Study Guides/Maths/tan 15° = 2−√3 ≈ 0.2679 — Value, Derivation & Proof
Study Guide · Maths

What is the Value of tan 15°?

The value of tan 15° = 2 − √3 ≈ 0.2679. It is derived using the subtraction formula for tangent: tan(A − B) = (tan A − tan B)/(1 + tan A · tan B). Setting A = 45° and B = 30° gives tan 15° = tan(45° − 30°) = (1 − 1/√3)/(1 + 1/√3) = 2 − √3.

Question (Click to Flip)

What is the value of tan 15°?

Answer

tan 15° = 2 − √3 ≈ 0.2679. It is derived using the tangent subtraction formula: tan(45° − 30°) = (1 − 1/√3)/(1 + 1/√3) = (√3 − 1)/(√3 + 1) = 2 − √3 after rationalisation.

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Key Facts

tan 15° = 2 − √3 ≈ 0.2679.

Derived using tan(45° − 30°) = (tan 45° − tan 30°)/(1 + tan 45° · tan 30°) = (1 − 1/√3)/(1 + 1/√3).

After rationalisation: tan 15° = (√3−1)²/2 = (4−2√3)/2 = 2−√3.

tan 75° = 2 + √3 ≈ 3.7321 (complement of tan 15°).

tan 15° × tan 75° = (2−√3)(2+√3) = 1.

sin 15° = (√6−√2)/4 ≈ 0.2588; cos 15° = (√6+√2)/4 ≈ 0.9659.

15° = 45° − 30° = 60° − 45°; both decompositions yield tan 15° = 2 − √3.

Value of tan 15° — Result

tan 15° = 2 − √3

Numerical value: √3 ≈ 1.7321 tan 15° = 2 − 1.7321 = 0.2679

Verification: tan 15° ≈ 0.26795 (from calculator) ✓

tan 15° is positive because 15° is in the first quadrant (0° to 90°), where all trigonometric functions are positive.

Note: 15° = 45° − 30° = 60° − 45°. Both decompositions can be used to derive the value.

Derivation Using tan(45° − 30°)

The tangent subtraction formula is: tan(A − B) = (tan A − tan B) / (1 + tan A · tan B)

Set A = 45° and B = 30°, so A − B = 15°.

Known values:

  • tan 45° = 1
  • tan 30° = 1/√3

Substitute into the formula: tan 15° = (tan 45° − tan 30°) / (1 + tan 45° · tan 30°) = (1 − 1/√3) / (1 + 1 · 1/√3) = (1 − 1/√3) / (1 + 1/√3)

Multiply numerator and denominator by √3: = (√3 − 1) / (√3 + 1)

Rationalise by multiplying by (√3 − 1)/(√3 − 1): = (√3 − 1)² / [(√3 + 1)(√3 − 1)] = (3 − 2√3 + 1) / (3 − 1) = (4 − 2√3) / 2 = 2 − √3

Therefore: tan 15° = 2 − √3 ≈ 0.2679 ✓

Alternative Derivation Using tan(60° − 45°)

We can also compute tan 15° as tan(60° − 45°):

tan(A − B) = (tan A − tan B) / (1 + tan A · tan B)

Set A = 60°, B = 45°:

  • tan 60° = √3
  • tan 45° = 1

tan 15° = (√3 − 1) / (1 + √3 · 1) = (√3 − 1) / (√3 + 1)

Rationalise: = (√3 − 1)(√3 − 1) / [(√3 + 1)(√3 − 1)] = (√3 − 1)² / (3 − 1) = (3 − 2√3 + 1) / 2 = (4 − 2√3) / 2 = 2 − √3

Same result: tan 15° = 2 − √3 ✓

Both methods give the same answer, confirming the result.

All Trig Values at 15°

Using the reference angle 15° (in the first quadrant, all values positive):

Starting from the known values for 45° and 30°:

FunctionExact ValueDecimal
sin 15°(√6 − √2)/40.2588
cos 15°(√6 + √2)/40.9659
tan 15°2 − √30.2679
cot 15°2 + √33.7321
sec 15°√6 − √21.0353
cosec 15°√6 + √23.8637

Key relationships:

  • cot 15° = 1/tan 15° = 1/(2−√3) = (2+√3) [by rationalisation]
  • tan 15° × cot 15° = (2−√3)(2+√3) = 4−3 = 1 ✓
  • tan 15° = tan(90° − 75°) = cot 75°
  • cot 15° = tan 75° = 2 + √3

Verification of sin²15° + cos²15°: [(√6−√2)/4]² + [(√6+√2)/4]² = [(6−2√12+2) + (6+2√12+2)] / 16 = [8−2√12+8+2√12] / 16 = 16/16 = 1 ✓

Rationalisation and tan 75°

tan 15° and tan 75° are complementary values: tan 75° = cot 15° = 2 + √3

Proof that 1/(2−√3) = 2+√3: Multiply numerator and denominator by (2+√3): 1/(2−√3) × (2+√3)/(2+√3) = (2+√3)/(4−3) = (2+√3)/1 = 2+√3 ✓

Derivation of tan 75°: tan 75° = tan(45°+30°) = (tan 45° + tan 30°) / (1 − tan 45° · tan 30°) = (1 + 1/√3) / (1 − 1/√3) = (√3 + 1) / (√3 − 1) = (√3 + 1)² / (3 − 1) = (4 + 2√3) / 2 = 2 + √3

Summary:

  • tan 15° = 2 − √3 ≈ 0.2679
  • tan 75° = 2 + √3 ≈ 3.7321
  • tan 15° × tan 75° = (2−√3)(2+√3) = 4−3 = 1 ✓ (product of complementary tangents = 1)

Questions and Answers

What is the value of tan 15°?+

tan 15° = 2 − √3 ≈ 0.2679. It is derived using the tangent subtraction formula: tan(45° − 30°) = (1 − 1/√3)/(1 + 1/√3) = (√3 − 1)/(√3 + 1) = 2 − √3 after rationalisation.

How do you derive tan 15° using the subtraction formula?+

Using tan(A−B) = (tan A − tan B)/(1 + tan A · tan B), set A = 45°, B = 30°: tan 15° = (1 − 1/√3)/(1 + 1/√3) = (√3−1)/(√3+1). Multiply by (√3−1)/(√3−1): = (√3−1)²/2 = (4−2√3)/2 = 2−√3.

What is the value of cot 15°?+

cot 15° = 1/tan 15° = 1/(2−√3). Rationalising: 1/(2−√3) × (2+√3)/(2+√3) = (2+√3)/(4−3) = 2+√3 ≈ 3.7321.

What is tan 75°?+

tan 75° = cot 15° = 2 + √3 ≈ 3.7321. It can be derived using tan(45°+30°) = (1 + 1/√3)/(1 − 1/√3) = (√3+1)/(√3−1) = (4+2√3)/2 = 2+√3.

What are sin 15° and cos 15°?+

sin 15° = (√6 − √2)/4 ≈ 0.2588 and cos 15° = (√6 + √2)/4 ≈ 0.9659. These are derived using sin(45°−30°) and cos(45°−30°) formulas.

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