Prove the Pythagorean identity sin²θ + cos²θ = 1 and use it with the quadrant to find the other trigonometric values

Learning goal: Prove the Pythagorean identity sin²(θ) + cos²(θ) = 1 from the unit circle, and use it to find sin(θ), cos(θ) or tan(θ) given one of them and the quadrant of the angle.

What an identity is. An identity is an equation that is true for every value of the variable for which both sides are defined, not just for particular solutions. The equation sin θ = 1/2 is true for some θ and false for others; the equation sin²θ + cos²θ = 1 is true for every real number θ without exception. Proving an identity means giving an argument that covers all θ at once, and the unit circle does exactly that.

The proof. Let θ be any real number and let P(θ) = (cos θ, sin θ) be its point on the unit circle, as defined in the second lesson of this unit. The unit circle is the set of points at distance 1 from the origin, and by the distance formula (which is the Pythagorean theorem) a point (x, y) is at distance 1 from the origin exactly when x² + y² = 1. Since P(θ) lies on the unit circle, its coordinates satisfy this equation: (cos θ)² + (sin θ)² = 1. Writing (sin θ)² as sin²θ and (cos θ)² as cos²θ, that is sin²θ + cos²θ = 1. The argument used nothing about θ except that P(θ) is on the circle, which is true for every real θ, so the identity holds for every real θ. For an acute angle you can see the right triangle: the legs are cos θ and sin θ and the hypotenuse is the radius 1, and the identity is the Pythagorean theorem for that triangle, which is why it carries that name. The unit-circle argument does better than the triangle picture, because it also covers angles in the other quadrants, on the axes, negative angles and angles beyond one turn, where there is no triangle to draw.

Two companion forms. Dividing every term of sin²θ + cos²θ = 1 by cos²θ (allowed when cos θ ≠ 0) gives tan²θ + 1 = 1/cos²θ. Dividing instead by sin²θ (when sin θ ≠ 0) gives 1 + cos²θ/sin²θ = 1/sin²θ, which can be written 1 + 1/tan²θ = 1/sin²θ only where tan θ is also defined and nonzero, that is, where cos θ ≠ 0 as well. The first of these is useful when tangent is the given value.

Using the identity: the role of the quadrant. Given sin θ, the identity produces cos²θ = 1 − sin²θ, and then cos θ = ±√(1 − sin²θ). The square root gives the size of cos θ but not its sign, and the identity alone cannot decide it: the points (a, b) and (−a, b) are both on the unit circle with the same sine. That is why the standard says "given … and the quadrant of the angle." The quadrant tells you the signs of the coordinates: cosine is positive in quadrants I and IV and negative in II and III; sine is positive in I and II and negative in III and IV; tangent is positive in I and III and negative in II and IV. Once the second value is known, tan θ = sin θ / cos θ follows.

A fully worked model. Suppose sin θ = 3/5 and θ is in quadrant II. Find cos θ and tan θ. From the identity, cos²θ = 1 − (3/5)² = 1 − 9/25 = 16/25, so cos θ = ±4/5. In quadrant II the x-coordinate is negative, so cos θ = −4/5. Then tan θ = (3/5)/(−4/5) = −3/4. Check the identity: (3/5)² + (−4/5)² = 9/25 + 16/25 = 25/25 = 1 ✓. Check the quadrant: sine positive, cosine negative, tangent negative, all as quadrant II requires. Had the angle been in quadrant I instead, the same arithmetic would give cos θ = +4/5 and tan θ = 3/4.

When tangent is given. Suppose tan θ = −2 and θ is in quadrant IV. Use the companion form: 1/cos²θ = 1 + tan²θ = 1 + 4 = 5, so cos²θ = 1/5 and cos θ = ±1/√5. In quadrant IV cosine is positive: cos θ = 1/√5 = √5/5 ≈ 0.4472. Then sin θ = tan θ × cos θ = −2 × √5/5 = −2√5/5 ≈ −0.8944, negative as quadrant IV requires. Check: (−2√5/5)² + (√5/5)² = 20/25 + 5/25 = 1 ✓. Note the reasoning: the divided form connects tangent to cosine in one step, which makes it a convenient shortcut. Starting from sin²θ + cos²θ = 1 works equally well: tan θ = −2 means sin θ = −2 cos θ, so 4cos²θ + cos²θ = 1, 5cos²θ = 1 and cos²θ = 1/5, the same result.

