1 − 2(4sin²x cos²x) = 1 − 8sin²x cos²x.
  3. 3.For the second result, use cos6x = 2cos²3x − 1. Write c = cos x and use cos3x = 4c³ − 3c.
  4. 4.Square: (4c³ − 3c)² = 16c⁶ − 24c⁴ + 9c². Multiply by 2 and subtract 1.
  5. 5.Thus cos6x = 32cos⁶x − 48cos⁴x + 18cos²x − 1. Every step used identities valid for all real x.

When proving a more complicated identity, inspect the structure before choosing a formula. Equal angle separations suggest sum-to-product; products suggest product-to-sum; squared terms suggest a fundamental or double-angle identity. Transform one side in justified steps toward the other. Do not begin by assuming the equality you are meant to prove.

Cancellation preserves the original domain

If a factor is cancelled in a quotient, it must be nonzero on the original domain. A simpler final expression may exist at additional inputs, but the original quotient remains undefined there. State the identity on its common valid domain.

Quiz

Quick check

Which formula correctly expresses cos(x + y)?

Quick check

Which value equals sin15°?

Quick check

Which formula holds for all real x?

Quick check

What is cos(π/2 + x)?

Quick check

What are the half-sum and half-difference of 7x and 5x?

Quick check

Which is the correct product-to-sum identity?

Quick check

For tan2x = 2tan x/(1 − tan²x), what must be checked?

Quick check

Which is the triple-angle cosine identity?

Practice Problems

Practice Problems
  1. Derive cos(x − y) and sin(x − y) from the sum formulas, explaining the negative-angle signs.
  2. Evaluate sin75°, tan15° and cos15° exactly, showing the angle decomposition and substitution.
  3. Prove cos(π/4 − x)cos(π/4 − y) − sin(π/4 − x)sin(π/4 − y) = sin(x + y).
  4. Prove tan(π/4 + x)/tan(π/4 − x) = [(1 + tan x)/(1 − tan x)]² on the common valid domain.
  5. Using related angles, simplify cos(π + x)cos(−x)/[sin(π − x)cos(π/2 + x)]. State every excluded input.
  6. Prove sin((n + 1)x)sin((n + 2)x) + cos((n + 1)x)cos((n + 2)x) = cos x.
  7. Prove cos(3π/4 + x) − cos(3π/4 − x) = −√2 sin x by sum-to-product.
  8. Prove cos²2x − cos²6x = sin4x sin8x by factoring or by double-angle identities.
  9. Prove sin2x + 2sin4x + sin6x = 4cos²x sin4x, grouping symmetric terms first.
  10. Prove cot4x(sin5x + sin3x) = cot x(sin5x − sin3x) where both cotangent factors are defined.
  11. Prove (cos9x − cos5x)/(sin17x − sin3x) = −sin2x/cos10x on the original quotient’s domain.
  12. Prove (sin5x + sin3x)/(cos5x + cos3x) = tan4x and (sin x − sin y)/(cos x + cos y) = tan((x − y)/2), with denominator checks.
  13. Prove (sin x − sin3x)/(sin²x − cos²x) = 2sin x wherever the quotient exists.
  14. Prove (cos4x + cos3x + cos2x)/(sin4x + sin3x + sin2x) = cot3x on its original domain.
  15. Use cotangent addition to prove cot x cot2x − cot2x cot3x − cot3x cot x = 1 where all cotangent values exist.
  16. Derive the identities for tan4x, cos4x and cos6x, and explain how successive substitutions reduce larger angles to known formulas.
  17. Prove cos(3π/2 + x)cos(2π + x)[cot(3π/2 − x) + cot(2π + x)] = 1 on its original domain. First reduce the related angles.
  18. Prove (sin x + sin3x)/(cos x + cos3x) = tan2x on the original quotient’s domain, explaining why each cancelled factor is nonzero.

Key Takeaways

Key Takeaways

• Sine and cosine addition formulas are the starting point for a connected family of identities. • Negative-angle and axis values generate difference, complementary and related-angle formulas. • Tangent and cotangent identities require the original functions and displayed denominators to be defined. • Double and triple angles follow by substituting x + x or 2x + x into addition identities. • Sum-to-product uses half-sum and half-difference; product-to-sum uses the combined angles. • A successful proof chooses a transformation that exposes structure and preserves the original domain.

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Trigonometric Functions, the Unit Circle and Graphs

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Mixed Applications and Half-Angle Reasoning