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.
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.
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.
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:
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 ✓
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 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.
Using the reference angle 15° (in the first quadrant, all values positive):
Starting from the known values for 45° and 30°:
| Function | Exact Value | Decimal |
|---|---|---|
| sin 15° | (√6 − √2)/4 | 0.2588 |
| cos 15° | (√6 + √2)/4 | 0.9659 |
| tan 15° | 2 − √3 | 0.2679 |
| cot 15° | 2 + √3 | 3.7321 |
| sec 15° | √6 − √2 | 1.0353 |
| cosec 15° | √6 + √2 | 3.8637 |
Key relationships:
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 ✓
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. 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.
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.
cot 15° = 1/tan 15° = 1/(2−√3). Rationalising: 1/(2−√3) × (2+√3)/(2+√3) = (2+√3)/(4−3) = 2+√3 ≈ 3.7321.
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.
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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