Decimal values work the same way. If sin θ = −0.6 and θ is in quadrant III, then cos²θ = 1 − 0.36 = 0.64, cos θ = −0.8 (negative in quadrant III), and tan θ = −0.6/−0.8 = 0.75, positive as expected in quadrant III. A calculator confirms: arcsin(−0.6) ≈ −0.6435, and the quadrant III angle with the same sine is π + 0.6435 ≈ 3.7851, whose cosine is −0.8.

The misconception. Students often write sin²θ + cos²θ = 1 as sin θ + cos θ = 1. That equation is false for almost every θ: at θ = π/4 the left side is √2/2 + √2/2 = √2 ≈ 1.414. The identity is about squares because it comes from distance, and distance comes from squares. A second error is to forget the ± when taking the square root and always report the positive value; the worked models above show that the quadrant, not the arithmetic, decides the sign.

Worked, cosine given: cos θ = −5/13 and θ is in quadrant III. sin²θ = 1 − 25/169 = 144/169, so sin θ = ±12/13; quadrant III makes sine negative, sin θ = −12/13. tan θ = (−12/13)/(−5/13) = 12/5. Check: 25/169 + 144/169 = 169/169 = 1, and tangent is positive in quadrant III ✓.

Worked, cosine given with an irrational sine: cos θ = 2/3 and θ is in quadrant IV. sin²θ = 1 − 4/9 = 5/9, so sin θ = ±√5/3; quadrant IV makes it negative, sin θ = −√5/3 ≈ −0.745. tan θ = (−√5/3)/(2/3) = −√5/2 ≈ −1.118. Check: 4/9 + 5/9 = 1, and a negative tangent is right for quadrant IV.

Worked, tangent given: tan θ = 5/12 and θ is in quadrant III. 1/cos²θ = 1 + 25/144 = 169/144, so cos²θ = 144/169 and cos θ = −12/13 (negative in quadrant III). sin θ = tan θ × cos θ = (5/12)(−12/13) = −5/13. Check: 25/169 + 144/169 = 1 ✓, and both coordinates negative fits quadrant III.

Worked, proving a related identity: Show that (1 − sin θ)(1 + sin θ) = cos²θ for all θ. Expand the left side: (1 − sin θ)(1 + sin θ) = 1 − sin²θ, a difference of squares. By the Pythagorean identity, 1 − sin²θ = cos²θ. The equality holds for every θ because each step is valid for every θ. Numerical spot check at θ = π/6: (1 − 1/2)(1 + 1/2) = 3/4 and cos²(π/6) = (√3/2)² = 3/4 ✓.

Representation task: Draw the unit circle and the point (0.6, 0.8) in quadrant I, with the right triangle formed by dropping a perpendicular to the x-axis. Label the legs 0.6 and 0.8 and the hypotenuse 1, and write 0.6² + 0.8² = 0.36 + 0.64 = 1 beside it. Then plot the three other points with the same coordinates up to sign, (−0.6, 0.8), (−0.6, −0.8) and (0.6, −0.8), and beside each write its quadrant and its values of sine, cosine and tangent (for example quadrant III: sin = −0.8, cos = −0.6, tan = 4/3). Record why all four satisfy the identity and why the quadrant is needed to pick one of them.

Independent performance: Write a complete proof of sin²θ + cos²θ = 1 from the definition of sine and cosine on the unit circle, in your own words, stating explicitly why the argument covers every real θ and not only acute angles. Then solve four problems you set yourself: given sine and a quadrant, given cosine and a quadrant, given tangent and a quadrant, and one with decimal data. For each, show the identity step, the ± step, the quadrant decision with a reason, the third value, and a check that the two coordinates satisfy x² + y² = 1. Success criteria: the proof names the distance formula or Pythagorean theorem and the unit radius; every sign is justified by the quadrant; every check comes out to 1; and at least one problem involves a radical that is simplified with a rational denominator.